Laws Of Motion Ques 2
2. 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 :
[2012 M]
(a) $68 \%$
(b) $41 \%$
(c) $200 \%$
(d) $100 \%$
Show Answer
Answer:
Correct Answer: 2.(b)
Solution: (b) Momentum $P=m v=m \sqrt{2 g h}$
$\left(\because v^2=u^2+2 g h ;\right.$ Here $\left.u=0\right)$
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, $\frac{\left(P^{\prime}-P\right)}{P} \times 100 \%$
$ \begin{aligned} & \Rightarrow\left(\frac{\sqrt{2} P-P}{P}\right) \times 100 \%=(\sqrt{2}-1) \times 100 \% \ & \Rightarrow(1.414-1) \times 100 \% \ & \Rightarrow 414 \times 100 \%=41.4 \% \end{aligned} $
Which is changed by $41 \%$ of initial.
BETA
