NEET Center of Mass , Momentum and Collision MCQs & PYQs

Centre of mass is the point which divides the line joining the masses in reverse ratio of masses. 

Here ratio of mass is 2:5 so distance from 2kg mass located at origin will be 5/7 of 7 cm so 

\[ x_{cm}= \frac{2(0)+5(7)}{5+2}= 5 \]

\[ x_{cm}= \frac{(m_{1}x_{1} + m_{2}x_{2})}{m_{1}+m_{2}} \]

\[ x_{cm}= \frac{12(0)+ 16 (1.2 × 10^{–10})}{12+16}= 0.69 \times 10^{–10} m \]

Since there is no external force acting on the system, the acceleration of the centre of mass is zero. Since the system was initially at rest, the velocity of the centre of mass remains zero.

In a one-dimensional elastic collision between a moving body of mass \(m\) and a stationary body of mass \(M\), maximum transfer of kinetic energy occurs when the masses are equal (\(m = M\)).

By conservation of momentum, \(m_1 v_1 = m_2 v_2\), which gives \(4 \times v_1 = 8 \times 6\), so \(v_1 = 12\text{ ms}^{-1}\). The kinetic energy of the \(4\text{ kg}\) body is \(K = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 4 \times 12^2 = 288\text{ J}\).

The ratio of masses is \(m' : 3m' : 3m'\). The combined momentum of the two perpendicular \(3m'\) masses is \(P = \sqrt{(3m' \times 15)^2 + (3m' \times 15)^2} = 45\sqrt{2} m'\). Conservation of momentum requires the third fragment \(m'\) to balance this: \(m' v_3 = 45\sqrt{2} m' ⇒ v_3 = 45\sqrt{2}\text{ ms}^{-1}\).

Let the total mass be \(2m\), so \(K = mv_0^2\). Under momentum conservation, \(2mv_0 = m(-v_0) + mv_B ⇒ v_B = 3v_0\). The kinetic energy of part B is \(K_B = \frac{1}{2}m(3v_0)^2 = \frac{9}{2}mv_0^2 = \frac{9}{2}K\).

By conservation of momentum, the two parts move in opposite directions with equal magnitude of momentum, \( \vec{p}_1 = -\vec{p}_2 \). Thus, they have different momenta (since momentum is a vector), so the Reason is false. However, since they have the same mass and same momentum magnitude, their kinetic energy \( K = \frac{p^2}{2m} \) is the same, so the Assertion is true.

If the kinetic energy of a system is zero, the speed of each particle must be zero, meaning the total linear momentum is also zero. If the total linear momentum is zero, particles can still be moving in opposite directions, resulting in a non-zero kinetic energy. Thus, the Assertion is true but the Reason is false.

The coefficient of restitution \( e \) for a vertical rebound is given by \( e = \sqrt{\frac{h_{\text{rebound}}}{h_{\text{initial}}}} \). Substituting the given values, \( e = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3} \).