A bullte weighing `5 g` and moving with a velocity `600 m//s` strikes a `5 kg` block of ice resting on a frictionless surface. The speed of the block after the collision is

Options

A. `6 cm //s`
B. `60 cm//s`
C. `6 m//s`
D. `0.6 cm//s`

Answer

B

Explanation

Here `m_(1)=5 g =0.005 kg, u_(1)=600 m//s`,
`m_(2)=5 kg, u_(2)=0`
After collision, they will move with the same velocity `upsilon`.
As momentum is conserved
`m_(1)u_(1)+ m_(2)u_(2)=(m_(1)+m_(2))upsilon`
`:. upsilon=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))`
`upsilon=(0.005xx600xx0)/((5+0.005))=0.6m//s=60 cm//s`