Laws of Motion - Result Question 3

3. 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: 3. (b)

Solution:

  1. (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.