CBSE Class 11 Physics Kinematics Basic Problems

BASIC LEVEL ASSIGNMENT
CLASS 12 PHYSICS : KINEMATICS

1. The displacement (x) of a particle moving in one dimension, under the action of a constant force is related to the time t by the equation t = √x +3 where x is in metres and t is in seconds. Find the displacement of the particle when its velocity is zero.

Answer:
1. From t = √x + 3, √x = t - 3, so x = (t - 3)².
2. Velocity v = dx/dt = 2(t - 3).
3. v = 0 when t = 3 s.
4. At t = 3 s, x = (3 - 3)² = 0.
5. Hence, the displacement of the particle is zero when its velocity is zero.

Teacher's Note:
a) Differentiate the displacement to get the velocity, then set it equal to zero.
b) The particle is at its starting position when it momentarily stops.

2. A car accelerates from rest at a constant rate α for sometime after which it decelerates at a constant rate β to come to rest. If the total time lapse is t seconds, evaluate
(a) the maximum velocity reached and
(b) the total distance travelled.

Answer:
1. Let the maximum velocity be v_max. The accelerating time is v_max/α and the decelerating time is v_max/β.
2. Total time: v_max(1/α + 1/β) = t, so v_max = αβt/(α + β).
3. (a) The maximum velocity is \( \frac{αβt}{α + β} \).
4. (b) The v-t graph is a triangle with base t and height v_max. Distance = (1/2) × t × v_max = \( \frac{αβt²}{2(α + β)} \).

Teacher's Note:
a) The distance is the area of the triangle under the velocity-time graph.
b) The maximum velocity is reached at the instant when acceleration changes to deceleration.

3. A rocket is fired vertically from rest and ascends with constant net vertical acceleration of 30 m/sec² for 1 minute. Its fuel is then all used up and it continues as a free particle in the gravitational field of the earth. Find.
(a) maximum height reached
(b) the total time elapsed from take-off until the rocket strikes the earth.

Answer:
1. Take g = 10 m/s². Powered stage, t = 60 s: v = 30 × 60 = 1800 m/s and height h₁ = (1/2)(30)(60)² = 54000 m.
2. Free stage: further rise h₂ = v²/(2g) = (1800)²/20 = 162000 m. The time to rise = v/g = 180 s.
3. (a) Maximum height = 54000 + 162000 = 216000 m = 216 km.
4. (b) Time to fall from 216000 m: t = √(2 × 216000/10) = √43200 = 207.8 s.
5. Total time = 60 + 180 + 207.8 = 447.8 s.

Teacher's Note:
a) After the fuel is used up the rocket continues upward until its velocity becomes zero.
b) The fall begins from rest at the maximum height, so use h = (1/2)gt².
c) Convert the minute to 60 seconds before substituting.

4. A parachutist after bailing out falls 50 m, without friction. When the parachute opens, he decelerates downwards with 2.0 m/s². He reaches the ground with a speed of 3 m/sec.
(a) How long is the parachutist in the air ?
(b) At what height did he bail out ?

Answer:
1. Take g = 9.8 m/s². Free fall of 50 m: v₁ = √(2 × 9.8 × 50) = 31.3 m/s and the time t₁ = 31.3/9.8 = 3.19 s.
2. After the parachute opens, the speed falls from 31.3 m/s to 3 m/s at 2.0 m/s²: t₂ = (31.3 - 3)/2 = 14.15 s.
3. Distance covered with the parachute open: s₂ = (31.3² - 3²)/(2 × 2) = (979.7 - 9)/4 = 242.7 m.
4. (a) Time in the air = 3.19 + 14.15 = 17.34 s.
5. (b) Height of bail-out = 50 + 242.7 = 292.7 m.

Teacher's Note:
a) Split the motion into two stages, free fall and retarded fall; the final speed of stage 1 is the initial speed of stage 2.
b) Use v² = u² - 2as with the deceleration taken as positive.

5. A man in a lift ascending with an acceleration f, throws a ball vertically upwards with a velocity v and catches it after time t₁. Afterwards when the lift is descending with the same acceleration, the man again throws a ball vertically upwards with the same velocity and catches it after time t₂. Determine v and f in terms of t₁, t₂ and g, the acceleration due to gravity.

Answer:
1. In the frame of the lift, the velocity of throwing is v. When the lift ascends with acceleration f, the effective acceleration is g + f, so t₁ = 2v/(g + f).
2. When the lift descends with acceleration f, the effective acceleration is g - f, so t₂ = 2v/(g - f).
3. Then g + f = 2v/t₁ and g - f = 2v/t₂. Adding: 2g = 2v(1/t₁ + 1/t₂), so v = \( \frac{g t1t2}{t1 + t2} \).
4. Subtracting: 2f = 2v(1/t₁ - 1/t₂), so f = \( \frac{(t2 - t1)g}{t1 + t2} \).

Teacher's Note:
a) In the lift frame a pseudo force acts on the ball, so g becomes g + f or g - f.
b) Take the velocity v as relative to the lift.

6. An open elevator is ascending with constant speed v = 10 m/sec. A ball is thrown vertically up by a boy on the elevator when he is at a height of h = 30 m from the ground. The velocity of the projection is v = 20 m/sec with respect to elevator. Find
(a) the maximum height attained by the ball
(b) time taken by the ball to meet the elevator again
(c) the time taken by the ball to reach the ground after crossing the elevator.

Answer:
1. Take g = 10 m/s². The velocity of the ball with respect to the ground = 20 + 10 = 30 m/s upward.
2. (a) Rise above the thrower = 30²/(2 × 10) = 45 m. Maximum height = 30 + 45 = 75 m.
3. (b) The elevator moves at constant speed, so in its frame the ball behaves like a ball thrown at 20 m/s under g. Time to meet the elevator = 2 × 20/10 = 4 s.
4. (c) At t = 4 s the ball is at height 30 + 30 × 4 - 5 × 16 = 70 m with velocity 30 - 40 = -10 m/s (downward).
5. Then 70 = 10t + 5t², so t² + 2t - 14 = 0, giving t = -1 + √15 = 2.87 s.

Teacher's Note:
a) Velocity relative to the ground is the sum of the velocity of the elevator and the relative velocity.
b) The elevator is not accelerating, so the relative time of flight is 2u/g.
c) For part (c), use the position and velocity of the ball at the instant of crossing.

7. A particle moves in the plane xy with constant acceleration a directed along the negative y axis. The equation of motion of the particle has the form y = αx - βx², where α and β are positive constants. Find the velocity of the particle at the origin of coordinates.

Answer:
1. The acceleration is only along -y, so the x-component of velocity is constant, v_x = u (say), and x = ut.
2. Then y = αut - βu²t². Differentiating twice with respect to t: a_y = -2βu².
3. Since a_y = -a, u² = a/(2β).
4. The y-component of velocity at the origin: v_y = dy/dt at t = 0 = αu.
5. Speed at the origin: v₀ = √(u² + α²u²) = u√(1 + α²) = \( \sqrt{\frac{a(1 + \alpha^2)}{2\beta}} \).

Teacher's Note:
a) This is projectile motion, with the horizontal velocity constant.
b) Compare the trajectory with y = x tanθ - gx²/(2u²cos²θ) to check: tanθ = α.

8. A particle moves along a horizontal path such that its velocity is given by v = (3t² -6t) m/s where t is the time in seconds. If it is initially located at origin O, determine the distance travelled by the particle during the time interval t = 0 to t = 3.5 seconds and the particle's average velocity and average speed during this time period.

Answer:
1. Position x = ∫v dt = t³ - 3t² (since x = 0 at t = 0).
2. The velocity is zero at t = 0 and t = 2 s. For t from 0 to 2 s the velocity is negative, so the particle moves backward, and for t greater than 2 s it moves forward.
3. x(2) = 8 - 12 = -4 m and x(3.5) = 42.875 - 36.75 = 6.125 m.
4. Distance travelled = 4 + (4 + 6.125) = 14.125 m, about 14.1 m.
5. Average velocity = displacement/time = 6.125/3.5 = 1.75 m/s.
6. Average speed = distance/time = 14.125/3.5 = 4.04 m/s (the answer key gives 4.03 m/s, using the rounded distance 14.1 m).

Teacher's Note:
a) The particle reverses direction at t = 2 s, so the distance is not the same as the displacement.
b) Average velocity uses displacement; average speed uses total path length.

9. How long will a plane take to fly around a square with side a with the wind blowing at a velocity u, in the two cases:
(a) the direction of the wind coincides with one of the sides
(b) the direction of the wind coincides with one diagonal of the square?
The velocity of the plane in still air v > u.

Answer:
1. (a) Two sides are along the wind: the times are a/(v + u) and a/(v - u). The other two sides are perpendicular to the wind: the plane heads upwind at an angle, so the ground speed is √(v² - u²), and each takes a/√(v² - u²).
2. Total time = a/(v + u) + a/(v - u) + 2a/√(v² - u²) = 2av/(v² - u²) + 2a/√(v² - u²).
3. So the time = \( \frac{2a\left[v + \sqrt{v^2 - u^2}\right]}{v^2 - u^2} \).
4. (b) With the wind along a diagonal, the wind makes 45° with every side. For each side, the wind component along it is ±u/√2 and across it u/√2.
5. The ground speed on a side is w ± u/√2, where w = √(v² - u²/2). Two sides have the plus sign and two have the minus sign.
6. Total time = 2a/(w + u/√2) + 2a/(w - u/√2) = 4aw/(w² - u²/2) = 4aw/(v² - u²).
7. So the time = \( \frac{4a\sqrt{v^2 - u^2/2}}{v^2 - u^2} \).

Teacher's Note:
a) Resolve the wind into components along and across the side being flown.
b) Cancel the cross-wind component by heading slightly into the wind; only the remaining component gives the ground speed.

10. A stone is thrown up from the top of a tower 20 metres with a velocity of 24 metres/second at an elevation of 30º above the horizontal. Find the horizontal distance from the foot of the tower to the point at which the stone hits the ground.

Answer:
1. Take g = 9.8 m/s². Horizontal velocity = 24 cos 30° = 20.78 m/s and vertical velocity = 24 sin 30° = 12 m/s upward.
2. Taking the top of the tower as the origin, the stone reaches the ground when -20 = 12t - 4.9t², that is, 4.9t² - 12t - 20 = 0.
3. t = [12 + √(144 + 392)]/9.8 = (12 + 23.15)/9.8 = 3.587 s.
4. Horizontal distance = 20.78 × 3.587 ≈ 74.5 m (the answer key gives 74.53 m).

Teacher's Note:
a) Reject the negative root of the quadratic because time cannot be negative.
b) The height of the tower is a displacement of -20 m in the vertical equation.

11. Two bodies were thrown simultaneously from the same point : one, straight up, and the other, at angle θ = 60º to the horizontal. The initial velocity of each body is equal to v₀ = 25 m/s. Neglecting the air drag, find the distance between the bodies t = 1.70 s later.

Answer:
1. Both bodies have the same acceleration g, so the distance between them is found from the relative velocity, which is constant.
2. Velocity of the first = (0, 25) m/s. Velocity of the second = (25 cos 60°, 25 sin 60°) = (12.5, 21.65) m/s.
3. Relative velocity = (12.5, -3.35), of magnitude √(156.25 + 11.2) = 12.94 m/s.
4. Distance = 12.94 × 1.70 = 22 m.

Teacher's Note:
a) In the frame of one body, the other moves in a straight line with constant velocity.
b) Check: the angle between the two velocities is 30°, so |v_rel| = 2v₀ sin 15° = 2 × 25 × 0.2588 = 12.94 m/s.

12. Two seconds after projection, a projectile is moving at 30º above the horizontal; after one more second it is moving horizontally. Find the magnitude and direction of its initial velocity. (g = 10 m/s²)

Answer:
1. The projectile moves horizontally at t = 3 s, so its vertical velocity is zero at t = 3 s: u_y = 10 × 3 = 30 m/s.
2. At t = 2 s: v_y = 30 - 20 = 10 m/s upward, and the direction is 30° above the horizontal, so v_x = 10/tan 30° = 10√3 m/s.
3. Horizontal velocity is constant, so u_x = 10√3 m/s.
4. Magnitude u = √(30² + (10√3)²) = √(900 + 300) = 20√3 m/s.
5. Direction: tan θ = 30/(10√3) = √3, so θ = 60° above the horizontal.

Teacher's Note:
a) At the highest point the vertical velocity is zero.
b) The horizontal component remains unchanged throughout the flight.

13. A ball is projected from O with an initial velocity 700 cm/s in a direction 37º above the horizontal. A ball B, 500 cm away from O on the line of the initial velocity of A, is released from rest at the instant A is projected. Find
(a) The height through which B falls, before it is hit by A.
(b) The direction of the velocity A at the time of impact
(Given g = 10 ms^-2, sin 37º = 0.6)

Answer:
1. Both balls have the same acceleration g, so in the frame of B the ball A moves in a straight line with the speed 7 m/s towards B. It reaches B when t = 5/7 s.
2. (a) Height through which B falls = (1/2)gt² = 5 × (5/7)² = 2.55 m.
3. (b) At t = 5/7 s: v_x = 7 cos 37° = 5.6 m/s and v_y = 7 sin 37° - 10 × 5/7 = 4.2 - 7.14 = -2.94 m/s.
4. tan φ = 2.94/5.6 = 0.525, so φ = 27.7° (about 27°43') below the horizontal.

Teacher's Note:
a) The time of impact follows from the relative motion, which is along the line of projection.
b) A negative vertical velocity means that the ball is already coming down at the time of impact.

14. Three particles are projected simultaneously and in the same vertical plane with different velocities and at different angles. Show that the area of the triangle formed by the particles at time t is proportional to t².

Answer:
1. Let the particles be projected from the same point (the origin) with velocities u₁, u₂ and u₃. At time t, r_i = u_it + (1/2)gt².
2. The sides of the triangle are the relative position vectors: r₂ - r₁ = (u₂ - u₁)t and r₃ - r₁ = (u₃ - u₁)t. The term (1/2)gt² cancels because it is the same for all.
3. Area = (1/2)|(r₂ - r₁) × (r₃ - r₁)| = (1/2)|(u₂ - u₁) × (u₃ - u₁)| t².
4. The vector part is constant, so the area is proportional to t². Hence proved.

Teacher's Note:
a) The particles are taken to start from a common point, which makes the sides of the triangle proportional to t.
b) Gravity acts equally on all three, so it does not change relative positions.

15. Two bodies are projected from the same point with equal velocities in such direction that they both strike the same point on a plane whose inclination is β. If α be the angle of projection of the first, show that the ratio of their time of flight is sin (α + β): cos α.

Answer:
1. Take the plane to slope downward from the point of projection at angle β to the horizontal, with α measured above the horizontal (this is the arrangement for which the given ratio holds). The angle of projection with the plane is α + β.
2. Along the plane: acceleration g sin β (down the slope). Perpendicular to the plane: acceleration g cos β. Time of flight T = 2u sin(α + β)/(g cos β).
3. Range along the plane R = u cos(α + β)T + (1/2)g sin β T² = 2u² sin(α + β) cos α/(g cos²β).
4. For the second body with angle α' and the same speed and same range: sin(α + β) cos α = sin(α' + β) cos α'. This gives sin(2α + β) = sin(2α' + β), so 2α' + β = π - (2α + β), that is, α' + β = π/2 - α.
5. So the time of flight of the second body is T' = 2u sin(α' + β)/(g cos β) = 2u cos α/(g cos β).
6. T/T' = sin(α + β)/cos α. Hence, the ratio of times of flight is sin(α + β) : cos α.

Teacher's Note:
a) For an upward-sloping plane, replace β by -β in the working.
b) Two different angles of projection give the same range when they add up to the complement of the plane's inclination.

16. Three points are located at the vertices of an equilateral triangle of side a. They all start moving simultaneously with speed v, with the first point heading continuously for the second, the second for the third, and the third for the first. How soon will the points meet.

Answer:
1. By symmetry the three points always form an equilateral triangle, which shrinks and rotates, and they meet at the centroid.
2. Each point moves with speed v directly towards the next point. The next point's velocity has a component along the line joining them, directed towards the first point, equal to v cos 60° = v/2.
3. So the distance between two neighbouring points decreases at the rate v + v/2 = 3v/2.
4. Time = a/(3v/2) = \( \frac{2a}{3v} \).

Teacher's Note:
a) Find the rate at which the separation of two adjacent points decreases.
b) The tangential motion does not change the separation along the line of sight.

17. A turn of radius 20 m is banked for vehicles going at speed of 36 km/h. If the co-officient of static friction between the road and the tyre is 0.4, what are possible speeds of a vehicle so that it neither slips down nor skids up.

Answer:
1. Take g = 10 m/s². The design speed is 36 km/h = 10 m/s, so tan θ = v²/(rg) = 100/200 = 0.5.
2. Maximum speed (friction acts down the slope): v_max² = rg(tan θ + μ)/(1 - μ tan θ) = 200 × 0.9/0.8 = 225, so v_max = 15 m/s = 54 km/h.
3. Minimum speed (friction acts up the slope): v_min² = rg(tan θ - μ)/(1 + μ tan θ) = 200 × 0.1/1.2 = 16.7, so v_min = 4.08 m/s = 14.7 km/h.
4. Hence, the possible speeds lie between 14.7 km/h and 54 km/h.

Teacher's Note:
a) The limiting friction acts down the slope at the highest safe speed and up the slope at the lowest.
b) Convert km/h into m/s by dividing by 3.6.
c) Check that the design speed of 36 km/h lies inside the range.

18. A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v₀. Due to the wind the balloon gathers the horizontal velocity component v_x = a y. where a is a constant and y is the height of ascent. Find how the following quantities depend on the height of ascent :
(a) the horizontal drift of the balloon x (y)
(b) the total, tangential, and normal accelerations of the balloon.

Answer:
1. (a) The vertical motion is uniform, so y = v₀t. Then dx/dt = ay = av₀t, so x = av₀t²/2 = \( \frac{a}{2v0} \) y².
2. (b) The velocity components are v_x = ay and v_y = v₀, so the acceleration components are a_x = a v₀ and a_y = 0.
3. Total acceleration w = a v₀, constant and horizontal.
4. The speed is v = √(v₀² + a²y²). Tangential acceleration w_t = (v · w)/v = (ay)(av₀)/v = \( \frac{a²y}{\sqrt{1 + (ay/v_0)^2}} \).
5. Normal acceleration w_n = √(w² - w_t²) = av₀v₀/v = \( \frac{a v0}{\sqrt{1 + (ay/v_0)^2}} \).

Teacher's Note:
a) The tangential acceleration is the component of the acceleration along the velocity.
b) The normal acceleration follows from w² = w_t² + w_n².
c) Express everything in terms of y by using t = y/v₀.

19. The acceleration-displacement graph of a particle moving in a straight line is as shown in figure. Initial velocity of particle is zero. Find the velocity of the particle when displacement of the particle is, s = 12 m.

Answer:
1. From v dv = a ds, we get (1/2)v² = area under the a-s graph, since v = 0 at s = 0.
2. Area from s = 0 to 2 m (triangle): (1/2) × 2 × 2 = 2.
3. From 2 to 8 m (rectangle): 6 × 2 = 12.
4. From 8 to 10 m (trapezium): (1/2)(2 + 4) × 2 = 6. From 10 to 12 m (triangle): (1/2) × 2 × 4 = 4.
5. Total area = 2 + 12 + 6 + 4 = 24, so (1/2)v² = 24 and v² = 48.
6. Hence, v = 4√3 m/s.

Teacher's Note:
a) The area under an a-s graph gives half the change in v².
b) Break the graph into simple shapes and add their areas.