Question

11. A bullet of mass 10 g is fired with an initial velocity 500 ms. It hits a 20 kg
wooden block at rest and gets embedded into the block.
(a) Calculate the velocity of the block after the impact
(b) How much energy is lost in the collision?​

Answer 1

Explanation:

we know that

m1u1+m2u2=m1v1+m2v2

here m1=0.01kg(mass of bullet)

m2=20kg(mass of wooden block)

u1=500m/s(speed of bullet)

u2=0m/s(speed of block)

v1=0m/s(speed of bullet)

v2=v(speed of wooden block after impact)

0.01*500+20*0=0.01*0+20*v

5+0=0+20v

5=20v

v=5/20=0.25m/s

II) energy lost=(m1+m2)*v/2=20.01*0.25/2=5/2=2.5J approx.