Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

96 of 130

Question 96 of 130

A body of mass 2kg2kg moving with a speed of 4ms−14m{s}^{-1} makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is x/10x/10. Find the value of x.

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Answer: 25

pi=pf{p}_{i}={p}_{f}
2×4=2×1+m2×v22\times 4=2\times 1+{m}_{2}\times {v}_{2}
m2v2=6…(1){m}_{2}{v}_{2}=6\ldots (1)
By coefficient of restitution,
1=v2−14⇒v2=5ms−11=\dfrac{{v}_{2}-1}{4}\Rightarrow {v}_{2}=5m{s}^{-1}
by (i)(i)
m2×5=6{m}_{2}\times 5=6
m2=1.2kg{m}_{2}=1.2kg
vcm=m1v1+m2v2m1+m2{v}_{cm}=\dfrac{{m}_{1}{v}_{1}+{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}
vcm=2×1+1.2×52+1.2=83.2=2510{v}_{cm}=\dfrac{2\times 1+1.2\times 5}{2+1.2}=\dfrac{8}{3.2}=\dfrac{25}{10}
x=25x=25

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