A 5.25-g bullet is fired into a 360-g block of wood suspended as a ballistic pendulum.The combined mass swings up to a height of 8.75 cm.What was the magnitude of the momentum of the combined mass immediately after the collision? A)0.338kg·m/s B)0.471 kg·m/s C)0.478 kg·m/s D)0.626 kg·m/s E)0.239 kg·m/s

Question

Quizplus · Accepted Answer

First, we need to find the initial velocity of the bullet using conservation of momentum:

m(bullet)v(bullet) = (m(bullet) + m(wood))v(final)

(0.00525 kg)v(bullet) = (0.00525 kg + 0.36 kg)v(final)

v(final) = 0.09375 m/s

Next, we can use conservation of energy to find the velocity of the combined mass at the top of its swing:

KE(initial) = PE(final)
(1/2)(m(bullet) + m(wood))v(final)^2 = (m(bullet) + m(wood))gh

(1/2)(0.00525 kg + 0.36 kg)(0.09375 m/s)^2 = (0.00525 kg + 0.36 kg)(9.81 m/s^2)(0.0875 m)
 
v(final) = 0.7203 m/s

Finally, we can find the momentum of the combined mass at the top of its swing:

p = (m(bullet) + m(wood))v(final)

p = (0.00525 kg + 0.36 kg)(0.7203 m/s)

p = 0.478 kg·m/s

Therefore, the answer is C.

A 525-G Bullet Is Fired into a 360-G Block of Wood