Rotational Motion MCQs for NEET — Physics Questions with Answers

According to the principle of conservation of momentum and kinetic energy, when an explosive body breaks into smaller parts, most of the kinetic energy is acquired by the smaller part due to its larger velocity. Hence, the 1 kg part has a higher kinetic energy of 1800 J.

In a perfectly elastic collision between two equal masses, the relative velocity after collision remains the same, but the direction is reversed. With coefficient of restitution 1/3, the relative velocity after collision is 1/3 of the initial relative velocity. Hence, each ball moves with a speed of 2 m/s.

About origin, angular momentum will be

    $$\begin{gathered} \mathrm{m}(\mathrm{r} \mathrm{x} \mathrm{v}) = 2\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 0\\ 2& -2& 2\end{array}\right| \\ = 4\hat{\mathrm{i}} - 4\hat{\mathrm{j}} - 8\hat{\mathrm{k}} \end{gathered}$$

Therefore, the component of angular momentum about the z-axis would be -8 kg-${\mathrm{m}}^2/\mathrm{s}$.

In an orbital motion, the angular momentum vector is perpendicular to the orbital plane. This is a fundamental result from the definition of angular momentum as the cross product of position and linear momentum vectors, which are mutually perpendicular for circular orbits.

$a=\frac{g \sin \theta }{1+{\displaystyle \frac{1}{m{R}^2}}} or a<g \sin \theta$

$$\begin{gathered} p_i=0 \\ \therefore p_r=0, \end{gathered}$$

i.e, both the pieces should have equal and opposite momentum.

For a particle moving in a horizontal circle with a fixed radius, the direction of the angular momentum vector changes continuously, but its magnitude remains constant. This is because the angular momentum is proportional to the product of mass, radius, and speed, all of which are constant.

Distance between bullet and block at this instant is (D-d).

Distance from bullet = $\frac{M}{M+m}(D-d)$

And the distance from block = $\frac{m}{M+m}(D-d)$