A bullet of mass `2 g` travelling at a speed of `500 m//s` is fired into a ballistic pendulum of mass `1.0 kg` suspended from a cord `1.0 m` long. The bullet penetrates the pendulum and emerges with a velocity of `100 m//s`. Through what vertical height will the pendulum rise?

Explanation

Let the masses of the bullet and the block be .m. and .M. respectively. Let their velocities after the impact be v and V respectively. Let the initial velocity of the bullet be .u..
According to the law of conservation of linear momentum. Mu = mv + MV
Here `m = 2xx10^(-3)kg, u=500 ms^(-1), v =100 ms^(-1)`
`(2xx10^(-3))xx500=(2xx10^(-3))xx100+(1xxV)`
`V=0.8 ms^(-1)`.
When the block rises to a height of .h., according to the law of conservation of energy.
`(M+m)gh=(1)/(2)(M+m)V^(2)`
i.e., `h=(1)/(2)(V^(2))/(g)=((0.8)^(2))/(2xx10)=0.032 m`