Question

A bullet of mass 10 g moving with a velocity of
400 m/s gets embedded in a freely suspended wooden block of mass
900 g. What is the velocity acquired by the block ?​

Answer 1

Here, Mass of the bullet,m1=10g

=101000kg

=0.01 kg

And, Velocity of the bullet, v1=400 m/s

So, Momentum of bullet=m1×v1

=0.01×400kg.m/s…….(1)

Now, this bullet of mass 10 g gets embedded into a wooden block of mass 900 g . So ,the mass of wooden blockk alongwith the embedded bullet will become 900+10=910 g. Thus,

Mass of wooden block+Bullet, m2=900+10

=910 g

=9101000 kg

=0.91 kg

And, Velocity of wooden block+bullet,v2=? (To be calculated)

So, Momentum of wooden block+bullet=m2×v2

=0.91×v2kg.m/s...(2)

Now, according to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal.

So, m1×v1=m2×v2

or 0.01×400=0.91×v2

And, v2=0.01×4000.91

=4.4 m/s

Thus, the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second.