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.

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 \(r_1 = \frac{m_2 d}{m_1+m_2} = \frac{6\sqrt{2}}{8} = \frac{3\sqrt{2}}{4}\text{ m}\).

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 = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).

By definition, the coefficient of restitution $e$ is equal to $1$ for a perfectly elastic collision where kinetic energy is fully conserved.

By conservation of momentum, total initial momentum vector is $\vec{P} = mv\hat{i} + (3m)(2v)\hat{j}$. Dividing by the total mass $4m$ gives the final velocity vector $\vec{v}_f = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$.

Using the collision velocity formulas $v_1 = \frac{(m_1 - em_2)u_1}{m_1+m_2}$ and $v_2 = \frac{(1+e)m_1 u_1}{m_1+m_2}$ with $m_1=m$, $m_2=2m$, $u_1=2$, and $e=0.5$, we get $v_1 = 0$ and $v_2 = 1\text{ m/s}$.