The ballistic pendulum is an apparatus used to measure the speed of a projectile. An 8.0 g bullet is fired into a 2.5 kg ballistic pendulum bob, which is initially at rest, and becomes embedded in the bob. The pendulum then rises to a vertical distance of 6.0 cm. What was the initial speed of the bullet (in m/s)?

1.242 m/s

Explanation:

Given that

Mass of the bullet, m(b) = 8 kg.

Mass of the pendulum, m(p) = 2.5 kg

Height of the pendulum, h = 6 cm

Initial speed of the bullet, v = ?

Assuming that Energy is conserved, using law of conservation of energy, we have

Potential Energy = Kinetic Energy

mgh = 1/2mv²

1/2mv² = [m(b) + m(p)] gh

1/2 * 8 * v² = (8 + 2.5) * 9.8 * 0.06

4 v² = 10.5 * 0.588

4 v² = 6.174

v² = 6.174 / 4

v² = 1.5435

v = √1.5435

v = 1.242 m/s

Thus, the initial speed of the bullet is 1.242 m/s

where is the circuit diagram ?