Question

a bullet of mass 20gm moving with a velocity of 300m/s gets embedded in a freely suspended wooden block of mass 880gm. what is the velocity acquired by the block?​

Answer 1

Answer:

applying conservation of momentum

momentum of system before = momentum of system after

(.02×300)+.88×0=(.02×0)+.88×v

6=.88v

v=6/.88

v=6.81m/s

Answer 2

Explanation:

This is completely inelastic collision. The linear momentum is conserved.

If a body of mass, m collides with a body of mass M and gets embedded in it, according to law of conservation of momentum,

mv =( m+M)V.

Here,

m is mass of colliding body. m= 20 g.

M is mass of target body. M=880 g.

v=300m/s is velocity of colliding body.

V= ? V is velocity of combined system of m and M.

Therefore 20(g).300(m/s)=(880+20)( g)V(m/s.)OR

V=6000/900=6.67 m/s.