A bullet of mass 2g travelling with a velocity of `500 ms^(-1)` is fired into a block of wood of mass 1kg suspended from a string of 1m 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= 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)= 0.032m`