JEE Main 2015 · 4 Apr · Shift 1 · Physics Q4

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A particle of mass m moving in the $x$ direction with speed $2v$ is hit by another particle of mass 2m moving in the $y$ direction with speed $v$ . If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to :

44%

50%

56%

62%

Answer and worked solution

Official answer: C

Initial KE: $\frac12m(2v)^2+\frac12(2m)v^2=3mv^2$ .

Momentum $(2mv,\,2mv)$ is shared by mass $3m$ , so the final KE is $\dfrac{(2mv)^2+(2mv)^2}{2\cdot3m}=\frac43mv^2$ .
$\text{Loss}=\frac{3-\frac43}3=\frac59\approx56\%.$

Topics:Elastic and inelastic collisions

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 4