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Question: A stone is dropped from a height, h. It hits the ground with a certain momentum P. If the same stone...

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
A. 6868 %
B. 4141 %
C. 200200 %
D. 100100 %

Explanation

Solution

Since, the stone is dropped its initial velocity is zero. The height through which it had fallen is given. It hits the ground with momentum p. We know momentum is given by p=mv, so we need to find the final velocity it has just before coming to rest. We can make use of the law of conservation of energy to find the velocity.

Formula used:
We will make use of the law of conservation of energy. Kinetic energy is given by mv22\dfrac{m{{v}^{2}}}{2} where mm is the mass of the body and vv is its velocity. The potential energy is given by mghmgh, where hh is the height measured from the surface of the earth.

Complete step by step answer:
Using law of conservation of energy:
mgh=\dfrac{m{{v}^{2}}}{2} \\\ \Rightarrow gh=\dfrac{{{v}^{2}}}{2} \\\
∴v=2gh\therefore v=\sqrt{2gh} -- (1)
So, momentum,
p=mv \\\ \Rightarrow p=m\sqrt{2gh} \\\
Now in the second case, the same stone is dropped from a height 100% more than the previous height, so the new height is 2h.
Using eq (1)
v′=2gh′ ⇒v′=4ghv'=\sqrt{2gh'} \\\ \Rightarrow v'=\sqrt{4gh}
New momentum, p′=mv′p'=mv'
⇒p′=m4gh\Rightarrow p'=m\sqrt{4gh}
∴p′=2p\therefore p'=\sqrt{2}p --(2)

Now percentage change in the momentum is given by:
p′−pp×100\dfrac{p'-p}{p}\times 100
⇒2p−pp×100 ⇒p(2−1)p×100 ⇒(2−1)×100 ⇒(1.414−1)×100 ⇒0.414×100 ∴41%\Rightarrow \dfrac{\sqrt{2}p-p}{p}\times 100 \\\ \Rightarrow \dfrac{p(\sqrt{2}-1)}{p}\times 100 \\\ \Rightarrow (\sqrt{2}-1)\times 100 \\\ \Rightarrow (1.414-1)\times 100 \\\ \Rightarrow 0.414\times 100 \\\ \therefore 41\%

So, the correct option is B.

Note: In this problem we have made use of the law of conservation of energy to find out the velocity at the impact. In the next part the height is just doubled and so we have put the new height in place of old to find out the new velocity. We have calculated the old and the new momentum and to find the percentage error we have to take the ratio of the change and divide it by the original quantity.