GSEB Class 9 Science Solutions Chapter 9 Force and Laws of Motion

 

Question 1. Which of the following has more inertia:
(a) rubber and a stone of the same size?
(b) a bicycle and a train?
(c) a five-rupees coin and a one-rupee coin?
Answer: (a) Stone has more mass and inertia
In simple words: A stone possesses more mass, which means it also has more inertia compared to a rubber of the same size. For option (b), a train has much greater inertia than a bicycle because the train's mass is significantly larger. For option (c), a five-rupees coin has more mass than a one-rupee coin, so it also has more inertia.

Exam Tip: Remember that inertia is directly proportional to mass. The more mass an object has, the harder it is to change its state of motion (i.e., the more inertia it possesses).

 

Question 2. A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team. Also, identify the agent supplying the force in each case.
Answer: The football's velocity changes four times in this scenario. First, when a football player kicks it to another player. Second, when that player kicks the football to the goalkeeper. Third, when the goalkeeper stops the football. Fourth, when the goalkeeper kicks the ball towards a player on his own team. The agent supplying the force in each case is:
First case – First player
Second case – Second player
Third case – Goalkeeper
Fourth case – Goalkeeper
In simple words: The ball changes speed four times. Each time, a player or the goalkeeper is the one making it change speed by kicking or stopping it.

Exam Tip: When analyzing forces, always identify the object whose motion is changing and the specific agent that applies the force to cause that change.

 

Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer: When we shake a tree branch with force, the branch moves quickly. However, the leaves tend to stay in their original position due to their inertia of rest. This difference in motion causes some leaves to break free and fall off the branch.
In simple words: When you shake a branch, the branch moves, but the leaves want to stay still. This sudden movement difference makes some leaves detach.

Exam Tip: This phenomenon is a classic example of Newton's first law of motion, or the law of inertia. Objects at rest tend to stay at rest.

 

Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall back when it accelerates from rest?
Answer: When a moving bus brakes suddenly, your feet come to rest with the bus. But the upper part of your body continues to move forward because of its inertia of motion. That is why you tend to fall forward. Similarly, when the bus starts moving from rest, the lower part of your body moves with the bus. However, the upper part of your body tries to stay at rest due to its inertia of rest. Therefore, you tend to fall backward.
In simple words: When a bus stops, your body keeps moving forward because of inertia. When it starts, your body wants to stay still, making you fall backward.

Exam Tip: Inertia causes resistance to changes in motion. Remember to clearly state which part of the body is affected by inertia of motion and which by inertia of rest for full marks.

 

Question 1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer: A horse pushes the ground in the backward direction. According to Newton's third law of motion, the Earth exerts a reaction force on the horse in the forward direction. This forward force allows the horse to move, and as a result, the cart moves forward too.
In simple words: A horse pushes backward on the ground. The ground pushes forward on the horse, making the horse and cart move.

Exam Tip: Emphasize that action and reaction forces act on *different* objects. The horse pushes the ground (action), and the ground pushes the horse (reaction).

 

Question 2. Explain, why is it difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity.
Answer: When a fireman holds a hose that ejects a large amount of water at high velocity, the exiting water creates a reaction force on the hose in the backward direction. This occurs due to Newton's third law of motion. As a result of this backward force, the fireman's stability is reduced, making it difficult for him to remain steady while holding the hose.
In simple words: When water shoots out fast from a hose, it pushes the hose backward. This push makes it hard for the fireman to stand still.

Exam Tip: This is a clear application of Newton's third law of motion. Explain the action (water ejected) and the equal and opposite reaction (force on the hose).

 

Question 3. From a rifle of mass 4 kg, a bullet of 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.
Answer: Given,
Mass of rifle, \( m_1 = 4 \) kg
Mass of bullet, \( m_2 = 50 \) g \( = 5 \times 10^{-3} \) kg
Velocity of bullet, \( u_2 = 35 \) m/s
Applying the principle of conservation of linear momentum,
Initial momentum \( P_i = \) Final momentum \( P_f \)
\( 0 = m_1u_1 + m_2u_2 \)
\( u_1 = - \frac{m_2 v_2}{m_1} \)
\( \implies u_1 = - \frac{5 \times 10^{-3} \times 35}{4} \)
\( \implies u_1 = - 0.044 \) m/s
The negative sign indicates that the rifle moves in a direction opposite to the direction of the bullet.
In simple words: We use the rule that momentum is saved. Before firing, the total momentum is zero. After firing, the rifle moves backward (recoils) to balance the forward momentum of the bullet.

Exam Tip: Remember to convert all masses to the same unit (kg) before performing calculations. The negative sign in the recoil velocity is crucial and indicates the opposite direction of motion.

 

Question 4. Two objects of masses 100 g and 200 g are moving along the same line in the same velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.
Answer: Given,
Mass of first body, \( m_1 = 100 \) g \( = 100 \times 10^{-3} \) kg
Mass of second body, \( m_2 = 200 \) g \( = 200 \times 10^{-3} \) kg
Initial velocity of first body, \( u_1 = 2 \) m/s
Initial velocity of second body, \( u_2 = 1 \) m/s
Final velocity of first body, \( v_1 = 1.67 \) m/s
Applying the conservation of linear momentum,
Initial momentum \( P_i = \) Final momentum \( P_f \)
\( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \)
\( (100 \times 10^{-3}) \times 2 + (200 \times 10^{-3}) \times 1 = (100 \times 10^{-3}) \times 1.67 + (200 \times 10^{-3}) \times v_2 \)
\( 0.2 + 0.2 = 0.167 + 0.2 v_2 \)
\( 0.4 = 0.167 + 0.2 v_2 \)
\( 0.2 v_2 = 0.4 - 0.167 \)
\( 0.2 v_2 = 0.233 \)
\( v_2 = \frac{0.233}{0.2} \)
\( v_2 = 1.165 \) m/s
In simple words: We used the principle that the total momentum before the collision equals the total momentum after the collision. We know the masses and most velocities, so we can calculate the unknown final velocity of the second object.

Exam Tip: It is crucial to use consistent units (e.g., kilograms for mass and meters per second for velocity). Clearly label each variable and show all steps in the conservation of momentum equation.

 

In-Text Activities Solved

 

Activity 9.1

 

Discussion and conclusion
Answer: The upper carrom coins fall vertically onto the table due to the inertia of rest. This means they tend to stay in their place unless a force acts on them, even if the lower coins are removed.
In simple words: The top carrom coins fall straight down because they stay still while the bottom ones are moved.

Exam Tip: Connect this observation directly to the concept of inertia of rest. Explain that the force applied to the bottom coin does not transfer sufficiently to the top coins to overcome their inertia.

 

Activity 9.2

 

Discussion and conclusion
Answer: The coin falls vertically into the glass tumbler because of its inertia of rest. When the card is flicked, the coin remains stationary for a moment, and gravity then pulls it straight down into the glass.
In simple words: The coin falls straight into the glass because it wants to stay still, even when the card under it is moved.

Exam Tip: Clearly identify the forces at play: the flicking force on the card, the coin's inertia of rest, and gravity acting on the coin. Explain how the card's movement is too quick for the coin's inertia to be overcome horizontally.

 

Activity 9.3

 

Discussion and conclusion
Answer: Water from the tumbler spills because of the inertia of rest of the water. When the tumbler is suddenly moved, the water tries to remain in its original position, causing it to spill out.
In simple words: The water spills when the tumbler moves suddenly because the water tries to stay in its place.

Exam Tip: Explain that liquids, like solids, exhibit inertia. When the container's motion changes abruptly, the liquid's inertia of rest (or motion) causes it to slosh or spill.

 

Activity 9.4

 

Discussion and conclusion
Answer: When two children standing on two different carts play catch with a sandbag, each child's cart moves outwards when they throw the bag. This movement is due to the instantaneous reaction force generated when the bag is thrown.
In simple words: When children on carts throw a sandbag, their carts move outward. This happens because throwing the bag creates a pushing force back on them.

Exam Tip: This activity demonstrates Newton's third law of motion and the conservation of momentum. When the bag is thrown (action), an equal and opposite force pushes the thrower (reaction).

 

Activity 9.5

Discussion and conclusion
Answer: The momentum of the inflated balloon is zero before it is pierced. When it is pierced with a pin, air rushes out with speed in one direction. To keep the total momentum the same, the balloon moves in the opposite, forward direction.
In simple words: When air shoots out of a balloon, the balloon gets pushed in the opposite direction to keep its total movement balanced.

Exam Tip: This experiment illustrates Newton's third law of motion (action-reaction) and the principle of conservation of momentum. The action is the air rushing out, and the reaction is the balloon moving forward.

 

Activity 9.6

Discussion and conclusion
Answer: This phenomenon happens due to the conservation of momentum. Initially, the total momentum of the system (cork and test tube) is zero. After the cork blows out, steam comes out in one direction. To maintain the conservation of momentum, the test tube then moves in the opposite direction.
In simple words: When the cork pops off, the steam goes one way. To keep things balanced, the test tube moves the other way.

Exam Tip: This is an example of a rocket propulsion system in miniature. It's vital to explain that momentum is conserved in an isolated system, and the total momentum remains zero even after the parts separate.

 

Gujarat Board Class 9 Science Force and Laws of Motion Textbook Questions and Answers

 

Question 1. An object experiences a net zero external unbalanced force. It is possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer: Yes, an object may travel with a non-zero velocity even when the net external force on it is zero.
The necessary conditions are:
• The object should already be moving with uniform speed along a straight line.
• All types of frictional resistance must be zero.
In simple words: Yes, something can keep moving even with no force acting on it, if it was already moving at a steady speed in a straight line and there's no friction.

Exam Tip: This question tests your understanding of Newton's First Law (Law of Inertia). Emphasize that zero net force means constant velocity (which includes zero velocity). The key is the *absence* of *change* in velocity.

 

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer: When the carpet is beaten, it is suddenly set into motion. The dust particles, however, tend to remain at rest due to their inertia. This tendency to stay put causes the dust to be dislodged and come out of the carpet.
In simple words: When you hit a carpet, the carpet moves, but the dust wants to stay still, so it shakes out.

Exam Tip: This is another clear example of inertia of rest. Explain how the carpet's sudden motion contrasts with the dust particles' resistance to change their state of rest.

 

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: Any luggage kept on the roof of a bus should be tied with a rope. This is because if the bus starts suddenly, the luggage may fall down due to its inertia of rest. Similarly, if the moving bus stops suddenly, the luggage may fall forward due to its inertia of motion.
In simple words: Luggage on a bus roof should be tied. If the bus starts, luggage can fall backward because it wants to stay still. If the bus stops, luggage can fall forward because it wants to keep moving.

Exam Tip: Use both types of inertia (rest and motion) in your explanation. Clearly state the scenario (starting, stopping) and how inertia causes the luggage to behave.

 

Question 4. A cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer: (c) The cricket ball comes to rest due to the opposition of friction force.
In simple words: A cricket ball stops rolling because a force called friction pushes against its movement, making it slow down and eventually stop.

Exam Tip: Understand that objects in motion will continue to move unless an unbalanced force acts upon them. Friction is a common opposing force in real-world scenarios.

 

Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes. (Hint: 1 metric tonne = 1000 kg.)
Answer: Given,
Distance, \( s = 400 \) m
Time, \( t = 20 \) s
Initial velocity, \( u = 0 \)
Mass, \( m = 7 \) metric tonnes \( = 7000 \) kg
Using the equation of motion, \( s = ut + \frac{1}{2}at^2 \)
\( 400 = (0 \times 20) + \frac{1}{2}a(20)^2 \)
\( 400 = 0 + \frac{1}{2}a(400) \)
\( 400 = 200a \)
\( a = \frac{400}{200} \)
\( a = 2 \) m/s\(^2\)
Force acting on the body, \( F = ma \)
\( F = 7000 \times 2 \)
\( F = 14000 \) N.
In simple words: First, we use the distance and time to find how fast the truck's speed is increasing (acceleration). Then, we use the truck's mass and its acceleration to figure out the total force pushing it.

Exam Tip: Always remember to convert units to the standard SI units (like kg for mass and m/s\(^2\) for acceleration). Clearly state the formula you are using for each step (equation of motion, Newton's second law).

 

Question 6. A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer: Given,
Mass, \( m = 1 \) kg
Initial velocity, \( u = 20 \) m/s
Distance, \( s = 50 \) m
Final velocity, \( v = 0 \) m/s
Applying the equation of motion, \( v^2 - u^2 = 2as \)
\( (0)^2 - (20)^2 = 2 \times a \times 50 \)
\( -400 = 100a \)
\( a = \frac{-400}{100} \)
\( a = -4 \) m/s\(^2\)
Force, \( F = ma \)
\( F = 1 \times (-4) \)
\( F = -4 \) N
The force of friction is 4 N.
In simple words: We first find how quickly the stone slows down (deceleration) using its starting speed, stopping speed, and distance. Then, we use the stone's mass and this slowing-down rate to calculate the friction force that stopped it. The negative sign simply shows the force is opposite to the motion.

Exam Tip: Pay close attention to the signs. Deceleration will result in a negative acceleration, and the opposing force (friction) will also be negative in relation to the initial direction of motion.

 

Question 7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
Mass of engine, \( m_1 = 8000 \) kg
Mass of all wagons, \( m_2 = 5 \times 2000 = 10,000 \) kg
Force exerted by engine, \( F_{engine} = 40,000 \) N
Friction force of track, \( F_{friction} = 5000 \) N

(a) Net accelerating force \( = \) Force exerted by engine \( - \) Frictional force
\( F_{net} = 40000 - 5000 = 35000 \) N

(b) Acceleration,
\( a = \frac{\text{Net force}}{\text{Total mass of train}} \)
\( a = \frac{F_{net}}{m_1 + m_2} \)
\( a = \frac{35000}{8000 + 10000} \)
\( a = \frac{35000}{18000} \)
\( a \approx 1.94 \) m/s\(^2\)

(c) Force on wagon 1 on wagon 2
\( = \) mass of 4 wagons \( \times \) acceleration
\( = 4 \times 2000 \times 1.94 \)
\( F = 15520 \) N
In simple words: First, we find the overall push by subtracting friction from the engine's pull. Then, we use this net push and the total weight of the engine and wagons to find how fast everything speeds up. Finally, for part (c), we figure out the force wagon 1 needs to push on the remaining four wagons to make them accelerate at that same rate.

Exam Tip: Break down the problem into smaller parts. For (a), always subtract opposing forces. For (b), remember to use the *total* mass of the system. For (c), consider the force acting on the *remaining* wagons *behind* wagon 1.

 

Question 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s\(^{-2}\)?
Answer: Given,
Mass of automobile, \( m = 1500 \) kg
Acceleration, \( a = -1.7 \) m/s\(^2\)
Force, \( F = ma \)
\( F = 1500 \times (-1.7) \)
\( F = -2550 \) N
In simple words: To find the stopping force, we multiply the car's weight by the speed at which it slows down. The negative sign means the force pushes backward, against the car's movement.

Exam Tip: Recognize that "negative acceleration" implies deceleration or slowing down. According to Newton's second law (\( F = ma \)), the force will also be in the direction of acceleration, hence negative.

 

Question 9. What is the momentum of an object of mass m, moving with a velocity y?
(a) \( (mu)^2 \)
(b) \( mu^2 \)
(c) \( \frac{1}{2} mu^2 \)
(d) \( mu \)
Answer: (d) momentum = mass x velocity
In simple words: Momentum is a measure of how much motion an object has. You calculate it by multiplying its mass (how heavy it is) by its velocity (how fast it's going and in what direction).

Exam Tip: Remember the fundamental definition and formula for momentum: \( p = mv \). Be careful not to confuse it with kinetic energy (\( \frac{1}{2}mv^2 \)) or other related concepts.

 

Question 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer: The cabinet will move with constant velocity if the net external force acting on it is zero. Since a horizontal force of 200 N acts on the cabinet in the forward direction, the net external force acting on it will be zero if a friction force of 200 N acts in the opposite direction. Thus, the friction force on the cabinet will be 200 N.
In simple words: To move something at a steady speed, the pushing force must be exactly equal to the friction force working against it. So, if you push with 200 N, friction must also be 200 N.

Exam Tip: "Constant velocity" is the key phrase here. It implies that acceleration is zero, and therefore, the net force acting on the object is also zero, according to Newton's first law.

 

Question 11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m/s before the collision during which they stick together. What will be the velocity of the combined object after the collision?
Answer: Given,
Mass of bodies, \( m_1 = m_2 = 1.5 \) kg
Initial velocity of first body, \( u_1 = 2.5 \) m/s
Initial velocity of second body, \( u_2 = -2.5 \) m/s (Negative sign for the opposite direction of motion).
Let \( v \) be the velocity of the combined object after collision.
Applying the principle of conservation of momentum,
\( (m_1 + m_2)v = m_1u_1 + m_2u_2 \)
\( (1.5 + 1.5)v = 1.5(2.5) + 1.5(-2.5) \)
\( 3v = 3.75 - 3.75 \)
\( 3v = 0 \)
\( v = 0 \)
In simple words: When two identical objects hit each other with the same speed but from opposite directions and then stick, their combined speed after the collision will be zero because their momentums cancel out.

Exam Tip: Crucially, assign a negative sign to the velocity of the object moving in the opposite direction. The total momentum before and after the collision must be conserved, especially in inelastic collisions where objects stick together.

 

Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer: The student's justification is not correct. Two equal and opposite forces cancel each other only if they act on the same body. According to the third law of motion, action and reaction forces are equal and opposite, but they both act on different bodies. Hence, they cannot cancel each other.
When we push a massive truck, the force applied on the truck is not enough to overcome the force of friction between the truck's tires and the ground. Therefore, the truck does not move. The truck will only move if the applied force is greater than the frictional force.
In simple words: The student is wrong because action and reaction forces push on different things, so they can't cancel each other out. A truck doesn't move when pushed lightly because the push isn't strong enough to beat the friction between its tires and the road.

Exam Tip: Clearly differentiate between forces acting on the same object (which can cancel) and action-reaction pairs (which always act on different objects and thus cannot cancel each other out). Also, remember to introduce the concept of static friction.

 

Question 13. A hockey ball of mass 200 g traveling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer: Given,
Mass of hockey ball, \( m = 200 \) g \( = 0.2 \) kg
Initial velocity, \( u = 10 \) m/s
Final velocity, \( v = -5 \) m/s (negative because it returns along its original path)
Change in momentum \( = \) Final momentum \( - \) Initial momentum
\( = mv - mu \)
\( = m(v - u) \)
\( = 0.2 (-5 - 10) \)
\( = 0.2 \times (-15) \)
\( = -3 \) kg m/s.
In simple words: To find the change in momentum, we subtract the starting momentum from the ending momentum. The ball moving backward means its final velocity is negative, which gives us a negative change in momentum.

Exam Tip: Be careful with the direction of velocity. If an object reverses direction, its final velocity must be assigned a negative sign to correctly calculate the change in momentum. Convert mass to kilograms.

 

Question 14. A bullet of mass 10 g traveling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer: Given,
Mass of bullet, \( m = 10 \) g \( = 10 \times 10^{-3} \) kg
Initial velocity of bullet, \( u = 150 \) m/s
Time, \( t = 0.03 \) s
Final velocity of bullet, \( v = 0 \)
Applying \( v = u + at \)
\( 0 = 150 + a(0.03) \)
\( -150 = 0.03a \)
\( a = \frac{-150}{0.03} \)
\( a = -5000 \) m/s\(^2\)
Applying \( s = ut + \frac{1}{2}at^2 \)
\( s = (150 \times 0.03) + \frac{1}{2}(-5000)(0.03)^2 \)
\( s = 4.5 + \frac{1}{2}(-5000)(0.0009) \)
\( s = 4.5 + (-2.25) \)
\( s = 2.25 \) m
Force exerted, \( F = ma \)
\( F = 10 \times 10^{-3} \times (-5000) \)
\( F = -50 \) N
In simple words: First, we find how fast the bullet slows down inside the block. Then, we use that deceleration and the bullet's initial speed to find how deep it went. Finally, we use the bullet's mass and deceleration to figure out the force the block pushed back with.

Exam Tip: This is a two-part kinematics and dynamics problem. Ensure correct unit conversions (grams to kilograms) and proper use of the kinematic equations. The negative signs for acceleration and force are important for direction.

 

Question 15. An object of mass 1 kg traveling in a straight line with a velocity of 10 m/s collides with and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer: Given,
Mass of object, \( m_1 = 1 \) kg
Velocity of object, \( u_1 = 10 \) m/s
Mass of wooden block, \( m_2 = 5 \) kg
Initial velocity of wooden block, \( u_2 = 0 \) m/s

Total momentum before impact \( P_i = m_1u_1 + m_2u_2 \)
\( P_i = (1 \times 10) + (5 \times 0) \)
\( P_i = 10 + 0 = 10 \) kg m/s

On the basis of conservation of momentum, the initial momentum and final momentum will be the same.
Total momentum after impact \( P_f = P_i = 10 \) kg m/s

Let \( v \) be the combined velocity of the object and block after collision.
\( P_f = (m_1 + m_2)v \)
\( 10 = (1 + 5)v \)
\( 10 = 6v \)
\( v = \frac{10}{6} \)
\( v \approx 1.67 \) m/s
In simple words: We calculate the total movement of the object before it hits the block. Because momentum is conserved, this total movement is the same after they stick together. Then, we use the combined weight and total movement to find their new speed.

Exam Tip: For inelastic collisions where objects stick together, the total momentum before collision equals the total momentum after. The masses combine for the final momentum calculation.

 

Question 16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer: Given,
Mass of object, \( m = 100 \) kg
Initial velocity, \( u = 5 \) m/s
Final velocity, \( v = 8 \) m/s
Time, \( t = 6 \) s

Initial momentum of object \( = mu \)
\( = 100 \times 5 = 500 \) kg m/s

Final momentum of the object \( = mv \)
\( = 100 \times 8 = 800 \) kg m/s

Force \( = \frac{\text{Change in momentum}}{\text{time}} \)
\( = \frac{800 - 500}{6} \)
\( = \frac{300}{6} \)
\( = 50 \) N
In simple words: First, we calculate the object's initial and final momentum by multiplying its mass by its starting and ending speeds. Then, we find the force by dividing the change in momentum by the time taken for that change.

Exam Tip: Remember that momentum is a vector quantity, but since motion is along a straight line here, direction can be handled by magnitude. Force is the rate of change of momentum, and it can also be found using \( F=ma \) after calculating acceleration.

 

Question 17. Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akbar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar because the change in the velocity of the insect was much more than that of the motorcar. Akhtar said that since the motorcar was moving with larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and change in their momentum. Comment on these suggestions.
Answer: Kiran's suggestion that the insect suffered a greater change in momentum compared to the motorcar is incorrect. Akhtar's suggestion that the motorcar exerted a larger force on the insect due to its high velocity is also incorrect.
Rahul's explanation is correct: on collision, the insect and the motorcar both experience the same force as action and reaction forces are always equal and opposite. Consequently, the change in momentum for both the motorcar and the insect is also the same in magnitude, although in opposite directions.
In simple words: Kiran and Akhtar were wrong. Rahul was right. When the insect hit the car, both felt the same push (force) and had the same amount of change in their movement, just in opposite directions.

Exam Tip: This question directly tests Newton's third law of motion and the conservation of momentum. Emphasize that action-reaction forces are equal in magnitude and opposite in direction, and thus the impulse (change in momentum) is also equal for both objects in a collision.

 

Question 18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10m/s\(^2\).
Answer: Given,
Mass of dumb-bell, \( m = 10 \) kg
Height, \( s = 80 \) cm \( = 0.8 \) m
Downward acceleration, \( a = 10 \) m/s\(^2\)
Initial velocity, \( u = 0 \) m/s (since it falls from rest)
Applying the equation of motion, \( v^2 - u^2 = 2as \)
\( v^2 - 0^2 = 2 \times 10 \times 0.8 \)
\( v^2 = 16 \)
\( v = \sqrt{16} \)
\( v = 4 \) m/s
Momentum transferred to the floor, \( P = mv \)
\( P = 10 \times 4 \)
\( P = 40 \) kg m/s.
In simple words: We first find how fast the dumb-bell is moving just before it hits the floor. Then, we multiply its mass by that final speed to find the momentum it transfers to the floor.

Exam Tip: First, use kinematics equations to find the velocity just before impact. Then, apply the momentum formula. Remember to convert height from centimeters to meters for consistent units.

 

Additional Exercises

 

Question 1. The following is the distance – time table of an object in motion:

1. What conclusion can you draw about the acceleration? It constant, increasing, decreasing, or zero?
2. What do you infer about the forces acting on the object?
Answer:
1. Initial velocity, \( u = 0 \)
From the second law of motion, \( s = ut + \frac{1}{2}at^2 \). Since \( u = 0 \), we have \( s = \frac{1}{2}at^2 \).
Rearranging for acceleration: \( a = \frac{2s}{t^2} \)
Let's calculate acceleration for each time interval:

Thus, the acceleration is increasing.
2. Since the acceleration is increasing, it means an unbalanced net force is continuously acting on the object, and this force is also increasing.
In simple words: By looking at the distance covered each second, we see that the object is speeding up faster and faster, which means a stronger pushing force is constantly acting on it.

Exam Tip: To determine acceleration from a distance-time table when starting from rest, calculate \( a = \frac{2s}{t^2} \) for each point. If 'a' increases, acceleration is increasing. Increasing acceleration implies an increasing net unbalanced force.

 

Question 2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort).
Answer: The motorcar's mass, \( m = 1200 \) kg. Its acceleration, \( a = 0.2 \) m/s². The force applied on the car by three people is \( F = ma = 1200 \times 0.2 = 240 \) N. The force put on the car by a single person is \( 80 \) N. Therefore, each person moves the motorcar with a force of \( 80 \) N.
In simple words: The car weighs 1200 kg and speeds up at 0.2 m/s². Three people together push with 240 N. So, each person pushes with 80 N.

Exam Tip: When calculating force per person, ensure to divide the total force by the number of people involved, assuming equal effort.

 

Question 3. A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer: Given, the hammer's mass is \( 500 \) g, which is \( 0.5 \) kg. Its initial velocity, \( u = 50 \) m/s. The final velocity of the hammer, \( \nu = 0 \) m/s, because it stops. The time taken is \( t = 0.01 \) s. Using Newton's second law of motion, the force of the nail on the hammer is the rate of change of momentum of the hammer.
\( F = \frac{\text{Change in momentum}}{\text{Time}} \)
\( F = \frac{\text{Final momentum - Initial momentum}}{\text{Time}} \)
\( F = \frac{m\nu - mu}{t} \)
\( F = \frac{0.5 \times 0 - 0.5 \times 50}{0.01} \)
\( F = \frac{0 - 25}{0.01} \)
\( F = \frac{-25}{0.01} = -2500 \) N
The negative sign shows that the force acts in the opposite direction to the hammer's motion. The force of the nail on the hammer is equal and opposite, as per Newton's third law.
In simple words: The hammer's mass is 0.5 kg, starting at 50 m/s and stopping in 0.01 s. We calculate the change in its motion, which gives us the force. The nail pushes back with a force of 2500 N, stopping the hammer.

Exam Tip: Remember to convert mass to kilograms and ensure all units are consistent (e.g., m/s, s) before calculations. A negative force indicates a decelerating or opposing force.

 

Question 4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.
Answer: Given, the mass of the motorcar, \( m = 1200 \) kg. Its initial velocity, \( u = 90 \) km/h, which is \( 90 \times \frac{5}{18} = 25 \) m/s. The final velocity, \( \nu = 18 \) km/h, which is \( 18 \times \frac{5}{18} = 5 \) m/s. The time taken, \( t = 4 \) s.
1. The acceleration, \( a \), is calculated using the formula \( a = \frac{\nu-u}{t} \).
\( a = \frac{5-25}{4} = \frac{-20}{4} = -5 \) m/s².
The negative sign shows that the velocity is decreasing, meaning it is deceleration.
2. The change in momentum is the final momentum minus the initial momentum.
Change in momentum \( = m\nu - mu = m(\nu-u) \).
Change in momentum \( = 1200 (5 - 25) = 1200 \times (-20) = -24000 \) kg-m/s.
3. The magnitude of the force required is the rate of change of momentum.
\( F = \frac{\text{Change in momentum}}{\text{Time}} \)
\( F = \frac{-24000}{4} = -6000 \) N.
The magnitude of the force is \( 6000 \) N, acting in the opposite direction of motion.
In simple words: The car, weighing 1200 kg, went from 90 km/h to 18 km/h in 4 seconds. First, we found its slowdown rate, which was 5 m/s². Then, we worked out how much its motion changed, which was -24000 kg-m/s. Finally, the force needed to do this was 6000 N.

Exam Tip: Always remember to convert velocities from km/h to m/s before applying kinematic equations or momentum calculations. A negative value for acceleration or force indicates deceleration or a force opposing motion.

 

Question 5. A large truck and a car, both moving with a velocity of magnitude V, have a head-on collision and both of them come to a halt after that. If the collision lasts for 1 s:
(a) Which vehicle experiences the greater force of impact?
(b) Which vehicle experiences the greater change in momentum?
(c) Which vehicle experiences the greater acceleration?
(d) Why is the car likely to suffer more damage than the truck?
Answer:
(a) Let the mass of the truck be \( M \) and the mass of the car be \( m \). The velocity of both the truck and the car is \( V \). The collision lasts for \( 1 \) s. During a collision, both vehicles experience the same force because action and reaction forces are always equal and opposite, according to Newton's third law.
(b) The change in the car's momentum is equal to the change in the truck's momentum. This means both vehicles experience the same change in momentum.
(c) Acceleration \( a = \frac{\text{Force}}{\text{Mass}} \). Since the force on both vehicles is the same, the vehicle with less mass will have a greater acceleration \( (a \propto \frac{1}{\text{mass}}) \). Therefore, the car's acceleration will be greater than the truck's acceleration.
(d) Because the car's acceleration (meaning its change in velocity) is larger than the truck's, the car is more likely to experience more damage as it is lighter.
In simple words: Both the truck and car hit each other with the same amount of force. Both also have the same total change in their motion. But since the car is much lighter, it slows down faster than the truck, which causes more harm to the car.

Exam Tip: Remember Newton's Third Law (equal and opposite forces) and Second Law (Force = mass x acceleration) when analyzing collisions. Even if forces are equal, the effect (acceleration and damage) will differ based on mass.