Centre of Mass 3 Question 12
15. 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: 15. (a)
Solution:
- From momentum conservation equation, we have,
$$ \begin{aligned} & \stackrel{\bullet}{m} \stackrel{\rightharpoonup}{m} _{u _0 \cos \alpha}^{\sqrt{u _0^{2}-2 g H}} \\ & \xrightarrow{\hat{\hat{j}}} \hat{i} \\ & \begin{array}{cc} \mathbf{p} _i=\mathbf{p} _f \\ \therefore \quad m\left(u _0 \cos \alpha\right) \hat{\mathbf{i}}+m\left(\sqrt{u _0^{2}-2 g H}\right) \hat{\mathbf{j}}=(2 m) \mathbf{v} \end{array} \\ & H=\frac{u _0^{2} \sin ^{2} \alpha}{2 g} \end{aligned} $$
From Eqs. (i) and (ii)
$$ \mathbf{v}=\frac{u _0 \cos \alpha}{2} \hat{\mathbf{i}}+\frac{u _0 \cos \alpha}{2} \hat{\mathbf{j}} $$
Since both components of $\mathbf{v}$ are equal. Therefore, it is making $45^{\circ}$ with horizontal.
