Question

A bullet, m = 0.500 kg, traveling with some initial velocity strikes and embeds itself in the bob of a ballistic pendulum, M = 9.50 kg. a) The combined masses rise to a height h = 1.28 m. The speed Vf of the combined masses immediately following impact is ____

Answer 1

5 m/s

Step-by-step explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet is u.

mass of pendulum, M = 9.50 kg

height raised, h = 1.28 m

The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system

Let vf be the final velocity after the impact

KE = PE

1/2 (M + m) Vf^2 = (M + m) x g x h

vf^2 = 2 x 9.8 x 1.28 = 25.088

vf = 5 m/s

Thus, the final speed of the combined mass system after the impact s 5 m/s.