JEE Physics Centre of Mass MCQs Set 12

Multiple Choice Questions (MCQs) for JEE Physics: Centre of Mass

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Multiple Choice Questions

Question 1: (n - 1) equal point masses, each of mass m, are placed at the vertices of a regular n-sided polygon, leaving one vertex vacant. The vacant vertex has position vector \( \vec a \) with respect to the centre of the polygon. The position vector of the centre of mass of the system is
(a) \( \dfrac{1}{n-1}\vec a \)
(b) \( -\dfrac{1}{n-1}\vec a \)
(c) \( (n-1)\vec a \)
(d) \( -(n-1)\vec a \)
Show Answer & Explanation

Answer: (b) \( -\dfrac{1}{n-1}\vec a \)

Explanation:
1. For all n masses at the vertices, the centre of mass is at the centre, so the sum of all position vectors is zero.
2. Hence the sum of the other (n - 1) vectors is \( -\vec a \), and \( \vec r_{cm} = -\vec a/(n-1) \).

Teacher's Note:
1. Use symmetry: a complete regular polygon has its centre of mass at its centre.
2. Removing one mass shifts the centre of mass opposite to the removed mass.

Question 2: A gas molecule of mass m strikes the wall of a container with speed v at angle \( \theta \) to the normal and rebounds elastically. The magnitude of the impulse on the wall is
(a) \( 3mv \)
(b) \( 2mv\cos\theta \)
(c) \( mv \)
(d) zero
Show Answer & Explanation

Answer: (b) \( 2mv\cos\theta \)

Explanation:
1. The component of velocity along the wall is unchanged; the normal component reverses.
2. Impulse \( = m v\cos\theta - (-mv\cos\theta) = 2mv\cos\theta \).

Teacher's Note:
1. Only the component of momentum along the normal changes in an elastic bounce off a smooth wall.
2. The angle is measured from the normal, so the normal component is v cos θ.

Question 3: A body at rest explodes into three pieces. Two pieces, each of mass M/4, fly off at right angles to each other with speeds 3 m s⁻¹ and 4 m s⁻¹. The third piece, of mass M/2, moves with speed
(a) 1.5 m s⁻¹
(b) 2 m s⁻¹
(c) 2.5 m s⁻¹
(d) 3.0 m s⁻¹
Show Answer & Explanation

Answer: (c) 2.5 m s⁻¹

Explanation:
1. Momenta of the first two pieces: \( 3M/4 \) and \( 4M/4 = M \), at right angles, giving resultant \( 5M/4 \).
2. The third piece must carry equal and opposite momentum: \( (M/2)v = 5M/4 \), so \( v = 2.5 \) m/s.

Teacher's Note:
1. Total momentum of an exploding body at rest stays zero.
2. Add perpendicular momenta by Pythagoras.

Question 4: A particle is projected at 200 m s⁻¹ at 60° to the horizontal. At the highest point it explodes into three particles of equal masses. One goes vertically upward at 100 m s⁻¹ and another goes vertically downward at the same speed. The velocity of the third particle is
(a) 120 m s⁻¹ at 60°
(b) 200 m s⁻¹ at 30°
(c) 50 m s⁻¹ vertically upward
(d) 300 m s⁻¹ horizontally
Show Answer & Explanation

Answer: (d) 300 m s⁻¹ horizontally

Explanation:
1. At the top the velocity is horizontal: \( 200\cos60^\circ = 100 \) m/s, so total momentum is \( 3m\times100 \) horizontally.
2. The two vertical pieces cancel each other, so \( m v_3 = 300m \) and \( v_3 = 300 \) m/s horizontally.

Teacher's Note:
1. Velocity at the highest point of a projectile is only the horizontal component.
2. Momentum is conserved in each direction separately during the explosion.

Question 5: A circular section of radius R/2 is cut out from a uniform circular disc of radius R. The centre of the hole is R/2 from the centre O of the original disc. The centre of mass of the remaining flat body is at
(a) \( R/6\text{ to the right of }O \)
(b) \( R/3\text{ to the right of }O \)
(c) \( R/3\text{ to the left of }O \)
(d) \( R/6\text{ to the left of }O \)
Show Answer & Explanation

Answer: (d) \( R/6\text{ to the left of }O \)

Explanation:
1. Mass ratio: hole is \( M/4 \), remainder is \( 3M/4 \). Taking the hole on the right at \( +R/2 \).
2. \( x = \dfrac{0\cdot M - (M/4)(R/2)}{M - M/4} = -\dfrac{R}{6} \), i.e. R/6 on the side opposite to the hole.

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

Question 6: Two blocks of masses 10 kg and 4 kg are connected by a massless spring on a frictionless horizontal surface. An impulse gives the 10 kg block a velocity of 14 m s⁻¹ towards the 4 kg block, which is at rest. The velocity of the centre of mass is
(a) 30 m s⁻¹
(b) 20 m s⁻¹
(c) 10 m s⁻¹
(d) 5 m s⁻¹
Show Answer & Explanation

Answer: (c) 10 m s⁻¹

Explanation:
1. \( v_{cm} = \dfrac{10\times14 + 4\times0}{10+4} \).
2. \( v_{cm} = 140/14 = 10 \) m/s.

Teacher's Note:
1. The internal spring force does not change the velocity of the centre of mass.
2. Velocity of CM = total momentum / total mass.

Question 7: A bullet of mass 0.01 kg travelling at 500 m s⁻¹ strikes a block of mass 2 kg suspended by a string of length 5 m. The centre of gravity of the block rises by 0.1 m. Taking g = 10 m s⁻², the speed of the bullet after it emerges from the block is nearly
(a) 580 m s⁻¹
(b) 220 m s⁻¹
(c) 1.4 m s⁻¹
(d) 7.8 m s⁻¹
Show Answer & Explanation

Answer: (b) 220 m s⁻¹

Explanation:
1. Block speed just after impact: \( v_b = \sqrt{2gh} = \sqrt{2\times10\times0.1} = \sqrt2 \) m/s.
2. Momentum: \( 0.01\times500 = 0.01v + 2\sqrt2 \), so \( v = 500 - 200\sqrt2 \approx 217 \) m/s, nearest 220 m/s.

Teacher's Note:
1. Use momentum conservation for the very short impact and energy conservation for the rise.
2. Do not use energy conservation across the impact.

Question 8: Two bodies of masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are tied to the ends of a massless string passing over a light frictionless pulley and released from rest. The acceleration of the centre of mass of the system is
(a) \( \left(\dfrac{m_1-m_2}{m_1+m_2}\right)^2 g \)
(b) \( \dfrac{m_1-m_2}{m_1+m_2}\,g \)
(c) \( g \)
(d) zero
Show Answer & Explanation

Answer: (a) \( \left(\dfrac{m_1-m_2}{m_1+m_2}\right)^2 g \)

Explanation:
1. Each block has acceleration \( a = \dfrac{m_1-m_2}{m_1+m_2}g \), \( m_1 \) downward and \( m_2 \) upward.
2. \( a_{cm} = \dfrac{m_1a - m_2a}{m_1+m_2} = \dfrac{(m_1-m_2)a}{m_1+m_2} = \left(\dfrac{m_1-m_2}{m_1+m_2}\right)^2 g \).

Teacher's Note:
1. Accelerations of the two blocks point in opposite directions, so subtract them when finding a_cm.
2. The CM accelerates because there is a net external force (tension and gravity).

Question 9: A spherical hollow is made in a lead sphere of radius R such that its surface touches the outside surface of the sphere and passes through the centre. The shift in the centre of mass of the lead sphere due to the hollowing is
(a) \( R/7 \)
(b) \( R/14 \)
(c) \( R/2 \)
(d) \( R \)
Show Answer & Explanation

Answer: (b) \( R/14 \)

Explanation:
1. Hollow has radius R/2 so its mass is \( M/8 \); its centre is at \( R/2 \) from O.
2. \( x = \dfrac{-(M/8)(R/2)}{M - M/8} = -\dfrac{R}{14} \), a shift of R/14.

Teacher's Note:
1. Mass scales as the cube of radius for spheres of the same density.
2. Hollow = negative mass at the centre of the hollow.

Question 10: Two identical blocks A and B, each of mass m, rest on a smooth floor connected by a light spring of spring constant k at its natural length. A third identical block C, moving with speed v along the line joining A and B, collides with A (collision is perfectly inelastic with A, which does not stick to B). Taking C to come to rest after striking A, the maximum compression of the spring is
(a) \( v\sqrt{\dfrac{m}{2k}} \)
(b) \( m\sqrt{\dfrac{v}{2k}} \)
(c) \( \sqrt{\dfrac{mv}{k}} \)
(d) \( \dfrac{mv}{2k} \)
Show Answer & Explanation

Answer: (a) \( v\sqrt{\dfrac{m}{2k}} \)

Explanation:
1. C stops and A moves with v, so A and B then share momentum; at maximum compression both move with \( v/2 \).
2. Energy: \( \tfrac12 mv^2 = \tfrac12(2m)(v/2)^2 + \tfrac12 kx^2 \), so \( kx^2 = \tfrac12 mv^2 \) and \( x = v\sqrt{m/2k} \).

Teacher's Note:
1. At maximum compression the two blocks have a common velocity.
2. Kinetic energy lost goes into spring potential energy.

Question 11: A bullet of mass M hits a block of mass M' in a head-on elastic collision, the block being at rest. The transfer of energy is maximum when
(a) \( M' = M \)
(b) \( M' = 2M \)
(c) \( M' \ll M \)
(d) \( M' \gg M \)
Show Answer & Explanation

Answer: (a) \( M' = M \)

Explanation:
1. For a head-on elastic collision the fraction of energy transferred is \( 4MM'/(M+M')^2 \).
2. This equals 1 only when \( M' = M \), in which case the bullet stops and gives all its energy.

Teacher's Note:
1. Equal masses exchange velocities in a one-dimensional elastic collision.
2. Very different masses transfer little energy.

Question 12: A nucleus ruptures into two nuclear parts whose velocities are in the ratio 2 : 1. The ratio of their nuclear sizes (radii, nuclear density being constant) is
(a) \( 2^{1/3}:1 \)
(b) \( 1:2^{1/3} \)
(c) \( 3^{1/2}:1 \)
(d) \( 1:3^{1/2} \)
Show Answer & Explanation

Answer: (b) \( 1:2^{1/3} \)

Explanation:
1. Momentum conservation: \( m_1v_1 = m_2v_2 \), so \( m_1:m_2 = 1:2 \).
2. Mass \( \propto r^3 \), so \( r_1:r_2 = 1:2^{1/3} \).

Teacher's Note:
1. The part with the greater speed has the smaller mass.
2. Nuclear density is the same for all nuclei, so mass is proportional to volume.

Question 13: Four identical spheres, each of mass M and radius 10 cm, rest on a horizontal surface touching one another so that their centres are at the corners of a square of side 20 cm. The distance of their centre of mass from the centre of any one sphere is
(a) 5 cm
(b) 10 cm
(c) 20 cm
(d) \( 10\sqrt2\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 10\sqrt2\text{ cm} \)

Explanation:
1. The centre of mass of the four equal masses is at the centre of the square.
2. Distance from a corner to the centre of the square is half the diagonal: \( 20\sqrt2/2 = 10\sqrt2 \) cm.

Teacher's Note:
1. For equal masses at the corners of a square the CM is at the intersection of the diagonals.
2. The radius of the spheres is only a distractor.

Question 14: A bullet of mass 50 g is fired from a gun of mass 2 kg. If the total kinetic energy produced is 2050 J, the kinetic energies of the bullet and the gun respectively are
(a) 2000 J and 50 J
(b) 1025 J and 1025 J
(c) 50 J and 2000 J
(d) 1500 J and 550 J
Show Answer & Explanation

Answer: (a) 2000 J and 50 J

Explanation:
1. Equal and opposite momenta: \( K \propto 1/m \), so \( K_b/K_g = m_g/m_b = 2/0.05 = 40 \).
2. \( K_b = 2050\times40/41 = 2000 \) J and \( K_g = 50 \) J.

Teacher's Note:
1. For equal momenta the lighter body carries more kinetic energy.
2. K = p²/2m.

Question 15: A small block of mass M moves at 5 m s⁻¹ towards another block of the same mass placed 2 m ahead on a rough horizontal surface. The coefficient of friction is 0.25 and g = 10 m s⁻². The collision is elastic. The separation between the blocks when both have come to rest is
(a) 3 m
(b) 4 m
(c) 2 m
(d) 1.5 m
Show Answer & Explanation

Answer: (a) 3 m

Explanation:
1. Before impact: \( v^2 = 25 - 2(2.5)(2) = 15 \).
2. Equal masses exchange velocities, so the first block stops and the second moves off at \( \sqrt{15} \) and travels \( 15/(2\times2.5) = 3 \) m before stopping.

Teacher's Note:
1. Equal masses exchange velocities in an elastic head-on collision.
2. Deceleration due to friction is μg.

Question 16: A thick uniform bar of mass 0.16 kg lies on a frictionless horizontal surface. Two particles of mass 0.08 kg each move on the same surface towards the bar, perpendicular to it, with speeds 10 m s⁻¹ and 6 m s⁻¹, and both strike the bar and stick to it simultaneously. The velocity of the centre of mass after collision is
(a) 2 m s⁻¹
(b) 4 m s⁻¹
(c) 10 m s⁻¹
(d) 167 m s⁻¹
Show Answer & Explanation

Answer: (b) 4 m s⁻¹

Explanation:
1. Total momentum \( = 0.08\times10 + 0.08\times6 = 1.28 \) kg m/s.
2. Total mass \( = 0.16 + 0.08 + 0.08 = 0.32 \) kg, so \( v_{cm} = 4 \) m/s.

Teacher's Note:
1. The velocity of the centre of mass is unaffected by internal impulses.
2. The particles move in the same direction, so momenta add.

Question 17: A ball moving with a certain velocity hits another identical ball at rest. If the plane is frictionless and the collision is elastic (oblique), the angle between the directions in which the balls move after collision is
(a) \( 30^\circ \)
(b) \( 60^\circ \)
(c) \( 90^\circ \)
(d) \( 120^\circ \)
Show Answer & Explanation

Answer: (c) \( 90^\circ \)

Explanation:
1. Momentum: \( \vec u = \vec v_1 + \vec v_2 \). Energy: \( u^2 = v_1^2 + v_2^2 \).
2. Squaring the first equation and comparing gives \( \vec v_1\cdot\vec v_2 = 0 \), so the angle is \( 90^\circ \).

Teacher's Note:
1. This is a standard result for equal masses, one initially at rest.
2. It does not hold for inelastic collisions.

Question 18: A body at rest breaks up into 3 parts. Two parts of equal mass fly off perpendicular to each other, each with a velocity of 12 m s⁻¹. The velocity of the third part, which has 3 times the mass of each of the other two, is
(a) \( 4\sqrt2\text{ m/s at }45^\circ\text{ from each body} \)
(b) \( 24\sqrt2\text{ m/s at }135^\circ\text{ from each body} \)
(c) \( 6\sqrt2\text{ m/s at }135^\circ\text{ from each body} \)
(d) \( 4\sqrt2\text{ m/s at }135^\circ\text{ from each body} \)
Show Answer & Explanation

Answer: (d) \( 4\sqrt2\text{ m/s at }135^\circ\text{ from each body} \)

Explanation:
1. Momentum of the two parts: \( 12m \) each, perpendicular, resultant \( 12\sqrt2\,m \) at \( 45^\circ \) to each.
2. Third part: \( 3m\,v = 12\sqrt2\,m \), so \( v = 4\sqrt2 \) m/s, opposite to the resultant, i.e. at \( 135^\circ \) from each body.

Teacher's Note:
1. Total momentum remains zero.
2. Opposite to a direction at 45° from each is 135° from each.

Question 19: Three identical blocks A, B and C lie on a horizontal frictionless surface. Blocks B and C are at rest and A moves towards B at 10 m s⁻¹. The coefficient of restitution for all collisions is 0.5. The speed of block C just after its collision with B is nearly
(a) 5.6 m s⁻¹
(b) 6 m s⁻¹
(c) 8 m s⁻¹
(d) 10 m s⁻¹
Show Answer & Explanation

Answer: (a) 5.6 m s⁻¹

Explanation:
1. For equal masses and a target at rest, the target speed is \( \tfrac{1+e}{2}\,u = 0.75u \).
2. B gets \( 7.5 \) m/s, then C gets \( 0.75\times7.5 = 5.625 \approx 5.6 \) m/s.

Teacher's Note:
1. For equal masses, v₂ = (1 + e)u/2 and v₁ = (1 − e)u/2.
2. Apply the same formula in succession.

Question 20: A smooth steel ball strikes a fixed smooth steel plate at an angle \( \theta \) with the vertical (normal). If the coefficient of restitution is e, the angle with the vertical at which the rebound takes place is
(a) \( \theta \)
(b) \( \tan^{-1}\!\left(\dfrac{\tan\theta}{e}\right) \)
(c) \( e\tan\theta \)
(d) \( \tan^{-1}\!\left(\dfrac{e}{\tan\theta}\right) \)
Show Answer & Explanation

Answer: (b) \( \tan^{-1}\!\left(\dfrac{\tan\theta}{e}\right) \)

Explanation:
1. Tangential component \( u\sin\theta \) is unchanged; normal component becomes \( e\,u\cos\theta \).
2. \( \tan\theta' = \dfrac{u\sin\theta}{eu\cos\theta} = \dfrac{\tan\theta}{e} \).

Teacher's Note:
1. The plate is smooth, so there is no impulse along its surface.
2. For e = 1 the angle of rebound equals the angle of incidence.

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