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)).

A bullet of mass 2 g travelling with a velocity of 500 ms^(-1) is fire

1.	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)).
Answer» Solution :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`