A 0.02kg bullet collides with a 5.75kg pendulum. After the collision, the pair swings up to a maximum height of 0.386m. Determine the velocity of the bullet just before impact.

Answer 1

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