A bullet of mass 2 g travellingwith 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 ofg to be" 10ms^(-2)).

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

1.	A bullet of mass 2 g travellingwith 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 ofg to be" 10ms^(-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 = 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`