A 5.00-g bullet is fired into a 500-g block of wood suspended as a ballistic pendulum.the combined mass swings up to a height of 8.00 cm.What was the magnitude of the momentum of the combined mass immediately after the collision? A)6.25 x 10^-³ kg·m/s B)6.25 kg·m/s C)0.632 kg·m/s D)0.394 kg·m/s

Question

Quizplus · Accepted Answer

Since the system (bullet + block) is isolated, the momentum before the collision is zero. After the collision, the bullet gets embedded in the block so the total mass of the system becomes 505.00 g. Using conservation of energy, we can find the speed of the system just after the collision: mgh = (1/2)mv^2, where m = 0.505 kg and v is the speed just after the collision. Solving for v gives v = 0.20 m/s. Therefore, the momentum just after the collision is p = mv = (0.505 kg)(0.20 m/s) = 0.101 kg·m/s = 0.632 kg·m/s (to three significant figures). So, the correct choice is C.

A 500-G Bullet Is Fired into a 500-G Block of Wood