JEE Physics Centre of Mass MCQs Set 10

Practice MCQs for JEE Physics Centre of Mass

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Multiple Choice Questions

Question 1: A disc of radius r is cut from a larger disc of radius 4r so that the edge of the hole touches the edge of the larger disc. The centre of mass of the residual disc is at a distance from the centre of the larger disc equal to
(a) \( \frac r4 \)
(b) \( \frac r5 \)
(c) \( \frac r2 \)
(d) \( \frac r3 \)
Show Answer & Explanation

Answer: (b) \( \frac r5 \)

Explanation:
1. The hole's centre is at 3r from the centre. Masses are proportional to area: removed \( \pi r^2 \), residual \( 15\pi r^2 \).
2. \( x = \frac{0\cdot15\pi r^2 - 3r\cdot\pi r^2}{15\pi r^2} = -\frac{r}{5} \), so the distance is r/5.

Teacher's Note:
1. Treat the hole as negative mass.
2. Mass is proportional to area for a uniform disc.

Question 2: Two particles of masses m₁ and m₂ are projected from the top of a tower with the same speed u; m₁ vertically downward and m₂ horizontally. Neglecting air resistance, the acceleration of the centre of mass of the system is
(a) \( \frac{m_1g}{m_1+m_2} \) downward
(b) \( \frac{m_2g}{m_1+m_2} \) downward
(c) cannot be predicted
(d) g downward
Show Answer & Explanation

Answer: (d) g downward

Explanation:
1. \( a_{cm} = \frac{m_1g+m_2g}{m_1+m_2} = g \) downward, since both particles have acceleration g.

Teacher's Note:
1. Only external force is gravity.
2. Both accelerations equal g.

Question 3: A cart of mass M is tied to one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M/2. The system is on a smooth horizontal surface. If the man pulls the cart by the rope, the distance moved by the man before the man and the cart meet is
(a) \( \frac{20}{3} \) m
(b) \( \frac{10}{3} \) m
(c) 0 m
(d) 10 m
Show Answer & Explanation

Answer: (a) \( \frac{20}{3} \) m

Explanation:
1. No external horizontal force, so the centre of mass stays fixed: (M/2)x_man = M x_cart.
2. x_man + x_cart = 10 m gives x_cart = 10/3 m and x_man = 20/3 m.

Teacher's Note:
1. They meet at the centre of mass.
2. The lighter body moves farther.

Question 4: Two point masses m₁ and m₂ at rest in gravity-free space are released from a distance d. The velocity of the centre of mass of the system at the time of collision of the particles is
(a) 2 m/s
(b) 3 m/s
(c) zero
(d) depends on m₁ and m₂
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. Only internal gravitational forces act, so total momentum remains zero.
2. Hence the velocity of the centre of mass is zero at all times.

Teacher's Note:
1. No external force.
2. Initial momentum is zero.

Question 5: Two blocks of masses 5 kg and 2 kg are connected by a spring of negligible mass on a frictionless horizontal surface. An impulse gives a velocity of 7 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
(a) 4 m/s
(b) 2 m/s
(c) 5 m/s
(d) 3 m/s
Show Answer & Explanation

Answer: (c) 5 m/s

Explanation:
1. \( v_{cm} = \frac{5\times7+2\times0}{5+2} = 5 \) m/s.

Teacher's Note:
1. Use total momentum over total mass.
2. Spring forces are internal.

Question 6: A shell at rest on a smooth horizontal surface explodes into two fragments of masses m₁ and m₂. If m₁ moves with speed u just after the explosion, the work done by internal forces during the explosion is
(a) \( \tfrac12m_1u^2\left(1+\frac{m_1}{m_2}\right) \)
(b) \( \tfrac12(m_1+m_2)\frac{m_2}{m_1}u^2 \)
(c) \( \tfrac12(m_1+m_2)u^2 \)
(d) \( \tfrac12(m_2-m_1)u^2 \)
Show Answer & Explanation

Answer: (a) \( \tfrac12m_1u^2\left(1+\frac{m_1}{m_2}\right) \)

Explanation:
1. Momentum conservation: \( m_2v = m_1u \Rightarrow v = \frac{m_1u}{m_2} \).
2. Work = gain in KE = \( \tfrac12m_1u^2+\tfrac12m_2\frac{m_1^2u^2}{m_2^2} = \tfrac12m_1u^2\left(1+\frac{m_1}{m_2}\right) \).

Teacher's Note:
1. Work done equals total kinetic energy gained.
2. Find the second speed first.

Question 7: A body of mass 2 kg moving with a velocity of 3 m/s collides head on with a body of mass 1 kg moving at 4 m/s in the opposite direction. After the collision the two bodies stick together. The common velocity is
(a) \( \frac14 \) m/s
(b) \( \frac13 \) m/s
(c) \( \frac34 \) m/s
(d) \( \frac23 \) m/s
Show Answer & Explanation

Answer: (d) \( \frac23 \) m/s

Explanation:
1. \( 2(3)+1(-4) = 3v \Rightarrow v = \frac23 \) m/s.

Teacher's Note:
1. Take opposite directions with opposite signs.
2. Perfectly inelastic collision conserves momentum only.

Question 8: A ball is dropped on the ground from a height of 1 m. The coefficient of restitution is 0.6. The height to which the ball rebounds is
(a) 0.6 m
(b) 0.36 m
(c) 0.4 m
(d) 0.16 m
Show Answer & Explanation

Answer: (b) 0.36 m

Explanation:
1. Rebound speed \( = e\times \) impact speed, so \( h' = e^2h = 0.36\times1 = 0.36 \) m.

Teacher's Note:
1. Height scales as speed squared.
2. e² multiplies the height.

Question 9: A massive ball moving with speed v collides elastically with a tiny ball initially at rest. Immediately after the impact the second ball moves with a speed approximately equal to
(a) v
(b) 2v
(c) v/2
(d) infinity
Show Answer & Explanation

Answer: (b) 2v

Explanation:
1. For an elastic collision \( v_2 = \frac{2m_1}{m_1+m_2}v \to 2v \) when \( m_2\ll m_1 \).

Teacher's Note:
1. The heavy ball's speed is almost unchanged.
2. Relative speed of separation equals approach speed.

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Practice MCQs for JEE Physics Centre of Mass

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FAQs

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