Question

A bullet of mass 4.8 g strikes a ballistic pendulum of mass 3.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

m = 4.8 grams = 0.0048 kg

M = 3.0 kg

Let V is the speed of bullet and pendulum after bullet embeded into the pendulum.

Apply Conservation of energy

final potentail energy = initial kinetic energy

(m+M)*g*h = (1/2)*(m+M)*V^2

2*g*h = V^2

==> V = sqrt(2*g*h)

= sqrt(2*9.8*0.18)

= 1.9 m/s

Let u is the initial speed of bullet.

Now Apply conservation of momentum

Momentum before the collision = momentum after the collsion.

m*u = (m+M)*V

u = (m+M)*V/m

= (0.0048 + 3.0)*1.6/0.0048

= 1189.4 m/s (Answer)