A bullet of mass 2 g travelling with a velocity of 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 10 ms^(-2)).

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

1.	A bullet of mass 2 g travelling with a velocity of 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 10 ms^(-2)).
Answer» Solution :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`