A 1 kg ball, moving at 12 m s⁻¹, collides head-on with a 2 kg ball moving in the opposite direction at 24 m s⁻¹. If the coefficient of restitution is 2/3, then the energy lost in the collision is 60 × x joule. The value of x is ...
Physics · Mixed Chapter
JEE Mains

Question

A 1 kg ball, moving at 12 m s⁻¹, collides head-on with a 2 kg ball moving in the opposite direction at 24 m s⁻¹. If the coefficient of restitution is 2/3, then the energy lost in the collision is 60 × x joule. The value of x is ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The approach speed is u₁−u₂=12−(−24)=36 m/s. The reduced mass is mᵣ=m₁m₂/(m₁+m₂)=2/3 kg. Step 2: For a one-dimensional collision, kinetic energy lost is ΔK=(1/2)mᵣ(1−e²)(u₁−u₂)². Step 3: ΔK=(1/2)(2/3)[1−(2/3)²](36)²=(1/3)(5/9)(1296)=240 J. Step 4: 60x=240, so x=4.