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Two balls of masses m(1), m(2) and speed...

Two balls of masses `m_(1), m_(2)` and speeds `v_(1)` and `v_(2)` collide at right angle . The maximum amount of kinetic energy loss due ot inelastic collision is ___.

A

`(1)/(2)(m_(1)v_(1)^(2) + m_(2) V_(2)^(2))`

B

`(1)/(2) (m_(1)m_(2))/(m_(1) + m_(2)) (v_(1)^(2) + v_(2)^(2))`

C

`(1)/(2) (1)/(m_(1) + m_(2)) (m_(1)^(2)v_(1)^(2) + m_(2)^(2)v_(2)^(2))`

D

`(1)/(2) (m_(1)m_(2))/(m_(1) + m_(2)) (v_(1) + v_(2))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B

Maximum energy loss happens when balls stick together (perfectly inelastic)

`F_(ext) = 0` both along x & y aixs
Aong x-axis :
`m_(1)v_(1) = (m_(1) + m_(2))v_(x)`
`v_(x) = (m_(1)v_(1))/(m_(1) + m_(2))`
Along y-axis :
`v_(v) = (m_(2)v_(2))/(m_(1) + m_(2))`
`v = sqrt(v_(x)^(2) + v_(y)^(2))`
`DeltaKE = (1)/(2) (m_(1) + m_(2))[((m_(1)v_(1))/(m_(1) + m_(2)))^(2) + ((m_(2)v_(2))/(m_(1) + m_(2)))^(2)]`
`-(1)/(2)m_(1)v_(1)^(2) - (1)/(2) m_(2)v_(2)^(2)`
`= -(1)/(2)(m_(1)m_(2))/(m_(1) + m_(2)) (v_(1)^(2) + v_(2)^(2))`
METHOD-2
For two body system
`DeltaK = -(1)/(2)mu|vec(v)_(rel)|^(2)`
`mu = (m_(1)m_(2))/(m_(1) + m_(2)), vec(v)_(rel) = vec(v)_(1) - vec(v)_(2)`
`vec(v)_(rel) = v_(1)hat(i) - v_(2)hat(j)`
`|vec(v)_(rel)| = sqrt(v_(1)^(2) + v_(2)^(2))`
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