A bullet of mass 20 g moving with a velocity of 100 ms−1 strikes a wooden block of mass 80 g and gets embedded into it. Calculate velocity of the combined system.
20 ms−1
Let mass of bullet be m1 =20g
Let mass of the wooden block be m2 =80g
Let initial velocity of the bullet be u1 = 100 ms−1 and final velocity of the bullet be v1.
Let the initial velocity of wooden block be u2, since it was at rest, the initial velocity is zero, i.e u2=0 ms−1
Let final velocity of the wooden block be v2.
By applying law of conservation of momentum,
intial momentum of the system = final momentum of the system
i.e, m1u1+m2u2=m1v1+m2v2
After the bullet is fired, the bullet gets embedded on the block.
Therefore, it acts as a system
Hence , the final velocity of the bullet and wooden block after collision will be equal as it works as a system
i.e v1=v2=v where v is the velocity of the system after collision.
Substituting on the equation, we get
(0.020×100)+(0.080×0)=v×(0.020+0.080)
⇒ 2=0.1×v
⇒ v = 20 ms−1
∴Velocity of system after collision = 20 ms−1