A bullet of mass 2 g travelling with a velocity f 500 `ms^(-1)` is fired into a block of wood of mass 1kg 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" 10ms^(-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 = 2 xx 10^(-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)=0.032m`