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Rotational Motion MCQs for NEET — Physics Questions with Answers

Practice free Rotational Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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A shell of mass m moving with velocity v suddenly breaks into 2 pieces. The part having mass m/4 remains stationary. The velocity of the other shell will be

$\frac{3}{4} v$

$\frac{4}{3} v$

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Two equal masses m₁ and m₂ moving along the same straight line with velocities +3 m/s and –5 m/s respectively collide elastically. Their velocities after the collision will be respectively

+4 m/s for both

–3 m/s and +5 m/s

–4 m/s and +4 m/s

–5 m/s and +3 m/s

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Explanation

In an elastic collision between two bodies, the total momentum before and after the collision remains the same, and the total kinetic energy is also conserved. Using these principles and the given initial velocities, we can calculate the final velocities to be -5 m/s and +3 m/s.

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A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m. Then the body rebounds to a height of

0.6 m

0.4 m

1 m

0.36 m

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Explanation

$h_{n} = h e^{2 n} = 1 \times e^{2 \times 1}$ = $1 \times {(0 .6)}^{2} = 0 .36 m$

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A body at rest breaks up into 3 parts. If 2 parts having equal masses fly off perpendicularly each after with a velocity of 12m/s, then the velocity of the third part which has 3 times mass of each part is

$4 \sqrt{2} m / s$ at an angle of 45° from each body

$24 \sqrt{2} m / s$ at an angle of 135° from each body

$6 \sqrt{2} m / s$ at 135° from each body

$4 \sqrt{2} m / s$ at 135° from each body

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Explanation

Using the conservation of momentum principle, we can equate the initial momentum (which is zero) to the final momentum, which is the vector sum of the momenta of the three parts. This gives the velocity of the third part as 4âˆš2 m/s, making an angle of 135Â° with each of the other two parts.

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A particle falls from a height h upon a fixed horizontal plane and rebounds. If e is the coefficient of restitution, the total distance travelled before rebounding has stopped is

$h (\frac{1 + e^{2}}{1 - e^{2}})$

$h (\frac{1 - e^{2}}{1 + e^{2}})$

$\frac{h}{2} (\frac{1 - e^{2}}{1 + e^{2}})$

$\frac{h}{2} (\frac{1 + e^{2}}{1 - e^{2}})$

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A body of mass 5 kg moving with a velocity 10m/s collides with another body of the mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision is

2.5 m/s

5 m/s

7.5 m/s

10 m/s

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Explanation

Momentum conservation

$5 \times 10 + 20 \times 0 = 5 \times 0 + 20 \times v$

⇒ v = 2.5 m/s

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A ball of mass m moving with velocity V, makes a head on elastic collision with a ball of the same mass moving with velocity 2V towards it. Taking direction of V as positive, velocities of the two balls after collision are-

–V and 2V

2 V and –V

V and –2V

–2V and V

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Explanation

Due to elastic collision of bodies having equal mass, their velocities get interchanged.

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A spacecraft of mass 'M' and moving with velocity 'v' suddenly breaks in two pieces. After the explosion one of the mass 'm' becomes stationary. What is the velocity of the other part of craft

$\frac{M v}{M - m}$

$\frac{M v}{m}$

$\frac{M - m}{m} v$

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Explanation

By the principle of conservation of linear momentum,

$M v = m v_{1} + m v_{2}$ ⇒ $M v = 0 + (M - m) v_{2}$ ⇒ $v_{2} = \frac{M v}{M - m}$

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A tennis ball is released from height h above ground level. If the ball makes inelastic collision with the ground, to what height will it rise after third collision

he⁶

e²h

e³h

None of these

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Explanation

$h_{n} = h e^{2 n}$ after third collision $h_{3} = h e^{6}$ [as n = 3]

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A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two spheres is

90°

0°

45°

Different from 90°

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Explanation

Angle will be 90° if collision is perfectly elastic.

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