A block of mass `0.50kg` is moving with a speed of `2.00m//s` on a smooth surface. It strikes another mass of `1 kg` at rest and they move as a single body. The energy loss during the collision is

Options

A. `0.16J`
B. `1.00J`
C. `0.67J`
D. `0.34J`

Answer

C

Explanation

Here, `m_(1)=0.50kg, u=2.00 m//s, m_(2)=1kg`
`u_(2)=0, upsilon_(1)=upsilon`
According to law of conservation of linear momentum
`m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))upsilon`
or`upsilon=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))`
`=(0.5xx2+1xx0)/(0.5+1)=(1)/(1.5)=(2)/(3)m//s=0.67m//s`
Loss of energy `=K_(i)-K_(f)`
`=(1)/(2)(0.5)(2)^(2)-(1)/(2)(1.5)(0.67)^(2)`
`=1-0.33=0.67J`