NEET Center of Mass , Momentum and Collision MCQs & PYQs

Assertion (A): Maximum kinetic energy loss occurs in a perfectly inelastic collision where particles stick together. So, (A) is true.

Reason (R): For a perfectly inelastic collision between mass (m\) (velocity (v\)) and stationary mass (M\), the energy loss is ( \Delta K = \frac{M}{(m+M)}\left(\frac{1}{2}mv^2\right)\). The given formula in (R) is incorrect.

So, (R) is false. Therefore, (A) is true and (R) is false. Option (3) is correct.

Reason (R): For the bullet-gun system, the forces causing the bullet to fire are internal. Thus, linear momentum of the system is conserved. So, (R) is true.

Assertion (A): Let (m\) and (M\) be masses of bullet and gun, (v\) and (V\) their velocities. By momentum conservation, (mv = MV\). The ratio of kinetic energies is \( \frac{K_g}{K_b} = \frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2} = \frac{M(mv/M)^2}{mv^2} = \frac{m}{M}\). So, (A) is true.

(R) correctly explains (A) as the kinetic energy ratio is derived directly from momentum conservation. Option (1) is correct.

For a two-particle system, the center of mass \( R_{CM} \) is defined such that the sum of mass moments about it is zero: \( m_1r_1 = m_2r_2 \). If \( m_1 > m_2 \), then \( r_1 < r_2 \), meaning the COM is closer to the heavier particle.

Thus both A and R are true, and R explains A.

The position of the centre of mass of a system of particles is defined as \( \vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i} \). It clearly depends on individual masses, their coordinates (positions), and consequently the relative distances between them.

Total linear momentum is conserved at each instant of collision because no external forces act. Kinetic energy, however, is not conserved during the period of deformation, and is conserved after only in perfectly elastic collisions. Thus, Statement 1 is true and Statement 2 is false.

Since no external force acts on the two-particle system, the acceleration of the centre of mass is zero. Since the system started from rest, the speed of the centre of mass remains zero.

Distance between masses is \(d = \sqrt{(2-1)^2+(2-1)^2+0^2} = \sqrt{2}\text{ m}\)
Distance of center of mass from \(m_1\) is \(r_1 = \frac{m_2 d}{m_1+m_2} = \frac{6\sqrt{2}}{8} = \frac{3\sqrt{2}}{4}\text{ m}\).

By conservation of momentum, \( P_{text{total}} = 0 \). Momentum of first part \( P_1 = 1 \text{ kg} \times 12 \text{ m/s} = 12 \text{ Ns} \). Momentum of second part \( P_2 = 2 \text{ kg} \times 8 \text{ m/s} = 16 \text{ Ns} \). As \( P_1 \) and \( P_2 \) are perpendicular, their resultant \( P_{12} = sqrt{12^2 + 16^2} = 20 \text{ Ns} \). For conservation, \( P_3 \) must be \( 20 \text{ Ns} \). \( m_3 = P_3 / v_3 = 20 \text{ Ns} / 4 \text{ m/s} = 5 \text{ kg} \).

Velocity of \( 1 \text{ kg} \) mass after \( 5 \text{ s} \): \( v = u - gt = 100 - 10 \times 5 = 50 \text{ m/s} \) (upward). Initial momentum before explosion \( P_i = 1 \text{ kg} \times 50 \text{ m/s} = 50 \text{ Ns} \) (upward). Mass of first part \( m_1 = 0.4 \text{ kg} \), \( v_1 = -25 \text{ m/s} \). Mass of second part \( m_2 = 0.6 \text{ kg} \). By conservation of momentum: \( P_i = m_1 v_1 + m_2 v_2 \). \( 50 = 0.4 \times (-25) + 0.6 v_2 \). \( 50 = -10 + 0.6 v_2 \implies v_2 = 100 \text{ m/s} \) (upward).

Initial momentum is zero. Two pieces of mass \(m\) move with velocity \(v\) perpendicular to each other. Their momenta are \(m\vec{v}_1 = mv\hat{i}\, m\vec{v}_2 = mv\hat{j}\). The third piece has mass \(m_3 = 4m - m - m = 2m\). By momentum conservation, \(m_3\vec{v}_3 = -(mv\hat{i} + mv\hat{j})\), so \(|\vec{v}_3| = \frac{\sqrt{(mv)^2 + (mv)^2}}{2m} = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}\). Total KE = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).