Centre Of Mass Ques 35
- A particle of mass $m$ is projected from the ground with an initial speed $u _0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u _0$. The angle that the composite system makes with the horizontal immediately after the collision is (2013 Adv.)
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{4}+\alpha$
(c) $\frac{\pi}{4}-\alpha$
(d) $\frac{\pi}{2}$
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Answer:
Correct Answer: 35.(a)
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
- From momentum conservation equation, we have,
$$
\begin{aligned}
& \stackrel{\bullet}{m} \stackrel{\rightharpoonup}{m} _{u_0 \cos \alpha}^{\sqrt{u_0^{2}-2 g H}} \\
& \xrightarrow{\hat{j}} \hat{i} \
& \begin{array}{cc}
\mathbf{p} _i=\mathbf{p} _f
\therefore \quad m\left(u _0 \cos \alpha\right) \hat{\mathbf{i}}+m\left(\sqrt{u _0^{2}-2 g H}\right) \hat{\mathbf{j}}=m \mathbf{v}
\end{array} \
& H=\frac{u _0^{2} \sin ^{2} \alpha}{2 g}
\end{aligned}
$$
From Eqs. (1) and (2)
$$ \mathbf{v}=\frac{u_0 \cos \alpha}{2} \hat{\mathbf{i}}+\frac{u_0 \cos \alpha}{2} \hat{\mathbf{j}} $$
Since both components of $\mathbf{v}$ are equal, it is making $45^{\circ}$ with the horizontal.
BETA
