Laws Of Motion Ques 29
29. A stone is dropped from a height $h$. It hits the ground with a certain momentum $P$. If the same stone is dropped from a height $100 \%$ more than the previous height, the momentum when it hits the ground will change by :
[2012M]
(a) $68 \%$
(b) $41 \%$
(c) $200 \%$
(d) $100 \%$
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Answer:
Correct Answer: 29.(b)
Solution:
- (b) Momentum $P=m v=m \sqrt{2 g h}$
$(\because v^{2}=u^{2}+2 g h ;.$ Here $.u=0)$
When stone hits the ground momentum,
$ P=m \sqrt{2 g h} $
when same stone dropped from $2 h(100 \%$ of initial) then momentum,
$ P^{\prime}=m \sqrt{2 g(2 h)}=\sqrt{2} P $
$50 \%$ change in momentum, $\frac{(P^{\prime}-P)}{P} \times 100 \%$ $\Rightarrow(\frac{\sqrt{2} P-P}{P}) \times 100 \%=(\sqrt{2}-1) \times 100 \%$
$\Rightarrow(1.414-1) \times 100 \%$
$\Rightarrow 414 \times 100 \%=41.4 \%$
Which is changed by $41 \%$ of initial.
BETA
