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

Take `g = 10 ms^(-2)` `m_(1)=50 g=0.05 kg` , `m_(2)=450 g=0.45 g`
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 xx 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 xx 10 xx 1.8)=sqrt(36)`
`v=6 ms^(-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) xx 6 = 10 xx 6`
`u_(1)=60 ms^(-1)`