A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take `g=10ms^(-2)`.

Explanation

`m_(1)=50g=0.05kg," "m_(2)=450g=0.45kg`

`" "`The speed of the bullet is `u_(1)`. The second body is at rest `(u_(2)=0)`. Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
`" "v=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2)))`
`" "v=(0.05u_(1)+(0.45 times 0))/(0.05+0.45)=(0.05)/(0.50)u_(1)`
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
`" "v=sqrt(2gh)`
`" "v=sqrt(2 times 10 times 1.8)=sqrt(36)`
`" "v=6ms^(-1)`
Substituting this in the above equation, the value of `u_(1)` is
`6=(0.05)/(0.50)u_(1)" or "u_(1)=(0.50)/(0.05) times 6=10 times 6`
`" "u=60ms^(-=1)`