pi=pf
2×4=2×1+m2×v2
m2v2=6…(1)
By coefficient of restitution,
1=4v2−1⇒v2=5ms−1
by (i)
m2×5=6
m2=1.2kg
vcm=m1+m2m1v1+m2v2
vcm=2+1.22×1+1.2×5=3.28=1025
x=25
JEE Main 2021 — Physics Mechanics
A body of mass 2kg moving with a speed of 4ms−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/10. Find the value of x.
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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