Centre Of Mass Ques 47
- If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point $B$, immediately after it strikes the second incline is
(a) $\sqrt{30} m / s$
(b) $\sqrt{15} m / s$
(c) zero
(d) $-\sqrt{15} m / s$
(2008, 4M)
Objective Questions II (One or more correct option)
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Answer:
Correct Answer: 47.(a, c)
Solution:
- In elastic collision, compound of $v _1$ parallel to $B C$ will remain unchanged, while component perpendicular to $B C$ will remain unchanged in magnitude but its direction will be reversed.
Just before
Just after
$$ \begin{aligned} v _{11} & =v _1 \cos 30^{\circ} \\ & =(\sqrt{60}) \frac{\sqrt{3}}{2}=\sqrt{45} ms^{-1} \end{aligned} $$
$$ v _{\perp}=v _1 \sin 30^{\circ}=(\sqrt{60}) \quad \frac{1}{2}=\sqrt{15} ms^{-1} $$
Now vertical component of velocity of block
$$ v=v _{\perp} \cos 30^{\circ}-v _{|} \cos 60^{\circ} $$
$$ =(\sqrt{15}) \frac{\sqrt{3} _2^{2}}{2}-(\sqrt{45}) \frac{1}{2}=0 $$
BETA
