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JEE Main 2014 · 19 Apr · Shift 1, Physics question 2

A bullet looses (1n)th\left(\dfrac1n\right)^{\text{th}} of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :

Choose one option
Answer and worked solution

Official answer: A

Each plank removes the same energy, ΔK=12mv2[1−(1−1n)2]=12mv2⋅2n−1n2\Delta K=\frac12mv^2\left[1-\left(1-\frac1n\right)^2\right]=\frac12mv^2\cdot\frac{2n-1}{n^2}.

The number of planks to remove all of 12mv2\frac12mv^2:
N=n22n−1.N=\frac{n^2}{2n-1}.

Topics:Work-energy theorem
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