Question

5.) A ballistic pendulum consists of a 10 gram projectile striking and sticking to a 1.5 kg mass. Both objects then rise 2.5 cms. A.) Calculate the velocity of the pendulum and projectile (stuck together) after the collision. B.) Determine the initial speed of the projectile as it hits the 1.5 kg mass. C.) How much non-conservative work was done during the process?

Answer 1

A) Immediately after the collision when the projectile and pendulum are stuck together, the combined system has a kinetic energy equal to (m₁+m₂)v²/2, where m₁ is the mass of the projectile and equal to 0.01 kg and m₂ is the mass of the pendulum which is 1.5 kg. As both rise to a height h of 0.025m, this kinetic energy is converted completely into potential energy given by (m₁+m₂)gh.

Thus, we have:

(m₁+m₂)v²/2 = (m₁+m₂)gh ⇒ v²/2 = gh ⇒ v²=2gh ⇒ v = √(2gh) = √(2×9.8×0.025) = 0.7m/s

∴ The velocity of the pendulum and projectile (stuck together) after the collision is 0.7m/s.

B) Now, we'll conserve momentum before and after the collision for the system consisting of the pendulum and the projectile.

Let u₁ be the speed of the projectile just before hitting the pendulum. The pendulum is at rest before the collision. Then the initial momentum is:

p_i (momentum before collision) = m₁u₁ + m₂×0 =m₁u₁ = 0.01u₁

Since after the collision both are stuck together we have a combined mass of m₁+m₂ having a speed of v calculated earlier.

p_f (momentum after collision) = (m₁+m₂)v = (0.01+1.5)×0.7 = 1.51×0.7 = 1.057 kg.m/s

Conserving momentum:

p_i = p_f ⇒ 0.01u₁ = 1.057 ⇒ u₁ = 1.057 /0.01 = 105.7 m/s

∴ The projectile was travelling with a speed of 105.7 m/s before the collision.

C) The difference in the total kinetic energies before and after the collision will give the amount of non-conservative work done in the process, since the work energy theorem states that the net work done is equal to the change in the kinetic energies, and since an inelastic collision is happening here, the work done will be dissipative and hence non-conservative. Let W be the non-conservative work.

W = Final Kinetic Energy - Initial Kinetic Energy = (m₁+m₂)v²/2 - m₁u₁²/2

⇒ W = (0.01+1.5)×0.7²/2 - 0.01×105.7²/2 = 1.51×0.49/2 - 0.01×11172.49/2 = 0.37 - 55.86 = -55.49 J

∴ The non-conservative work done is -55.49 J.