Question

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

Answer 1

$m_1v_1=m_2v_2\\\frac{1}{50}300=\frac{88}{100}v_2\\\frac{100}{50*88}300=v_2\\v_2=6.81 m/s$

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.