Question

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 different height \(h'\) such that percentage change in momentum is \(41.4\%\), then the height from which the stone is dropped is \(h' = xh\), where \(x\) is:

• A. 4
• B. 2
• C. 6
• D. 3

Hint:

Momentum in free fall varies as \( \sqrt{h} \). Percentage change questions become easy using proportional relation.

Answer 1

The Correct Option is B

Step 1: Relation between velocity and height.
For a body falling freely from height \(h\), velocity just before hitting ground is:
\[ v = \sqrt{2gh} \]

Step 2: Expression for momentum.
Momentum is given by:
\[ p = mv = m\sqrt{2gh} \]
Thus, momentum is proportional to \( \sqrt{h} \).

Step 3: Write relation for new height.
\[ p' = m\sqrt{2gh'} \] \[ \frac{p'}{p} = \sqrt{\frac{h'}{h}} \]

Step 4: Use percentage increase.
Given increase in momentum is \(41.4\%\):
\[ p' = 1.414 \, p \]

Step 5: Substitute in ratio.
\[ 1.414 = \sqrt{\frac{h'}{h}} \]

Step 6: Square both sides.
\[ \frac{h'}{h} = (1.414)^2 = 2 \]

Step 7: Final conclusion.
\[ x = 2 \] \[ \boxed{2} \] Hence, correct answer is option (B).