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Step-by-Step Solution
Step 1: Express the momentum for the first drop
When a stone of mass $m$ is dropped from a height $h$, it acquires a velocity $v$ just before hitting the ground. From energy conservation (or from kinematic equations),
$v = \sqrt{2gh}.
Hence, the momentum $P$ with which it hits the ground is
$P = m \sqrt{2gh}.$
Step 2: Express the momentum for the second drop
Now, the same stone is dropped from a height that is 100% more than the original height, i.e., from $2h$. Its velocity $v'$ just before hitting the ground becomes
$v' = \sqrt{2g \cdot 2h} = \sqrt{4gh} = \sqrt{2}\,\sqrt{2gh}.
Thus, its momentum $P'$ is
$P' = m \sqrt{2g(2h)} = m \sqrt{2} \sqrt{2gh} = \sqrt{2}\,P.
Step 3: Calculate the percentage change in momentum
The percentage change in momentum is given by
$
\text{% change} = \frac{P' - P}{P} \times 100\%.
$
Substitute $P' = \sqrt{2}\,P$ into the formula:
$
\text{% change} = \frac{\sqrt{2} P - P}{P} \times 100\% = \left(\sqrt{2} - 1\right) \times 100\%.
$
Since $\sqrt{2} \approx 1.414$, $\sqrt{2} - 1 \approx 0.414.$ Thus,
$
\text{% change} \approx 0.414 \times 100\% \approx 41\%.
$
Final Answer
The momentum increases by approximately 41%.
