GSEB Class 11 Physics Solutions Chapter 5 Laws of Motion

Question 1. Give the magnitude and direction of the net force acting on:
1. a drop of rain falling down with a constant speed.
2. a cork of mass 10 g floating on water.
3. a kite skillfully held stationary in the sky.
4. a car moving with a constant velocity of 30 km/h on a rough road.
5. a high-speed electron in space far from all gravitating (material) objects, and free of electric and magnetic fields.
Answer:
1. Since the rain drop is falling downward at a steady speed, its acceleration is zero. According to Newton's first law of motion, the net force on the drop is therefore zero (i.e., \(a = 0\)), which means \(F = ma = 0\).
2. When a cork is floating in water, its weight is balanced by the upward buoyant force (which equals the weight of the water it displaces). Consequently, the total force on the cork is zero.
3. Since the kite is held still, its acceleration is approximately zero. Based on Newton's first law of motion, there is no net force acting on the kite.
4. Because the car is moving at a constant velocity, its acceleration is zero. By Newton's first law of motion, the net force acting on the car is zero.
5. As there are no electric or magnetic fields, and no massive (gravitating) objects nearby, no force (gravitational, electric, or magnetic) acts on the electron, resulting in a net force of zero.
In simple words: For all these situations, if an object is moving at a constant speed or is completely still, its acceleration is zero. This means the total (net) force acting on it is also zero, following Newton's first law. If nothing is pulling or pushing it, or if all forces cancel out, the net force is zero.

Exam Tip: Remember Newton's First Law: an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Constant speed means zero acceleration, which implies zero net force.

Question 2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble:
(a) during its upward motion
(b) during its downward motion
(c) at the highest point where it is momentarily at rest.
Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance?
Answer:
We understand that whenever an object is thrown straight up or moves straight down, Earth's gravitational pull gives it a uniform acceleration of \(a = +g = 9.8 \text{ ms}^{-2}\) in the vertically downward direction.
If \(m\) represents the mass of the object, then the net force on it is \(F = mg\).
Here, the mass of the pebble is \(m = 0.05 \text{ kg}\).

(a) Therefore, the force on the pebble is \(F = mg\) (since \(a = g\)):
\(F = 0.05 \times 9.8\)
\(F = 0.49 \text{ N}\) (This force acts vertically downwards).

(b) The net force on the pebble is also \(F = mg\) (since \(a = g\)):
\(F = 0.05 \times 9.8\)
\(F = 0.49 \text{ N}\) (This force also acts vertically downwards).

(c) Even when the stone reaches its highest point and is momentarily at rest, the net force (which is \(mg\)) still acts vertically downwards.

If the pebble was thrown at a 45° angle to the horizontal, it would have both horizontal and vertical components of velocity. However, these components do not change the force acting on the pebble, which remains its weight. Hence, our answers for the net force will not change in any situation. Nevertheless, in case (c), the pebble would not be completely at rest at its highest point; it would still have a horizontal velocity component at that moment.
In simple words: No matter if the pebble is going up, down, or is at its highest point, the only force acting on it (ignoring air) is gravity, pulling it down. This force is its mass times gravity, which is 0.49 N. Even if you throw it at an angle, gravity still pulls it straight down, so the force stays the same. The only difference is at the highest point, it would still be moving sideways if thrown at an angle.

Exam Tip: In projectile motion (ignoring air resistance), the net force acting on an object is always its weight (mg), directed vertically downwards, at every point in its trajectory, including the peak.

Question 3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg?
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 ms\(^{-2}\),
(d) lying on the floor of a train which is accelerating with 1 ms\(^{-2}\), the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
Mass of stone, \(m = 0.1 \text{ kg}\).

(a) When the stone is dropped from the window of a stationary train, it freely falls with an acceleration \(a = +g\).
Therefore, the net force on the stone is \(F = mg\)
\(F = 0.1 \times 9.8\)
\(F = 0.98 \text{ N}\)
\(F \approx 1 \text{ N}\) and it acts vertically downwards.

(b) When the train is running at a constant velocity of \(36 \text{ kmh}^{-1}\), the acceleration of the train in the horizontal direction is \(0\). So, no horizontal force acts on the stone because of this motion. When the stone is dropped from the window of this train, it is still falling freely.
The net force on the stone equals its own weight:
\(F = mg = 0.98 \text{ N}\)
\(F \approx 1 \text{ N}\) and it acts vertically downwards.

(c) When the train is accelerating with \(1 \text{ ms}^{-2}\), two forces then act on the stone:
(i) A force due to the initial acceleration of \(1 \text{ ms}^{-2}\) from the train, acting in the horizontal direction.
(ii) A force due to its acceleration of \(9.8 \text{ ms}^{-2}\) when dropped from the train, acting vertically downward.
If \(F_1\) and \(F_2\) are the forces acting on the stone along the horizontal and vertical directions, respectively:
Then \(F_1 = ma = 0.1 \times 1 = 0.1 \text{ N}\)
And \(F_2 = mg = 0.1 \times 9.8 = 0.98 \text{ N}\)
If \(F\) is the net force acting on the stone, then
\(F = \sqrt{F_{1}^{2}+F_{2}^{2}}\)
\(F = \sqrt{(0.1)^{2}+(0.98)^{2}}\)
\(F = \sqrt{0.01 + 0.9604}\)
\(F = \sqrt{0.9704} \approx 0.985 \text{ N}\)
\(F \approx 1 \text{ N}\)
Direction of \(F\): Let \( \theta \) be the angle made by \(F\) with \(F_1\).
\( \tan \theta = \frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} = \frac{0.98}{0.1} = 9.8 \)
\( \theta = \tan^{-1}(9.8) \approx 84^\circ 12'\) with the horizontal (i.e., \(5^\circ 48'\) with the vertical direction).

Aliter: Here, once the stone is dropped from the train, the force \(F_1\) from the train's motion no longer affects the stone; \(F_1\) becomes zero. So, the net force on the stone is just \(F_2 = 0.98 \text{ N} \approx 1 \text{ N}\). It acts vertically downward.

(d) For the stone lying on the floor of a train accelerating with \(1 \text{ ms}^{-2}\), and if the stone is at rest relative to the train, the only force causing it to accelerate with the train is the friction force. The magnitude of this force is:
\(F = ma = 0.1 \times 1 = 0.1 \text{ N}\)
This force acts along the direction of the train's motion.
In simple words: If you drop a stone from a train, usually gravity is the main force, pulling it down at about 1 N. This is true whether the train is stopped or moving at a steady speed. But if the train is speeding up, the stone will also feel a small push sideways (0.1 N) along with the gravity pull, making the total force a bit diagonal. If the stone just sits on the floor of a speeding-up train, only the sideways push of 0.1 N acts on it, in the direction the train is going.

Exam Tip: Carefully identify the reference frame and all forces acting on the object. Remember that horizontal motion components do not affect the vertical gravitational force unless air resistance is considered.

Question 4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on smooth horizontal table. If the particle moves in a circle with speed v, the net force on the particle (directed towards the centre) is:
(a) T
(b) T-\(\frac{{mv}^{2}}{l}\)
(c) T + \(\frac{{mv}^{2}}{l}\)
(d) 0
Answer: (a) T
In simple words: When an object spins in a circle on a smooth table, the string pulls it towards the middle. This pull, called tension (T), is the only force making it move in a circle. It's exactly the force needed to keep it from flying away in a straight line.

Exam Tip: For an object moving in a circle, the net force pointing towards the center is the centripetal force. In this case, the tension in the string directly provides this force.

Question 5. A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15ms\(^{-1}\). How long does the body take to stop?
Answer:
We are given the following values:
Force acting on the body, \(F = -50 \text{ N}\) (The negative sign indicates it is a retarding force, opposing motion).
Mass of the body, \(m = 20 \text{ kg}\)
Initial velocity of the body, \(u = 15 \text{ ms}^{-1}\)
Final velocity of the body, \(v = 0\) (since the body comes to a stop).
We need to find the time, \(t\).

First, we use Newton's second law, \(F = ma\), to find the acceleration (\(a\)):
\(a = \frac{F}{m}\)
\(a = \frac{-50}{20}\)
\(a = -2.5 \text{ ms}^{-2}\) (This is a retardation, or deceleration).

Now, we use the kinematic relation, \(v = u + at\), to find the time (\(t\)):
\(0 = 15 + (-2.5)t\)
\(2.5t = 15\)
\(t = \frac{15}{2.5}\)
\(t = 6 \text{ s}\)
So, the body takes \(6\) seconds to stop.
In simple words: We know the force pushing against the body and its mass, so we can figure out how quickly it slows down. Once we know the slowing-down rate, we use the initial speed and the fact that it stops (final speed is zero) to calculate how much time it will take to come to a complete halt.

Exam Tip: Remember to use a negative sign for retarding (braking) forces and acceleration. Break down the problem: first find acceleration from force and mass, then use kinematics to find time or distance.

Question 6. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms\(^{-1}\) to 3.5 ms\(^{-1}\) in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
Given values:
Mass of the body, \(m = 3 \text{ kg}\)
Initial speed, \(u = 2 \text{ ms}^{-1}\)
Final speed, \(v = 3.5 \text{ ms}^{-1}\)
Time taken, \(t = 25 \text{ s}\)
We need to find the constant force (\(F\)) acting on the body.

First, let's find the acceleration (\(a\)) produced in the body using the kinematic relation:
\(v = u + at\)
\(a = \frac{v-u}{t}\)
\(a = \frac{3.5 - 2}{25}\)
\(a = \frac{1.5}{25}\)
\(a = \frac{15}{250} \text{ ms}^{-2}\)

Now, using Newton's second law, \(F = ma\), we can find the magnitude of the force:
\(F = 3 \times \frac{15}{250}\)
\(F = \frac{45}{250}\)
\(F = 0.18 \text{ N}\)
The direction of the force is along the direction of motion, as the speed increased and the direction of motion remained unchanged.
In simple words: We're given how much a 3 kg object speeds up over 25 seconds. First, we calculate its acceleration, which is how fast its speed changes. Then, using Newton's law (Force = mass × acceleration), we can easily find the strength of the push that made it speed up. Since it speeds up in the same direction, the force also acts in that direction.

Exam Tip: When a problem asks for both magnitude and direction, ensure both are provided in your final answer. Acceleration is positive if the object speeds up, and the force acts in the direction of acceleration.

Question 7. A body of mass 5 kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body?
Answer:
Mass of the body, \(m = 5 \text{ kg}\)
First force acting on the body, \(F_1 = 8 \text{ N}\)
Second force acting on the body, \(F_2 = 6 \text{ N}\)
The angle between \(F_1\) and \(F_2\) is \( \theta = 90^\circ \).
We need to find the acceleration (\(a\)) of the body.

To find the acceleration, we first need to determine the net force acting on the body. Since the two forces are perpendicular, we can find the resultant force (\(F\)) using the parallelogram law of vector addition:

\(F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2F_{1}F_{2}\cos\theta}\)
Since \( \theta = 90^\circ \), \( \cos 90^\circ = 0 \). So, the equation simplifies to:
\(F = \sqrt{F_{1}^{2} + F_{2}^{2}}\)
\(F = \sqrt{8^{2} + 6^{2}}\)
\(F = \sqrt{64 + 36}\)
\(F = \sqrt{100}\)
\(F = 10 \text{ N}\)

Now, we find the acceleration using Newton's second law, \(F = ma\):
\(a = \frac{F}{m}\)
\(a = \frac{10}{5}\)
\(a = 2 \text{ ms}^{-2}\)

To find the direction of the resultant force (and thus acceleration), we can find the angle \( \alpha \) it makes with \(F_1\). Using the triangle formed by the vectors:
\( \tan \alpha = \frac{F_2}{F_1} = \frac{6}{8} = 0.7500 \)
\( \alpha = \tan^{-1}(0.7500) \approx 36^\circ 53' \)
This is the direction of the resultant force and, consequently, the direction of the body's acceleration.
The magnitude of acceleration is \(2 \text{ ms}^{-2}\) and it acts in the direction of the resultant force.
In simple words: When two forces push on an object at a right angle, we can find the total force by using a special square root calculation (like the Pythagorean theorem). This total force then helps us find how fast the object speeds up, using its mass. The direction of this speeding up is slanted between the two original pushes.

Exam Tip: For perpendicular forces, use the Pythagorean theorem to find the magnitude of the resultant force. The direction can be found using the tangent function relative to one of the forces.

Question 8. The driver of a three wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg?
Answer:
Given:
Initial speed of the three-wheeler, \(u = 36 \text{ kmh}^{-1}\). We need to convert this to \(\text{ms}^{-1}\):
\(u = 36 \times \frac{5}{18} = 10 \text{ ms}^{-1}\)
Final speed of the three-wheeler, \(v = 0\) (since it comes to rest).
Time taken, \(t = 4 \text{ s}\)
Total mass of the system, \(m = \text{mass of three-wheeler} + \text{mass of driver}\)
\(m = 400 \text{ kg} + 65 \text{ kg} = 465 \text{ kg}\)
We need to find the average retarding force (\(F\)).

First, let's find the acceleration (\(a\)) using the kinematic relation:
\(v = u + at\)
\(0 = 10 + a(4)\)
\(4a = -10\)
\(a = -\frac{10}{4} = -2.5 \text{ ms}^{-2}\)
The negative sign shows that it is a retardation (deceleration).

Now, we can find the retarding force using Newton's second law, \(F = ma\). Since it's a retarding force, we consider the magnitude of acceleration:
\(F = m(-a)\)
\(F = 465 \times (2.5)\)
\(F = 1162.5 \text{ N}\)
This can also be expressed as \(F \approx 1.16 \times 10^3 \text{ N}\) or approximately \(1.2 \times 10^3 \text{ N}\).
In simple words: We know how fast the vehicle was going, how long it took to stop, and its total weight. First, we calculate how quickly it slowed down (its acceleration). Then, using the total weight and this slowing-down rate, we can figure out the average braking force applied.

Exam Tip: Always convert all units to SI units (meters, kilograms, seconds) before starting calculations to avoid errors. Remember that a retarding force is associated with negative acceleration.

Question 9. A rocket with a lift – off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms\(^{-2}\)N. Calculate the initial thrust (force) of the blast?
Answer:
Given:
Mass of the rocket, \(m = 20,000 \text{ kg}\)
Acceleration of the rocket, \(a = 5.0 \text{ ms}^{-2}\) (in the upward direction)
Acceleration due to gravity, \(g = 9.8 \text{ ms}^{-2}\)
Let \(T\) be the initial thrust acting upward, and \(W\) be the weight of the rocket acting downward.

The weight of the rocket is \(W = mg\).
According to Newton's second law, the net upward force equals mass times acceleration. The net upward force is the thrust minus the weight (\(T - W\)).
Therefore, the equation of motion becomes:
\(T - mg = ma\)
So, the thrust \(T\) is:
\(T = mg + ma\)
\(T = m(g + a)\)
\(T = 20,000 \text{ kg} (9.8 \text{ ms}^{-2} + 5.0 \text{ ms}^{-2})\)
\(T = 20,000 (14.8)\)
\(T = 296,000 \text{ N}\)
\(T = 2.96 \times 10^5 \text{ N}\)
This can be rounded to \(T \approx 3.0 \times 10^5 \text{ N}\).
In simple words: To lift a rocket up, the engine's push (thrust) needs to be strong enough to overcome its own weight AND make it speed up. We add the force needed to fight gravity to the force needed to accelerate it upwards to find the total push from the blast.

Exam Tip: When calculating thrust for upward motion, remember that the thrust must overcome both the weight of the rocket (\(mg\)) and provide the necessary force (\(ma\)) for upward acceleration.

Question 10. A body of mass 0.40 kg moving initially with a constant speed of 10 ms\(^{-1}\) to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the particle at that time to be x = 0 and predict its position at t = -5s, 25s, 100s?
Answer:
Given:
Mass of particle, \(m = 0.40 \text{ kg}\)
Initial velocity, \(u = 10 \text{ ms}^{-1}\) (towards North)
Constant force applied on the particle, \(F = 8.0 \text{ N}\) (towards South, so we can denote it as \(F = -8 \text{ N}\) if North is positive).
Time for which force is applied, \(t = 30 \text{ s}\)

We can find the acceleration (\(a\)) produced in the particle:
\(a = \frac{F}{m} = \frac{-8}{0.40} = \frac{-80}{4} = -20 \text{ ms}^{-2}\) (This acceleration is towards the South, opposing the initial Northward motion). This acceleration acts for \(0 \le t \le 30 \text{ s}\).

(i) Position of the particle at \(t = -5 \text{ s}\)
For \(t < 0\), the force has not yet been applied, so the particle moves with a constant speed of \(10 \text{ ms}^{-1}\) towards North. The displacement (\(x_1\)) covered by the particle for \(t = -5 \text{ s}\) (i.e., 5 seconds before \(t=0\), still moving North) is:
\(x_1 = ut = 10 \times (-5) = -50 \text{ m}\)
(The negative sign indicates a position 50m North of the origin if we consider \(t=0, x=0\) as the reference, or 50m south of where it would be if it continued from \(t=-5\)). Interpreting this as "its position relative to x=0 at t=0", this means at t=-5s, it was 50m south of x=0. However, the wording "from t = 0 to t = -5s, the particle moves towards north" suggests we trace back. If \(x=0\) at \(t=0\), then at \(t=-5s\), moving north, its position was 50m south. So, \(x = -50 \text{ m}\).

(ii) Position of the particle at \(t = 25 \text{ s}\)
For \(0 \le t \le 30 \text{ s}\), the acceleration is \(a = -20 \text{ ms}^{-2}\).
Using the relation, \(s = ut + \frac{1}{2}at^{2}\), we find the displacement (\(x_{25}\)) from \(t=0\) to \(t=25 \text{ s}\):
\(x_{25} = (10)(25) + \frac{1}{2}(-20)(25)^2\)
\(x_{25} = 250 - 10(625)\)
\(x_{25} = 250 - 6250\)
\(x_{25} = -6000 \text{ m}\)
So, \(x_{25} = -6 \text{ km}\). This position is 6 km South of the starting point.

(iii) Position of the particle at \(t = 100 \text{ s}\)
The force acts for \(30\) seconds (from \(t=0\) to \(t=30 \text{ s}\)). Let's first find the position at \(t = 30 \text{ s}\) (\(x_{30}\)) and the velocity at \(t = 30 \text{ s}\) (\(v_{30}\)).

Position at \(t = 30 \text{ s}\):
\(x_{30} = (10)(30) + \frac{1}{2}(-20)(30)^2\)
\(x_{30} = 300 - 10(900)\)
\(x_{30} = 300 - 9000\)
\(x_{30} = -8700 \text{ m}\)

Velocity at \(t = 30 \text{ s}\):
\(v_{30} = u + at\)
\(v_{30} = 10 + (-20)(30)\)
\(v_{30} = 10 - 600\)
\(v_{30} = -590 \text{ ms}^{-1}\) (The negative sign indicates velocity towards the South).

After \(t = 30 \text{ s}\), the force stops acting, so the particle will move with a constant speed of \(-590 \text{ ms}^{-1}\) (i.e., \(590 \text{ ms}^{-1}\) towards the South) because there is no more acceleration.
We need the position at \(t = 100 \text{ s}\). This means we need the displacement from \(t=30 \text{ s}\) to \(t=100 \text{ s}\), which is a duration of \(70\) seconds.
The distance covered in these \(70\) seconds (\(x_{70}\)) at a constant velocity of \(-590 \text{ ms}^{-1}\) is:
\(x_{70} = v_{30} \times (100 - 30) = (-590)(70)\)
\(x_{70} = -41300 \text{ m}\)

The total position at \(t=100 \text{ s}\) (\(x_{100}\)) will be the position at \(t=30 \text{ s}\) plus the displacement from \(t=30 \text{ s}\) to \(t=100 \text{ s}\):
\(x_{100} = x_{30} + x_{70}\)
\(x_{100} = -8700 \text{ m} + (-41300 \text{ m})\)
\(x_{100} = -50000 \text{ m}\)
\(x_{100} = -50 \text{ km}\)
So, the position of the particle at \(t = -5 \text{ s}\), \(t = 25 \text{ s}\), and \(t = 100 \text{ s}\) are \(-50 \text{ m}\), \(-6 \text{ km}\), and \(-50 \text{ km}\) respectively (where negative indicates South of the origin).
In simple words: First, we find how fast the object is slowing down because a force is pushing it opposite to its motion. Then, for each requested time, we figure out its position. Before the force starts, it moves at a steady speed. While the force is on, it slows down and might even turn around. After the force stops, it moves at a new steady speed. We sum up these movements to find its final position.

Exam Tip: Break down complex motion problems into segments where acceleration is constant. Pay close attention to signs (positive for North/up, negative for South/down) and the duration for which each force/acceleration acts. Remember that after the force ceases, the object moves at a constant velocity.

Question 11. A truck starts from rest and accelerates uniformly with 2ms\(^{-2}\). At t = 10s, a stone is dropped by a person standing on the top of the truck (6m high from the ground). What are the (a) velocity and (b) acceleration of the stone at t = 11s? (Neglect air resistance)?
Answer:
Given:
Initial velocity of the truck (and stone before it is dropped), \(u = 0\)
Acceleration of the truck, \(a = 2.0 \text{ ms}^{-2}\)
Time when the stone is dropped, \(t = 10 \text{ s}\)
Height of the truck, \(h = 6 \text{ m}\)

First, let's find the velocity of the truck at \(t = 10 \text{ s}\) (which is the initial horizontal velocity of the stone when dropped).
Using the relation, \(v = u + at\):
\(v = 0 + (2.0 \text{ ms}^{-2})(10 \text{ s})\)
\(v = 20 \text{ ms}^{-1}\)

(a) Velocity of the stone at \(t = 11 \text{ s}\):
The stone is dropped at \(t = 10 \text{ s}\). We are interested in its velocity at \(t = 11 \text{ s}\), which is \(1\) second after being dropped.

Horizontal velocity (\(v_x\)):
When the stone is dropped, its initial horizontal velocity is the truck's velocity at that moment, which is \(v_x = 20 \text{ ms}^{-1}\). Since air resistance is ignored, there is no horizontal force, so the horizontal velocity remains constant.
Therefore, \(v_x = 20 \text{ ms}^{-1}\) at \(t = 11 \text{ s}\).

Vertical velocity (\(v_y\)):
The initial vertical velocity of the stone when dropped is \(u_y = 0\) (it's simply let go, not thrown down or up).
The vertical acceleration is due to gravity, \(a_y = g = 9.8 \text{ ms}^{-2}\) (downwards).
The time the stone has been falling is \( \Delta t = 11 \text{ s} - 10 \text{ s} = 1 \text{ s}\).
Using the kinematic relation \(v_y = u_y + a_y \Delta t\):
\(v_y = 0 + (9.8 \text{ ms}^{-2})(1 \text{ s})\)
\(v_y = 9.8 \text{ ms}^{-2}\)

The resultant velocity (\(V\)) of the stone at \(t = 11 \text{ s}\) is the vector sum of its horizontal and vertical components:
\(V = \sqrt{v_x^2 + v_y^2}\)
\(V = \sqrt{(20)^2 + (9.8)^2}\)
\(V = \sqrt{400 + 96.04}\)
\(V = \sqrt{496.04} \approx 22.27 \text{ ms}^{-1}\)
\(V \approx 22.4 \text{ ms}^{-1}\)

Let \( \theta \) be the angle which the resultant velocity \(V\) (OC in a general diagram, here the blue vector) makes with the horizontal direction (OA, or \(v_x\)).
\( \tan \theta = \frac{v_y}{v_x} = \frac{9.8}{20} = 0.49 \)
\( \theta = \tan^{-1}(0.49) \approx 26.1^\circ \)
So, \( \theta \approx 26.2^\circ \) below the horizontal.

(b) Acceleration of the stone at \(t = 11 \text{ s}\):
At the moment the stone is dropped from the truck, the horizontal force on the stone becomes zero (neglecting air resistance).
So, horizontal acceleration, \(a_x = 0\).
The vertical acceleration (\(a_y\)) is due to gravity:
\(a_y = +g = 9.8 \text{ ms}^{-2}\) (acting vertically downward).
The resultant acceleration (\(a\)) of the stone is:
\(a = \sqrt{a_x^2 + a_y^2}\)
\(a = \sqrt{0^2 + (9.8)^2}\)
\(a = 9.8 \text{ ms}^{-2}\)
This acceleration acts vertically downward.
In simple words: First, we find the truck's speed when the stone is dropped. This speed becomes the stone's sideways speed. One second later, the stone still has that sideways speed. But it also falls down for one second, gaining a downward speed due to gravity. We combine these two speeds to get its total speed and direction. For acceleration, once the stone leaves the truck, only gravity pulls it down; there's no sideways push, so its acceleration is just the downward pull of gravity.

Exam Tip: For projectile motion (like a dropped stone), horizontal and vertical motions are independent. Horizontal velocity remains constant (if no air resistance), while vertical motion is subject to gravitational acceleration. The resultant velocity is a vector sum, but resultant acceleration is purely vertical (g) once free from the vehicle's influence.

Question 12. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of the bob at its mean position is 1 ms\(^{-1}\). What is the trajectory of the bob if the string is cut when the bob is:
(a) at one of its extreme positions.
(b) at its mean position.
Answer:
(a) We understand that at each extreme position, the instantaneous velocity of the bob is zero. If the string is cut at an extreme position, the bob is only under the action of 'g' (gravity). Consequently, the bob will simply fall vertically downwards.
(b) When the bob is at its mean position, it is moving at its maximum speed. At the mean position, the bob has a velocity of \(1 \text{ ms}^{-1}\) along the tangent to the arc, which is in the horizontal direction. If the string is cut at this point, the bob will act like a horizontal projectile. Therefore, it will follow a parabolic path.
In simple words: (a) If you cut the string when the pendulum bob is at its highest swing point, it stops moving sideways for a tiny moment. So, it will just fall straight down because only gravity is acting on it. (b) If you cut the string when the pendulum bob is at the bottom (middle) of its swing, it's moving fastest sideways. So, it will fly off sideways and then curve downwards, following a path like a thrown ball.

Exam Tip: The trajectory of an object depends crucially on its velocity at the moment forces change. If velocity is zero, it falls straight down. If there is a horizontal velocity component, it will follow a projectile path.

Question 13. A man of mass 70 kg stands on a weighing scale in a lift which is moving:
(a) upwards with uniform speed of 10 ms\(^{-1}\).
(b) downwards with a uniform acceleration of 5 ms\(^{-2}\).
(c) upwards with uniform acceleration of 5 ms\(^{-2}\).
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Answer:
Given:
Mass of the man, \(m = 70 \text{ kg}\)
Acceleration due to gravity, \(g = 9.8 \text{ ms}^{-2}\)
The weighing machine in each case measures the reaction force (\(R\)) exerted by the scale on the man, which is his apparent weight.

(a) When the lift moves upwards with uniform speed of \(10 \text{ ms}^{-1}\):
Uniform speed means the acceleration (\(a\)) is \(0\). In this situation, the lift acts as an inertial frame of reference.
The apparent weight (\(R\)) will be equal to the man's actual weight (\(W\)):
\(R = W = mg\)
\(R = 70 \text{ kg} \times 9.8 \text{ ms}^{-2}\)
\(R = 686 \text{ N}\)
So, the reading on the scale is \(686 \text{ N}\) or approximately \(70 \text{ kg}\) (since \(70 \times 9.8 \approx 686\), if the scale reads in kg, it divides by \(g\)).

(b) When the lift moves downward with a uniform acceleration of \(5 \text{ ms}^{-2}\):
Acceleration, \(a = 5 \text{ ms}^{-2}\) (downwards).
In this case, the resultant force acts vertically downward. The equation of motion is:
\(mg - R = ma\)
\(R = mg - ma\)
\(R = m(g - a)\)
\(R = 70 \text{ kg} (9.8 \text{ ms}^{-2} - 5 \text{ ms}^{-2})\)
\(R = 70 \times 4.8\)
\(R = 336 \text{ N}\)
So, the reading on the scale is \(336 \text{ N}\). If the scale shows mass, it would be \(\frac{336}{9.8} \approx 34.29 \text{ kg}\).

(c) When the lift moves upwards with a uniform acceleration of \(5 \text{ ms}^{-2}\):
Acceleration, \(a = 5 \text{ ms}^{-2}\) (upwards).
In this case, the resultant force acts vertically upward. The equation of motion is:
\(R - mg = ma\)
\(R = ma + mg\)
\(R = m(g + a)\)
\(R = 70 \text{ kg} (9.8 \text{ ms}^{-2} + 5 \text{ ms}^{-2})\)
\(R = 70 \times 14.8\)
\(R = 1036 \text{ N}\)
So, the reading on the scale is \(1036 \text{ N}\). If the scale shows mass, it would be \(\frac{1036}{9.8} \approx 105.7 \text{ kg}\).

(d) If the lift mechanism failed and it hurtled down freely under gravity:
In this situation, the lift accelerates downwards with \(a = g = 9.8 \text{ ms}^{-2}\).
Using the formula from part (b):
\(R = m(g - a)\)
\(R = m(g - g)\)
\(R = m(0)\)
\(R = 0 \text{ N}\)
Thus, the reading on the scale is zero. This condition is known as weightlessness.
In simple words: The scale measures how hard the floor pushes back on you. (a) If the lift moves at a steady speed, the scale shows your normal weight. (b) If the lift speeds up going down, you feel lighter, so the scale shows less than your normal weight. (c) If the lift speeds up going up, you feel heavier, and the scale shows more than your normal weight. (d) If the lift falls freely, you would float, and the scale would read zero because it's not pushing back on you at all.

Exam Tip: Apparent weight in a lift depends on the lift's acceleration. Remember the formulas: \(R = mg\) for constant velocity (or at rest), \(R = m(g-a)\) for downward acceleration, and \(R = m(g+a)\) for upward acceleration. Weightlessness occurs when \(a=g\) downwards.

Question 14. Figure below shows the position-time graph of a particle of mass 4 kg. What is the
(a) force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) impulse at t = 0 and t = 4s? (Consider one dimensional motion only).
Answer:
Mass of particle, \(m = 4 \text{ kg}\)

(a) Force on the particle:
(i) For \(t < 0\), the position-time graph is the line 0'0 (from \(x=0\) at negative time to \(x=0\) at \(t=0\)). This signifies that the displacement of the particle is zero, meaning it is at rest at the origin. So, the acceleration is zero, and consequently, the force on the particle is \(F = ma = 0\).
(ii) For \(t > 4 \text{ s}\), the position-time graph is the line AA'. This line is parallel to the time axis and located at \(x=3 \text{ m}\). This means the particle remains at a constant distance of \(3 \text{ m}\) from the origin and is at rest. Therefore, the acceleration is zero, and no force acts on the particle during this interval; \(F = 0\).
(iii) For \(0 < t < 4 \text{ s}\), the particle continuously changes its position. The position-time graph (OA) during this interval is a straight line with a positive slope. This indicates uniform motion of the particle, meaning it moves with a constant speed, and thus its acceleration is zero. Hence, \(F = ma = 0\) during this interval.

(b) Impulse at \(t = 0\) and \(t = 4 \text{ s}\) (one-dimensional motion only):
Impulse is defined as the change in linear momentum, \(J = \Delta p = mv - mu\).

(i) Impulse at \(t = 0\):
Before \(t = 0\) (e.g., at \(t = -0.001 \text{ s}\)), the particle is at rest, so its initial velocity \(u_{t<0} = 0\).
Immediately after \(t = 0\) (e.g., at \(t = 0.001 \text{ s}\)), the particle starts moving with a constant velocity. From the graph for \(0 < t < 4 \text{ s}\), the velocity is the slope of the line OA:
\(v_{0So, at \(t = 0\), the velocity changes from \(0\) to \(0.75 \text{ ms}^{-1}\).
Impulse at \(t = 0\) = \(m(v_{0Impulse = \(4 \text{ kg} (0.75 \text{ ms}^{-1} - 0 \text{ ms}^{-1})\)
Impulse = \(4 \times 0.75 = 3 \text{ kg ms}^{-1}\).

(ii) Impulse at \(t = 4 \text{ s}\):
Before \(t = 4 \text{ s}\) (e.g., at \(t = 3.999 \text{ s}\)), the particle is moving with a constant velocity \(u_{t<4} = 0.75 \text{ ms}^{-1}\).
Immediately after \(t = 4 \text{ s}\) (e.g., at \(t = 4.001 \text{ s}\)), the particle is at rest (as shown by the horizontal line AA'), so its final velocity \(v_{t>4} = 0\).
Impulse at \(t = 4 \text{ s}\) = \(m(v_{t>4} - u_{t<4})\)
Impulse = \(4 \text{ kg} (0 \text{ ms}^{-1} - 0.75 \text{ ms}^{-1})\)
Impulse = \(4 \times (-0.75) = -3 \text{ kg ms}^{-1}\).
The negative sign indicates that the impulse is in the opposite direction to the initial motion.
In simple words: (a) The graph shows if the particle is moving or still. If the position line is flat, it's still (no force). If it's a straight slope, it's moving at a steady speed (still no force). So, for all these time periods, the force is zero. (b) Impulse is a sudden change in motion. At t=0, it suddenly starts moving from rest, so there's an impulse. At t=4s, it suddenly stops, so there's an impulse in the opposite direction. We calculate these changes in motion using its mass and the speeds before and after.

Exam Tip: For a position-time graph, a horizontal line means the object is at rest, and a straight sloping line means constant velocity. The slope of a position-time graph gives velocity. Impulse always equals the change in momentum (mass × change in velocity).

Question 15. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Answer: We are given that the force \( F = 600N \). Let \( m_1 = 10 \) kg and \( m_2 = 20 \) kg be the masses resting on a frictionless horizontal table.
(a) When the force is applied to the heavier mass (mass \( m_2 \)):
The equations of motion for mass A and mass B are:
\( m_1a = T \) (i)
\( m_2a = F - T \) (ii)
Dividing equation (ii) by equation (i), we get:
\( \frac{m_2a}{m_1a} = \frac{F - T}{T} \)
\( \frac{m_2}{m_1} = \frac{F}{T} - 1 \)
\( \frac{20}{10} = \frac{F}{T} - 1 \)
\( 2 = \frac{F}{T} - 1 \)
\( 3 = \frac{F}{T} \)
So, \( T = \frac{F}{3} = \frac{600}{3} = 200 \,N \).
(b) When the force is applied to the lighter mass (mass \( m_1 \)):
Let \( T' \) be the tension in the string in this situation. The equations of motion for mass A and mass B are:
\( F - T' = m_1a \) (iii)
\( T' = m_2a \) (iv)
Dividing equation (iii) by equation (iv), we get:
\( \frac{F - T'}{T'} = \frac{m_1a}{m_2a} \)
\( \frac{F}{T'} - 1 = \frac{m_1}{m_2} \)
\( \frac{F}{T'} - 1 = \frac{10}{20} \)
\( \frac{F}{T'} - 1 = \frac{1}{2} \)
\( \frac{F}{T'} = 1 + \frac{1}{2} = \frac{3}{2} \)
So, \( T' = \frac{2}{3}F = \frac{2}{3} \times 600 = 400 \,N \).
In simple words: We used Newton's second law to write equations for both masses. When the force pulls the heavier mass, the tension is 200N. When the force pulls the lighter mass, the tension in the string becomes 400N.

Exam Tip: Remember to correctly identify which mass the external force is applied to, and draw clear free-body diagrams to set up the equations of motion for each mass. The tension will change depending on where the force is applied.

Question 16. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released?
Answer: Let \( m_1 = 8 \) kg and \( m_2 = 12 \) kg be the masses suspended at the ends of a light inextensible string passing over the pulley. We need to find the tension \( T \) in the string and the common acceleration \( a \) of the masses. Assume \( m_1 \) moves upward and \( m_2 \) moves downward because \( m_2 > m_1 \). We use \( g = 10 \,ms^{-2} \).
The equations of motion for mass \( m_1 \) (moving upward) and mass \( m_2 \) (moving downward) are:
For \( m_1 \): \( T - m_1g = m_1a \) (i)
For \( m_2 \): \( m_2g - T = m_2a \) (ii)
To find acceleration \( a \), add equations (i) and (ii):
\( (T - m_1g) + (m_2g - T) = m_1a + m_2a \)
\( m_2g - m_1g = (m_1 + m_2)a \)
\( (m_2 - m_1)g = (m_1 + m_2)a \)
\( \implies a = \frac{(m_2 - m_1)g}{m_1 + m_2} \)
Substitute the given values: \( m_1 = 8 \) kg, \( m_2 = 12 \) kg, \( g = 10 \,ms^{-2} \).
\( a = \frac{(12 - 8) \times 10}{8 + 12} = \frac{4 \times 10}{20} = \frac{40}{20} = 2 \,ms^{-2} \).
Now, to find tension \( T \), substitute the value of \( a \) into equation (i):
\( T - m_1g = m_1a \)
\( T = m_1g + m_1a = m_1(g + a) \)
\( T = 8(10 + 2) = 8 \times 12 = 96 \,N \).
Alternatively, from equation (ii):
\( T = m_2g - m_2a = m_2(g - a) \)
\( T = 12(10 - 2) = 12 \times 8 = 96 \,N \).
The acceleration of the masses is \( 2 \,ms^{-2} \) and the tension in the string is \( 96 \,N \).
In simple words: When two different masses hang over a pulley, the heavier one pulls the lighter one up. We used physics rules to find how fast they accelerate (2 meters per second squared) and how much pull the string has (96 Newtons).

Exam Tip: For Atwood machine problems, always draw a free-body diagram for each mass and set up separate equations of motion using Newton's second law, then solve them simultaneously for acceleration and tension.

Question 17. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must be emitted in opposite directions?
Answer: Let \( m \) be the mass of the nucleus at rest. Its initial velocity \( u = 0 \). Therefore, its initial momentum \( P_i = mu = 0 \).
When the nucleus disintegrates, it forms two smaller nuclei with masses \( m_1 \) and \( m_2 \) and respective velocities \( v_1 \) and \( v_2 \). The final momentum \( P_f \) of the system is the sum of the momenta of the two product nuclei, so \( P_f = m_1v_1 + m_2v_2 \).
According to the law of conservation of linear momentum, the total momentum of a closed system remains constant if no external forces act on it. In this case, since the initial nucleus was at rest, the total momentum before disintegration is zero. Thus, the total momentum after disintegration must also be zero.
\( P_i = P_f \)
\( 0 = m_1v_1 + m_2v_2 \)
\( \implies m_1v_1 = -m_2v_2 \)
\( \implies v_2 = -\frac{m_1v_1}{m_2} \)
The negative sign in the equation \( v_2 = -\frac{m_1v_1}{m_2} \) clearly indicates that the velocity vector \( v_2 \) is in the opposite direction to the velocity vector \( v_1 \). This means that the two smaller nuclei must be emitted in opposite directions to conserve momentum.
In simple words: When a nucleus breaks apart while standing still, the pieces must fly off in exact opposite directions. This happens because the total motion (momentum) has to stay zero, just like it was before it broke.

Exam Tip: The core concept here is the conservation of linear momentum. Always start by stating that the initial momentum equals the final momentum, especially when the system is initially at rest, meaning the total momentum is zero.

Question 18. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms\(^{-1}\) collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer: Given data:
Mass of each billiard ball, \( m = 0.05 \) kg.
Speed of each ball, \( v = 6 \,ms^{-1} \).
Let's consider one ball. Its initial velocity is \( u = +6 \,ms^{-1} \) (taking one direction as positive). After colliding and rebounding, its final velocity is \( v = -6 \,ms^{-1} \) (since it moves in the opposite direction with the same speed).
Initial momentum of the ball, \( p_i = mu = (0.05)(6) = 0.30 \,kg \,ms^{-1} \).
Final momentum of the ball, \( p_f = mv = (0.05)(-6) = -0.30 \,kg \,ms^{-1} \).
Impulse imparted to the ball is equal to the change in momentum of the ball:
Impulse \( J = p_f - p_i = (-0.30) - (0.30) = -0.60 \,kg \,ms^{-1} \).
The magnitude of the impulse imparted to this ball is \( 0.60 \,kg \,ms^{-1} \).
Since the forces involved in the collision are action-reaction pairs, the impulse imparted to the other ball will have the same magnitude but an opposite direction (i.e., \( +0.60 \,kg \,ms^{-1} \)). The two impulses are equal in magnitude but opposite in direction.
In simple words: Each billiard ball, after hitting the other and bouncing back with the same speed, experiences a change in motion of 0.60 kg ms\(^{-1}\). This "change in motion" is called impulse, and for the other ball, it's the same amount but in the reverse direction.

Exam Tip: Remember that impulse is the change in momentum. Pay close attention to the signs of velocity, especially when objects rebound or change direction, as they are crucial for correctly calculating the change in momentum.

Question 19. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms\(^{-1}\), what is the recoil speed of the gun?
Answer: Given data:
Mass of the gun, \( M = 100 \) kg.
Mass of the shell, \( m = 0.02 \) kg.
Muzzle speed of the shell, \( v = 80 \,ms^{-1} \).
Let \( V \) be the recoil speed of the gun.
Initially, both the gun and the shell are at rest before firing, so the initial momentum of the system \( P_i = 0 \).
After firing, the final momentum of the system \( P_f = MV + mv \). Here, \( V \) and \( v \) will be in opposite directions.
According to the law of conservation of linear momentum:
\( P_i = P_f \)
\( 0 = MV + mv \)
Let's take the direction of the shell's velocity as positive. Then \( v = +80 \,ms^{-1} \). The gun's recoil velocity \( V \) will be negative.
\( 0 = (100)V + (0.02)(80) \)
\( 0 = 100V + 1.6 \)
\( 100V = -1.6 \)
\( V = -\frac{1.6}{100} = -0.016 \,ms^{-1} \).
The negative sign shows that the gun recoils in the opposite direction to the shell's motion. The magnitude of the recoil speed of the gun is \( 0.016 \,ms^{-1} \). This can also be expressed as \( 1.6 \,cms^{-1} \).
In simple words: When a gun shoots a shell, the shell goes one way, and the gun kicks back the other way. Because of momentum rules, the gun's kick-back speed is much slower than the shell's speed, around 0.016 meters per second.

Exam Tip: Recoil problems are classic examples of momentum conservation. Always remember that the total momentum before and after an interaction (like firing a gun) must remain the same, especially if the initial momentum is zero.

Question 20. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 kmh\(^{-1}\). What is the impulse imparted to the ball? (mass of the ball is 0.15 kg).
Answer: Given data:
Mass of the ball, \( m = 0.15 \) kg.
Initial speed of the ball, \( u = 54 \,kmh^{-1} \).
First, convert the speed from km/h to m/s:
\( u = 54 \times \frac{5}{18} = 15 \,ms^{-1} \).
The ball deflects by an angle of 45° without changing its speed.
Let the initial velocity be along AO and the final velocity be along OB, with the angle \( \angle AOB = 45^\circ \).
Consider ON as the normal to the bat. The angle of incidence equals the angle of reflection relative to the normal.
So, \( \angle AON = \angle NOB = \theta = \frac{45^\circ}{2} = 22.5^\circ \).
The initial and final velocities of the ball can be resolved into rectangular components.
Component of initial velocity along the normal NO: \( u \cos \theta \).
Component of final velocity along the normal ON: \( -u \cos \theta \) (negative sign indicates opposite direction).
The components of velocity along the tangent (perpendicular to NO) remain constant.
Initial momentum along NO: \( p_i = mu \cos \theta \).
Final momentum along ON: \( p_f = -mu \cos \theta \).
Impulse imparted to the ball is the change in linear momentum:
Impulse \( J = p_f - p_i = (-mu \cos \theta) - (mu \cos \theta) = -2mu \cos \theta \).
Now, substitute the values:
\( J = -2 \times (0.15) \times (15) \times \cos(22.5^\circ) \)
\( \cos(22.5^\circ) \approx 0.9239 \)
\( J = -2 \times 0.15 \times 15 \times 0.9239 \)
\( J = -4.5 \times 0.9239 \approx -4.15755 \,kg \,ms^{-1} \).
The magnitude of the impulse imparted to the ball is approximately \( 4.16 \,kg \,ms^{-1} \) (or \( 4.2 \,kg \,ms^{-1} \) as rounded in the source).
In simple words: A batsman hits a ball, changing its direction by 45 degrees but not its speed. By calculating the change in the ball's motion along the direction perpendicular to the bat, we find the impulse delivered to the ball is about 4.16 kg ms\(^{-1}\).

Exam Tip: For deflection problems, resolve the initial and final velocities into components perpendicular and parallel to the surface of impact. The change in momentum (impulse) occurs only along the perpendicular direction, while the parallel component usually remains unchanged.

Question 21. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with speed 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200N?
Answer: Given data:
Mass of the stone, \( m = 0.25 \) kg.
Radius of the circle, \( r = 1.5 \) m.
Speed of whirling, \( 40 \) revolutions per minute (rev./min).
First, convert the frequency of revolution to radians per second to find angular speed \( \omega \).
Frequency \( \nu = \frac{40}{60} = \frac{2}{3} \) revolutions per second (r/s).
Angular speed \( \omega = 2\pi\nu = 2\pi \times \frac{2}{3} = \frac{4\pi}{3} \,rad \,s^{-1} \).
The tension \( T \) in the string provides the necessary centripetal force for circular motion:
\( T = m r \omega^2 \)
\( T = (0.25) \times (1.5) \times \left(\frac{4\pi}{3}\right)^2 \)
\( T = 0.25 \times 1.5 \times \frac{16\pi^2}{9} \)
Using \( \pi^2 \approx 9.87 \):
\( T = 0.25 \times 1.5 \times \frac{16 \times 9.87}{9} \)
\( T = 0.375 \times \frac{157.92}{9} \)
\( T = 0.375 \times 17.546 \approx 6.58 \,N \).
Rounded to one decimal place, \( T \approx 6.6 \,N \).
Now, for the maximum speed \( V_{max} \) the string can withstand, the maximum tension \( T_{max} = 200 \,N \).
The centripetal force is also given by \( T = \frac{mV^2}{r} \). So, for maximum tension:
\( T_{max} = \frac{mV_{max}^2}{r} \)
\( \implies V_{max}^2 = \frac{T_{max} \times r}{m} \)
\( V_{max} = \sqrt{\frac{T_{max} \times r}{m}} \)
\( V_{max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{\frac{300}{0.25}} = \sqrt{1200} \)
\( V_{max} \approx 34.64 \,ms^{-1} \).
Rounded, the maximum speed is \( 35.0 \,ms^{-1} \).
In simple words: We calculated the pull (tension) in the string while whirling a stone to be about 6.6 Newtons. If the string can handle up to 200 Newtons, the stone can be whirled much faster, reaching a maximum speed of about 35 meters per second.

Exam Tip: For circular motion, remember the formulas for centripetal force \( (F_c = mr\omega^2 \) or \( F_c = \frac{mv^2}{r}) \). Pay attention to unit conversions (e.g., rev./min to rad/s or km/h to m/s).

Question 22. If in 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Answer: (b) the stone flies off tangentially from the instant the string breaks,
The stone flies off tangentially from the instant the string breaks. This happens because, in circular motion, the velocity vector is always tangent to the circle at any given point. When the string breaks, the centripetal force (which kept the stone moving in a circle) is removed. According to Newton's first law of motion (the law of inertia), the stone will continue to move in a straight line in the direction of its velocity at the moment the string breaks, which is tangential to the circular path.
In simple words: When the string breaks, the stone will shoot off in a straight line. This line will be exactly along the direction it was moving at the moment the string snapped, which is always a line that just touches the circle (tangential).

Exam Tip: Remember Newton's First Law (Inertia): an object in motion will stay in motion in a straight line unless acted upon by an external force. In circular motion, the velocity is always tangential, so when the confining force is removed, the object continues tangentially.

Question 23. Explain why:
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it.
(d) a cricketer moves his hands backwards while holding a catch.
Answer:
(a) A horse cannot pull a cart and run in empty space because it needs a surface to push against. To move forward, the horse pushes the ground backward with its hooves. The ground then exerts an equal and opposite reaction force on the horse, pushing it forward. This forward component of the reaction force is what propels the horse and cart. In empty space, there is no surface to push against, so no reaction force is generated, and thus the horse cannot move forward or pull the cart.
(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly due to inertia of motion. When the bus is moving, the passengers are also moving at the same speed as the bus. When the bus suddenly applies brakes, it decelerates, but the passengers, due to their inertia, tend to continue moving forward at their original speed. This tendency causes them to be thrown forward.
(c) It is easier to pull a lawn mower than to push it. Let's consider the forces involved:
**When Pushing the Lawn Mower:** The applied force (F) has a horizontal component that moves the mower forward and a vertical downward component \( (F \sin \theta) \) that adds to the weight (mg) of the mower. This increases the total downward force on the ground, leading to a greater normal reaction force (R = mg + \( F \sin \theta \)). Since the force of friction (f) is proportional to the normal reaction (\( f = \mu R \)), the friction resisting motion is greater. Thus, a larger horizontal force is needed to overcome this increased friction.
**When Pulling the Lawn Mower:** The applied force (F) still has a horizontal component that moves the mower forward, but its vertical component \( (F \sin \theta) \) is directed upward. This upward component reduces the effective weight pressing down on the ground. The normal reaction force (R' = mg - \( F \sin \theta \)) is reduced. Consequently, the frictional force opposing motion (\( f' = \mu R' \)) is also reduced, making it easier to pull the mower than to push it.
(d) A cricketer moves his hands backward while holding a catch to increase the time of impact. The impulse imparted to the ball (change in its momentum) is constant, regardless of how the catch is made. Impulse is also equal to the average force multiplied by the time duration of impact (\( J = F_{avg} \times \Delta t \)). By moving his hands backward, the cricketer increases the time \( \Delta t \) over which the ball's momentum changes. Since the impulse is constant, increasing \( \Delta t \) effectively decreases the average force \( F_{avg} \) exerted on the hands, thereby reducing the impact and preventing injury.
In simple words: (a) A horse needs to push against the ground to move, which isn't possible in empty space. (b) People in a bus lurch forward when it stops because their bodies want to keep moving. (c) Pulling a lawnmower is easier than pushing it because pulling lifts it slightly, reducing friction. (d) A cricketer moves hands back to catch a ball gently, spreading the impact force over a longer time, making it less painful.

Exam Tip: These questions test fundamental concepts of Newton's laws: action-reaction for (a), inertia for (b), force components and friction for (c), and impulse-momentum theorem for (d). Provide clear, concise explanations linked to the relevant physical principles.