Center of Mass and Collision — NEET UG practice

1. Notice that mass $A$ and bob $B$ have equal mass $m$, and they collide elastically in one dimension. Recall the key result for an elastic collision between two equal masses: the moving mass comes to rest, and its entire velocity is transferred to the stationary mass. Hence, after the collision, bob $B$ moves off with the full initial speed of $A$. $v_B=10\ \mathrm{ms}^{-1}$

2. Since the bob is attached to a string and swings upward like a pendulum, use conservation of energy to relate its speed just after collision to the height $h$ it rises. Given that all kinetic energy converts into gravitational potential energy at the highest point, $mgh=\frac{1}{2}mv_B^2$

3. Cancel the mass $m$ from both sides and solve for $h$. $h=\frac{v_B^2}{2g}$

4. Substitute $v_B=10\ \mathrm{ms}^{-1}$ and $g=10\ \mathrm{ms}^{-2}$. $h=\frac{(10)^2}{2\times10}=\frac{100}{20}=5\ \mathrm{m}$

Hence, the bob rises to a height of 5 m, matching Option D.

1. Since both particles move with identical uniform angular speeds but in opposite directions, at any instant they are located symmetrically about the vertical axis passing through the center, at angle $\theta$ measured from the starting configuration.

2. Write the velocity components of each particle. For a unit mass moving with speed $v$ tangent to a unit circle, the two particles' momenta have components that are mirror images of each other in the horizontal direction and identical in the vertical direction. $\vec{P}_{net}=-mv\sin\theta\,\hat{i}-mv\cos\theta\,\hat{j}+mv\sin\theta\,\hat{i}-mv\cos\theta\,\hat{j}$

3. Notice that the horizontal ($\hat{i}$) components cancel exactly due to the symmetry, leaving only the vertical component. $\vec{P}_{net}=-2mv\cos\theta\,\hat{j}$

4. Evaluate the magnitude of this net momentum at the two endpoints of the motion, $\theta=\frac{\pi}{2}$ and $\theta=-\frac{\pi}{2}$, where $\cos\theta=0$. $\text{At }\theta=\pm\frac{\pi}{2},\quad \vec{P}_{net}=0$

5. Between these endpoints, the magnitude $|P|=2mv|\cos\theta|$ rises smoothly from zero, reaches a maximum somewhere in between, and returns to zero — a smooth bump-shaped curve, not a straight line or discontinuous jump.

Hence, the graph that matches this smooth zero-to-max-to-zero variation is shown in Option D.

1. First find the speed of the ball just before hitting the ground, using the kinematic relation for a free fall from height $h_1=40\ m$. $v_1=\sqrt{2gh_1}=\sqrt{2\times9.8\times40}=\sqrt{784}=28\ \mathrm{ms}^{-1}$ (directed downward)

2. Similarly, find the speed of the ball just after bouncing off the ground, since it rises to a height $h_2=10\ m$. $v_2=\sqrt{2gh_2}=\sqrt{2\times9.8\times10}=\sqrt{196}=14\ \mathrm{ms}^{-1}$ (directed upward)

3. Recall that impulse equals the change in momentum, and taking the upward direction as positive, the downward velocity $v_1$ becomes negative. $\text{Impulse}=\Delta p=m\left(v_2-(-v_1)\right)=m(v_2+v_1)$

4. Substitute $m=0.5\ kg$, $v_1=28\ \mathrm{ms}^{-1}$, and $v_2=14\ \mathrm{ms}^{-1}$. $\text{Impulse}=0.5\times(14+28)=0.5\times42=21\ \mathrm{Ns}$

Hence, the impulse imparted to the ball is 21 Ns, matching Option C.