A bullet of mass m = 1 g and speed v = 400 m/s passes completely through a pendu
ID: 1469041 • Letter: A
Question
A bullet of mass m = 1 g and speed v = 400 m/s passes completely through a pendulum bob of mass M = 1 kg. The bullet emerges with a speed of v/2. Find the velocity of the bob just after impact with the bullet using the principle of conservation of momentum. Calculate the energy of the system before the collision (E_o = K_o bullet) and the energy of the system just after the collision (E_f = K_fbullet + K_bob). Show that energy is not conserved during the collision by finding the amount of energy lost during the collision? Write this as a percent energy lost.Explanation / Answer
by conservation of momentum
initial momentum = final momentum
mass of bullet * velocity of bullet = (mass of bullet + mass of bob) * velocity of system
1 * 10^-3 * 400 = (1 * 10^-3 + 1) * v
v = 0.399 m/s
velocity of bob just after impact = 0.399 m/s
also
(1 * 10^-3 + 1) * 0.399 = 1 * 10^-3 * 400 / 2 + 1 * v1
v1 = 0.199 m/s
velocity of the bob after bullrt emerges out = 0.199 m/s
energy of the bullet before collision = 0.5 * 1 * 10^-3 * 400^2
energy of the bullet before collision = 80 J
energy of bullet + bob system after collision = 0.5 * 1 * 10^-3 * (400/2)^2 + 0.5 * 1 * 0.199^2
energy of bullet + bob system after collision = 20.019 J
since initial energy is not equal to final energy so energy is not conserved
energy lost = (80 - 20.019) * 100 / 80
energy lost = 74.97625%
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