Center of Mass , Momentum and Collision - NEET Physics Chapterwise MCQs & PYQs

NEET Center of Mass , Momentum and Collision MCQs & PYQs

Practice NEET Center of Mass , Momentum and Collision Questions

Question 21:

moderate

A boat of length \( 12\text{ m} \) and mass \( 840\text{ kg} \) is floating without motion in still water. A man of mass \( 60\text{ kg} \) standing at one end of it walks to the other end of it and stops. The magnitude of displacement of the boat relative to the ground is:

Since no external horizontal force acts on the boat-man system, the center of mass does not move. The displacement of the boat is \( x = \frac{m L}{m + M} = \frac{60 \times 12}{60 + 840} = 0.8\text{ m} = 80\text{ cm} \).

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Question 22:

moderate

Two bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:

The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).

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Question 23:

moderate

A ball of mass \( m \) approaches a wall of mass \( M \) (\( M \gg m \)) with speed \( 4\text{ m/s} \) along the normal to the wall. The speed of the wall is \( 1\text{ m/s} \) towards the ball. The speed of the ball after an elastic collision with the wall is:

Using the coefficient of restitution \( e = 1 \), the relative velocity of separation equals the relative velocity of approach. The approach velocity is \( 4 - (-1) = 5\text{ m/s} \). After collision, the relative separation velocity is \( v' - 1 = 5 \implies v' = 6\text{ m/s} \) away from the wall.

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Question 24:

easy

A bullet of mass \( m \) leaves the barrel of a gun of mass \( M \) with a velocity \( v \). The gun is known to recoil with a velocity \( V \). If \( k \) and \( K \) respectively denote the kinetic energies of the bullet and the gun respectively; then

By conservation of momentum, the bullet and gun have equal momentum magnitude, \( p \). Since kinetic energy is \( K_{\text{E}} = \frac{p^2}{2\text{mass}} \), we have \( k = \frac{p^2}{2m} \) and \( K = \frac{p^2}{2M} \). Thus, \( K = \left(\frac{m}{M}\right) k \).

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Question 25:

moderate

A block of mass \( m \) moving with a velocity \( v \) collides with another block of mass \( M \) at rest. The two blocks stick together due to the collision. The loss of K.E. expressed as a fraction of total initial kinetic energy is:

By conservation of momentum, the final velocity after a completely inelastic collision is \( v_f = \frac{mv}{m+M} \). The fractional loss of kinetic energy is \( \frac{K_i - K_f}{K_i} = 1 - \frac{\frac{1}{2}(m+M)v_f^2}{\frac{1}{2}mv^2} = \frac{M}{m+M} \).

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Question 26:

easy

Given below are two statements:

Statement I: Centre of mass of any object always coincide with centre of gravity.

Statement II: Centre of gravity is the point where total gravitational torque on the body is zero.

In the light of the above statements, choose the most appropriate answer from the options given below.

Statement I is false because the center of mass and center of gravity only coincide in a uniform gravitational field. Statement II is true because the center of gravity is defined as the point about which the net gravitational torque is zero.

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Question 27:

easy

Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration \( a \). Their centre of mass has an acceleration

The acceleration of the centre of mass is given by \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \). Since \( m_1 = m_2 = m \), \( a_1 = 0 \), and \( a_2 = a \), we get \( a_{\text{cm}} = \frac{m(0) + m(a)}{2m} = \frac{a}{2} \).

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Question 28:

easy

A body of mass 2 kg moving with velocity \(5 \text{ m s}^{-1}\) collides with a body at rest of mass 3 kg and sticks to it. Now the combined mass starts moving. The final velocity of whole mass is

Using the law of conservation of linear momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f\). Substituting the values: \(2(5) + 3(0) = (2 + 3) v_f ⇒ 10 = 5 v_f ⇒ v_f = 2 \text{ m s}^{-1}\).

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Question 29:

easy

Read the statements marked as assertion (A) and reason (R) and choose the correct option.

Assertion (A): If no external force acts on a system, the velocity of the centre of mass remains constant.

Reason (R): If there is no external force on system, then momentum of system is conserved.

 

If the net external force on a system is zero, its linear momentum is conserved (\(vec{P} = M\vec{v}_{\text{cm}} = \text{constant}\)). Consequently, the velocity of the centre of mass \(\vec{v}_{\text{cm}}\) remains constant.

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Question 30:

easy

Assertion (A): A body with negative energy cannot have linear momentum.

Reason (R): Magnitude of linear momentum can be negative.

Total mechanical energy \(E = K + U\) can be negative if potential energy \(U\) is negative and larger in magnitude than kinetic energy \(K\).

However, kinetic energy \(K = \frac{1}{2}mv^2\) is always non-negative, implying momentum exists. The magnitude of linear momentum \(|\vec{p}| = mv\) is always non-negative.

Therefore, both Assertion and Reason are false.

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