Question

A bullet of mass moving horizontally with velocity v strikes a block of mass 'm' and gets embedded
n
into it. The velocity of the block finally is
[ ]
ny
d) n(n+1)
c)
V
b)
n+1
n+1
a)
n
​

Answer 1

Answer:

The situation is as shown in the figure.

Let V be velocity of the block- bullet system just after collision. Then by the law of conservation of linear momentum, we get

mv=(m+M)V

V=mvm+M

Let the block rises to a height h.

According to law of conservation of mechanical energy, we get

12(m+M)V2=(m+M)gh

h=V22g=v22g(mm+m)2