Asked by xavier
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.
Answers
Answered by
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
=>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
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