A bullet of mass 2 g travelling with a velocity of `500 ms^(-1)` is fired into a block of wood of mass 1 kg suspended from a string of 1 m length. If the bullet penetrates the block of wood and comes out with a velocity of `100 ms^(-1)`, find the vertical height through which the block of wood will rise (assuming the value of g to be `10 ms^(-2)`).

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`