Answers by visitors named: oloef

Let the velocity of the (bullet+pendulum) is v m/s after the collision, =>By the law of energy conservation:- =>PE(final) = KE(initial) =>mgh = 1/2mv^2 =>v = √2gh =>v = √[2 x 9.8 x 0.386] =>v = 2.75 m/s Let the velocity of the bullet before the impact was u m/s, By the law of momentum conservation:- =>m1u1+m2u2 = (m1+m2)v =>0.02 x u + 0 = (0.02+5.75) x 2.75 =>u = 793.38 m/s