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 velcities after the impact be v and V respectively. Let the initial velcity of the bullet be .u..
According to the law of conservation of linear momentum mu = mv + MV
Here `m = 2 xx 10 ^(-3)kg, u = 500 ms ^(-1), v = 100 ms ^(-1)`
`(2 xx 10 ^(-3)) xx 500 = (2 xx 10^(-3)) xx 100 + (1 xx V)`
`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))/( 2 xx 10) = 0.032 m`