JEE Physics Kinematics MCQs Set 07

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Chapter-wise Objective Questions: Kinematics

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Multiple Choice Questions

Question 1: An electron moving along the x-axis has a position given by x = 20te⁻ᵗ m, where t is in seconds. How far is the electron from the origin when it momentarily stops?
(a) 20 m
(b) \( \frac{20}{e} \) m
(c) 20e m
(d) zero
Show Answer & Explanation

Answer: (b) \( \frac{20}{e} \) m

Explanation:
1. v = 20e⁻ᵗ(1 - t) = 0 at t = 1 s.
2. x = 20(1)e⁻¹ = 20/e m.

Teacher's Note:
1. The electron stops when v = 0.
2. Differentiate using the product rule.

Question 2: A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. The time when he will fall into a pit 13 m away from him is:
(a) 42 s
(b) 40 s
(c) 13 s
(d) 37 s
Show Answer & Explanation

Answer: (d) 37 s

Explanation:
1. Each cycle of 8 steps (8 s) gives a net gain of 2 m.
2. After 4 cycles (32 s) he is at 8 m; the next 5 steps take him to 13 m, taking 5 s more.
3. Total = 37 s.

Teacher's Note:
1. Find the net advance per cycle and when the forward run reaches the pit.
2. Check that the pit is reached during a forward run.

Question 3: A car moves on a straight track from station A to station B, with an acceleration a = (b - cx), where b and c are constants and x is the distance from station A. The maximum velocity between the two stations is:
(a) \( c/\sqrt a \)
(b) \( b/\sqrt c \)
(c) \( b/c \)
(d) \( \sqrt{b/c} \)
Show Answer & Explanation

Answer: (b) \( b/\sqrt c \)

Explanation:
1. v dv/dx = b - cx gives v²/2 = bx - cx²/2.
2. Maximum v occurs when a = 0, i.e. x = b/c, where v² = b²/c.
3. v_max = b/√c.

Teacher's Note:
1. The velocity is maximum when the acceleration is zero.
2. Integrate v dv = a dx.

Question 4: Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 4.9 m, the number of balls thrown per minute will be:
(a) 120
(b) 40
(c) 60
(d) 50
Show Answer & Explanation

Answer: (c) 60

Explanation:
1. Time to reach the maximum height = √(2 × 4.9/9.8) = 1 s.
2. One ball is thrown every second, so 60 per minute.

Teacher's Note:
1. The throwing interval equals the time to reach the top.
2. Use h = (1/2)gt².

Question 5: A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of the iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into the iron. A similar bullet fired with the same velocity from the opposite direction goes through the iron first and then penetrates 2 cm into the wood. If a1 and a2 are the retardations offered to the bullet by wood and iron respectively, then:
(a) \( a_1=a_2 \)
(b) \( a_1=2a_2 \)
(c) data insufficient
(d) \( a_2=2a_1 \)
Show Answer & Explanation

Answer: (d) \( a_2=2a_1 \)

Explanation:
1. v0² = 2a1(4) + 2a2(1) = 2a2(2) + 2a1(2).
2. 8a1 + 2a2 = 4a2 + 4a1, so 4a1 = 2a2 and a2 = 2a1.

Teacher's Note:
1. Equate the squares of the initial speed in the two cases.
2. Use v² = u² - 2as for each layer.

Question 6: A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height did he bail out? (g = 9.8 m/s²)
(a) 182 m
(b) 91 m
(c) 111 m
(d) 293 m
Show Answer & Explanation

Answer: (d) 293 m

Explanation:
1. Speed after the free fall: v² = 2(9.8)(50) = 980.
2. After the parachute opens: 3² = 980 - 2(2)h, so h = 971/4 = 242.75 m.
3. Total height = 50 + 242.75 ≈ 293 m.

Teacher's Note:
1. Treat the two stages separately.
2. The final speed of stage 1 is the initial speed of stage 2.

Question 7: For a particle in one-dimensional motion:
(i) with zero speed at an instant, it may have non-zero acceleration at that instant
(ii) with zero speed it may have non-zero velocity
(iii) with constant speed it must have zero acceleration
(iv) with a positive value of acceleration it must be speeding up
Which of the following is/are correct?
(a) i, iii
(b) i, ii
(c) ii, iv
(d) iii, iv
Show Answer & Explanation

Answer: (a) i, iii

Explanation:
1. At the top of a vertical throw, speed is zero but acceleration is g.
2. Zero speed means zero velocity.
3. In one dimension constant speed means constant velocity.
4. A positive acceleration slows a body moving in the negative direction.

Teacher's Note:
1. Speed zero implies velocity zero.
2. Acceleration and velocity signs decide speeding up or slowing down.

Question 8: Let v and a denote the velocity and acceleration of a body. Select the wrong statement(s):
(i) a can be non-zero when v = 0
(ii) a must be zero when v = 0
(iii) a may be zero when v is non-zero
(iv) the direction of a must have some correlation with the direction of v
Which of the above statements are wrong?
(a) i, iii
(b) i, ii
(c) iii, iv
(d) ii, iv
Show Answer & Explanation

Answer: (d) ii, iv

Explanation:
1. A body at the top of a vertical throw has v = 0 and a = g, so (i) is true and (ii) is wrong.
2. A body with uniform velocity has a = 0, so (iii) is true.
3. The directions of a and v can be unrelated, so (iv) is wrong.

Teacher's Note:
1. Check each with a simple example.
2. Wrong statements are those with 'must'.

Question 9: Let v and a denote the velocity and acceleration of a body in one-dimensional motion. Select the wrong statements:
(i) |v| must decrease when a < 0
(ii) speed must increase when a > 0
(iii) speed will increase when both v and a are < 0
(iv) speed will decrease when v < 0 and a > 0
Which of the above statements are wrong?
(a) ii, iv
(b) i, ii
(c) iii, iv
(d) i, iii
Show Answer & Explanation

Answer: (b) i, ii

Explanation:
1. If v > 0 and a < 0 the speed falls, but if v < 0 and a < 0 it rises; so (i) is wrong.
2. Similarly (ii) is wrong.
3. (iii) and (iv) are correct because the sign of a relative to v decides.

Teacher's Note:
1. Compare the signs of v and a.
2. Same signs: speeding up; opposite signs: slowing down.

Question 10: Consider the motion of the tip of the minute hand of a clock in one hour:
(i) the displacement is zero
(ii) the distance covered is zero
(iii) the average speed is zero
(iv) the average velocity is zero
Which of the following is/are correct?
(a) iii, iv
(b) i, iv
(c) ii, iii
(d) i, ii
Show Answer & Explanation

Answer: (b) i, iv

Explanation:
1. After one hour the tip returns to its starting point, so displacement and average velocity are zero.
2. The distance is 2πr and the average speed is non-zero.

Teacher's Note:
1. Displacement is the shortest vector from start to end.
2. Distance is the path length.

Question 11: The velocity (v) of a particle is plotted against time (t) as a straight line which is negative at t = 0, crosses zero at t = T and is positive and equal in magnitude to its initial value at t = 2T. Then:
(i) the particle changes its direction of motion at some point
(ii) the acceleration of the particle remains constant
(iii) the displacement of the particle is zero
(iv) the initial and final speeds of the particle are the same
Which of the following is/are correct?
(a) i, ii, iii, iv
(b) i, ii, iii
(c) ii, iii, iv
(d) i, iii, iv
Show Answer & Explanation

Answer: (a) i, ii, iii, iv

Explanation:
1. The velocity changes sign at t = T, so the direction reverses.
2. A straight v-t line means constant acceleration.
3. The area below the axis (0 to T) equals the area above it (T to 2T), so the net displacement is zero.
4. The initial and final velocities are equal in magnitude.

Teacher's Note:
1. Check the sign change of v.
2. The two triangles have equal areas.

Question 12: The velocity of a particle is zero at t = 0. Then:
(i) the acceleration may be zero at t = 0
(ii) the acceleration must be zero at t = 0
(iii) if the acceleration is zero from t = 0 to t = 2 s, the speed is also zero in this interval
(iv) if the speed is zero from t = 0 to t = 2 s, the acceleration is also zero in this interval
Which of the following is/are correct?
(a) i, iii, iv
(b) i, ii, iv
(c) ii, iii, iv
(d) i, ii, iii
Show Answer & Explanation

Answer: (a) i, iii, iv

Explanation:
1. A body at rest may or may not be accelerating at an instant.
2. With a = 0 throughout and v = 0 at the start, v stays zero.
3. If the speed stays zero, the acceleration is zero throughout.

Teacher's Note:
1. Use v = u + at over the interval.
2. Zero speed throughout means no change in velocity.

Question 13: Assertion (A): A body may be accelerated even when it is moving uniformly.
Reason (R): When the direction of motion of the body is changing, the body must have acceleration.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Show Answer & Explanation

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Explanation:
1. A body moving with constant speed along a circle is accelerated because its direction changes.
2. The reason states exactly why.

Teacher's Note:
1. Acceleration is the rate of change of the velocity vector.
2. A direction change is a velocity change.

Question 14: Assertion (A): The average velocity of an object over an interval of time is either smaller than or equal to the average speed of the object over the same interval.
Reason (R): Velocity is a vector quantity and speed is a scalar quantity.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Show Answer & Explanation

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Explanation:
1. Displacement is never greater than the path length, so |average velocity| ≤ average speed.
2. The vector nature of velocity (and the scalar nature of speed) is the reason.

Teacher's Note:
1. Displacement ≤ distance.
2. A vector magnitude can cancel; a scalar cannot.

Question 15: Assertion (A): A body can have acceleration even if its velocity is zero at a given instant of time.
Reason (R): A body is momentarily at rest when it reverses its direction.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Show Answer & Explanation

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Explanation:
1. At the top of a vertical throw, v = 0 but a = g.
2. The body reverses its direction there, being momentarily at rest while still accelerating.

Teacher's Note:
1. Reversal of direction passes through v = 0.
2. The acceleration need not be zero at that instant.

Question 16: Assertion (A): For one-dimensional motion, the angle between acceleration and velocity must be zero.
Reason (R): One-dimensional motion is always along a straight line.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Show Answer & Explanation

Answer: (d) Assertion is false but Reason is true.

Explanation:
1. The angle can be 0° or 180° (a decelerating body), so the assertion is false.
2. The reason is true.

Teacher's Note:
1. In 1-D the angle can be 180°.
2. Deceleration has acceleration opposite to velocity.

Question 17: Assertion (A): For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear stationary.
Reason (R): If the observer and the object are moving at velocities v1 and v2 respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is v2 - v1.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Show Answer & Explanation

Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Explanation:
1. Nearby objects have a large angular velocity relative to the observer, distant ones a small one.
2. The relative velocity formula is correct, but it does not explain the effect of distance (angular speed).

Teacher's Note:
1. The apparent motion depends on angular displacement.
2. Both statements are true but unrelated as explanation.

Question 18: From the top of a multi-storeyed building 39.2 m tall, a boy projects a stone vertically upwards with an initial velocity of 9.8 m/s such that it finally drops to the ground (g = 9.8 m/s²). The stone reaches the ground in:
(a) 1 s
(b) 3 s
(c) 2 s
(d) 4 s
Show Answer & Explanation

Answer: (d) 4 s

Explanation:
1. Taking upward as positive: -39.2 = 9.8t - 4.9t², so t² - 2t - 8 = 0.
2. t = 4 s.

Teacher's Note:
1. Use s = ut + (1/2)at² with s = -39.2 m.
2. Reject the negative root.

Question 19: From the top of a multi-storeyed building 39.2 m tall, a boy projects a stone vertically upwards with an initial velocity of 9.8 m/s such that it finally drops to the ground (g = 9.8 m/s²). The stone will pass through the point of projection again after:
(a) 3 s
(b) 5 s
(c) 2 s
(d) 4 s
Show Answer & Explanation

Answer: (c) 2 s

Explanation:
1. On the way back it passes the point of projection with the same speed after t = 2u/g = 2 s.

Teacher's Note:
1. Time of flight to the same level = 2u/g.
2. The motion is symmetric.

Question 20: From the top of a multi-storeyed building 39.2 m tall, a boy projects a stone vertically upwards with an initial velocity of 9.8 m/s such that it finally drops to the ground (g = 9.8 m/s²). The velocity of the stone just before striking the ground is:
(a) 14.2 m/s
(b) 22.4 m/s
(c) 34.2 m/s
(d) 29.4 m/s
Show Answer & Explanation

Answer: (d) 29.4 m/s

Explanation:
1. v = u - gt = 9.8 - 9.8(4) = -29.4 m/s.
2. The speed is 29.4 m/s.

Teacher's Note:
1. Use v = u + at with t = 4 s.
2. Or v² = u² + 2gh.

Question 21: The velocity-time graph of a body moving along a straight line is made of straight segments: v rises uniformly from 0 to +10 m/s during 0 to 2 s, stays at +10 m/s from 2 s to 6 s, falls uniformly through zero at 8 s to -10 m/s at 10 s, and then rises uniformly to 0 at 12 s. The acceleration from 10 s to 12 s is:
(a) 12 m/s²
(b) 10 m/s²
(c) 5 m/s²
(d) zero
Show Answer & Explanation

Answer: (c) 5 m/s²

Explanation:
1. a = (0 - (-10))/(12 - 10) = 5 m/s².

Teacher's Note:
1. Acceleration is the slope of the v-t graph.
2. Take the signed change in velocity.

Question 22: A ball is thrown from a point with a speed v0 at an elevation angle θ. From the same point and at the same instant, a person starts running with a constant speed v0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?
(a) Yes, 45°
(b) Yes, 30°
(c) Yes, 60°
(d) No
Show Answer & Explanation

Answer: (c) Yes, 60°

Explanation:
1. The person can catch the ball if his speed equals the ball's horizontal velocity: v0 cos θ = v0/2.
2. So cos θ = 1/2 and θ = 60°.

Teacher's Note:
1. The ball's horizontal velocity stays constant.
2. The person catches it at landing if the horizontal speeds match.

Question 23: The relation between time t and distance x is t = αx² + βx, where α and β are constants. The retardation is (v is the velocity):
(a) \( 2\alpha v^3 \)
(b) \( 2\alpha\beta v^3 \)
(c) \( 2\beta^3v^3 \)
(d) \( 2\beta v^3 \)
Show Answer & Explanation

Answer: (a) \( 2\alpha v^3 \)

Explanation:
1. dt/dx = 2αx + β, so v = 1/(2αx + β).
2. a = dv/dt = v dv/dx = -v · 2α/(2αx + β)² = -2αv³.
3. The retardation is 2αv³.

Teacher's Note:
1. Differentiate t with respect to x to get 1/v.
2. Use a = v dv/dx.

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