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Vertical circular motion and elastic/inelastic collisions in one and two dimensions: NEET MCQs with answers

73 questions on this topic of Work, Energy and Power (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A small body of mass m is tied to a light inextensible string of length L and whirled in a vertical circle. What is the minimum speed that must be imparted to the body at the lowest point so that it completes a full vertical circular loop?

  1. √(3gL)
  2. √(4gL)
  3. √(6gL)
  4. √(5gL)
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Answer: (D) √(5gL)

For the body to complete the vertical circle, the tension at the top must be greater than or equal to zero, which gives a minimum speed at the top of v_top = √(gL). By conservation of mechanical energy between the bottom and the top, (1/2)m(v_bottom)² = (1/2)m(v_top)² + mg(2L), yielding v_bottom = √(gL + 4gL) = √(5gL).

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2. A mass m moving at speed u collides head-on elastically with a stationary target of mass M. If the incoming mass rebounds with half of its initial speed, the ratio M/m is equal to:

  1. 2
  2. 4
  3. 3
  4. 5
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Answer: (C) 3

In a 1D elastic collision with target at rest, the velocity of the incident particle after collision is v₁ = [(m - M) / (m + M)] u. Since it rebounds with half its initial speed, v₁ = -u/2. Thus, (m - M) / (m + M) = -1/2 ⇒ 2(m - M) = -(m + M) ⇒ 2m - 2M = -m - M ⇒ 3m = M. Therefore, M/m = 3.

hard

3. A bob of mass m hangs from a fixed ceiling by a light string of length L. It is projected horizontally with speed u = √(4gL). When the string becomes horizontal, what is the magnitude of the total acceleration of the bob?

  1. 2g
  2. g√10
  3. 3g
  4. g√5
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Answer: (D) g√5

When the string is horizontal, the bob has climbed a vertical height of L. By conservation of energy, (1/2)m u² = (1/2)m v² + mgL, which gives v² = u² - 2gL = 4gL - 2gL = 2gL. At this position, the radial (centripetal) acceleration is a_r = v²/L = 2g pointing horizontally towards the center, while the tangential acceleration is purely due to gravity, a_t = g pointing vertically downward. Since these two accelerations are mutually perpendicular, the magnitude of the net acceleration is a_net = √[(a_r)² + (a_t)²] = √[(2g)² + g²] = g√5.

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4. A small block slides down from rest along the inside of a frictionless hemispherical bowl of radius R from the rim. The magnitude of the normal reaction exerted by the bowl on the block at the lowest point is:

  1. mg
  2. 2mg
  3. 3mg
  4. 4mg
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Answer: (C) 3mg

By conservation of mechanical energy, the speed at the bottom is v = √(2gR). At the lowest point, the equation of motion is N - mg = m v² / R. Substituting v² = 2gR gives N - mg = m(2gR)/R = 2mg, which yields N = 3mg.

easy

5. Given below are two statements: Statement I: In a two-dimensional elastic collision of two isolated bodies, both the linear momentum along any direction and the total kinetic energy of the system are conserved. Statement II: In any collision, the total mechanical energy of an isolated system always remains constant. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because in an elastic collision, total kinetic energy and linear momentum (in all directions) are conserved. Statement II is incorrect because in inelastic collisions, mechanical energy is converted into non-mechanical forms of energy such as heat, sound, or deformation energy, so total mechanical energy is not conserved.

hard

6. A pendulum bob of mass m is suspended by a string of length L. The bob is given a horizontal speed u at its lowest point such that the string becomes slack when it makes an angle of 60° with the upward vertical. The speed u is given by:

  1. √(1.5 gL)
  2. √(2.5 gL)
  3. √(4.5 gL)
  4. √(3.5 gL)
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Answer: (D) √(3.5 gL)

At an angle θ = 60° with the upward vertical, the vertical height gained from the bottom is h = L + L cos 60° = 1.5 L. The string becomes slack when tension T = 0, which implies m v² / L = m g cos 60° = 0.5 m g, so v² = 0.5 gL. From conservation of energy: (1/2) m u² = (1/2) m v² + m g h = (1/2) m (0.5 gL) + m g (1.5 L) = 1.75 m gL. Thus u² = 3.5 gL, giving u = √(3.5 gL).

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7. A body of mass M moving with velocity v collides elastically and head-on with another stationary body of mass m. What fraction of its initial kinetic energy is transferred to the stationary body?

  1. 2Mm / (M + m)²
  2. 4Mm / (M + m)²
  3. (M - m)² / (M + m)²
  4. 4M² / (M + m)²
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Answer: (B) 4Mm / (M + m)²

The velocity acquired by the target body of mass m is v₂ = 2Mv / (M + m). The kinetic energy transferred is K₂ = (1/2)m v₂² = 2m M² v² / (M + m)². Dividing by initial kinetic energy K₁ = (1/2)M v² gives the fraction ΔK/K₁ = 4Mm / (M + m)².

easy

8. A particle of mass m moving with speed u undergoes a head-on elastic collision with a stationary particle of mass 3m. What fraction of its initial kinetic energy is retained by the incident particle after the collision?

  1. 1/2
  2. 1/4
  3. 3/4
  4. 1/16
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Answer: (B) 1/4

For a 1D elastic collision with a target initially at rest, the final velocity of the incident mass is v₁ = [(m₁ - m₂)/(m₁ + m₂)]u. Here, v₁ = [(m - 3m)/(m + 3m)]u = -u/2. The kinetic energy retained is (1/2)m(v₁)² = (1/2)m(-u/2)² = (1/4)[(1/2)mu²], which is 1/4th of the initial kinetic energy.

medium

9. A small stone of mass m tied to an inextensible string of length L is rotated in a vertical circle. If the ratio of the maximum tension to the minimum tension in the string during complete circular motion is 4, then the speed of the stone at the lowest point is:

  1. √(3gL)
  2. 3√(gL)
  3. √(7gL)
  4. √(5gL)
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Answer: (C) √(7gL)

Let u be the speed at the lowest point. T_max = m u²/L + mg and T_min = m u²/L - 5mg. Given T_max / T_min = 4. Thus (u²/L + g) = 4(u²/L - 5g), which gives 3 u²/L = 21g, so u² = 7gL and u = √(7gL).

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10. A 3 kg body moving at 4 m/s collides with a stationary 5 kg body and they stick together. Their common velocity after collision is:

  1. 4 m/s
  2. 2 m/s
  3. 1.5 m/s
  4. 2.5 m/s
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Answer: (C) 1.5 m/s

Momentum is conserved in a perfectly inelastic collision: v = m₁v₁/(m₁ + m₂) = 3 × 4/8 = 1.5 m/s.

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11. A block of mass m moving with speed u collides with an identical stationary block. The collision is partially elastic with coefficient of restitution e. The fraction of initial kinetic energy lost during the collision is:

  1. (1 - e²)/2
  2. (1 - e)/2
  3. (1 + e²)/2
  4. (1 - e²)/4
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Answer: (A) (1 - e²)/2

For head-on collision of two equal masses m with one initially stationary, the initial kinetic energy is K_i = (1/2)m u². The loss in kinetic energy is given by ΔK = (1/4) m (1 - e²) u². Therefore, the fractional loss in kinetic energy is ΔK / K_i = [(1/4) m (1 - e²) u²] / [(1/2) m u²] = (1 - e²)/2.

medium

12. A 3 kg body moving at 4 m/s collides with a stationary 6 kg body and they stick together. Their common velocity after collision is:

  1. 2.67 m/s
  2. 1.33 m/s
  3. 4 m/s
  4. 2 m/s
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Answer: (B) 1.33 m/s

Momentum is conserved in a perfectly inelastic collision: v = m₁v₁/(m₁ + m₂) = 3 × 4/9 = 1.33 m/s.

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