Question

A bullet of mass 3.3 g strikes a ballistic pendulum of mass 1.3kg. The center of mass of the pendulum rises a vertical distance of6.6 cm. Assuming that the bullet remains embedded in the pendulum,calculate the bullet's initial speed.

Answer 1

Momentum initial = momentum final

bullet is mass 1 (m₁) and velocity 1(v₁)

Pendulum is mass 2 (m₂) and Velocity 2(v₂)

the bullet sticks in the pendulum after collision there isonly one velocity with the two masses for the final momentum. But since you're not given the final velocity you'll have tocalculate it based on conservation of energy methods ΔKE =ΔPE

The trick is treating this part seperately, set initial PE to0 and initial KE to 0.

Thus KE_f = PE_f

.5*(m₁+m₂)*v² =(m₁+m₂)*g*d

m₁=3.3g = 3.3*10^-3kg   but ifyou look closely above masses cancel

d = 6.6 cm = 6.6*10^-2 m

So basically you get an equation v= √(2*g*d)

v=√ (2*9.81*.066) = √1.29492 = 1.1379 m/s this isfinal velocity in the next part

so now look at the momentum

m₁v₁ + m₂v₂ =(m₁+m₂)v_f      v₂= 0 to begin with

m₁v₁=(m₁+m₂)v_f

.0033v₁ = (.0033+1.3)*1.1379  

v₁ = 1.4827/.0033

V₁ = 449.30 m/s this is initial bullet velocityjust before it hits the pendulum