Ball 1 Collides head-on with another identical ball 2 at rest. The velocity of ball 2 after collision becomes two times to that of ball 1 after a collision. The coefficient of restitution between the two balls is
A
e=1/3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
e=1/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
e=1/4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
e=2/3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ae=1/3 Both balls have mass 'm'. Initially ball 1 moves with v. After collision ball 1 moves with v2 After collision ball 2 moves with 2v2 Momentum is conserved. i.e. mv=mv2+2mv2=3mv2 i.e. v2=v3 Velocity of approach =v−0=v Velocity of separation =23v−13v=13v Coefficient of restitution is velocity of approach by velocity of separation. e=v/3v=13