Question

Revisiting the ballistic pendulum. In lab we used both conservation of momentum and conservation of energy to relate the launch speed of a projectile to the maximum height of the swing of a pendulum. Here we will study the parts of this problem in a bit more detail.

(a) Briefly explain why momentum is conserved during the collision of the projectile and the pendulum, but mechanical energy is not conserved.

(b) Briefly explain why mechanical energy (kinetic plus potential energy) is conserved during the swing of the pendulum, but momentum is not conserved.

(c) Since the motion of the pendulum is circular motion, it makes sense that we could also approach the problem using angular momentum and/or rotational kinetic energy. Use conservation of angular momentum to show that vlaunch = m+M m vf , where vlaunch is the initial speed of the projectile and vpendulum is the speed of the pendulum and projectile just after they collide and stick together. m is the mass of the projectile and M is the mass of the pendulum. This is the same answer that ordinary momentum conservation gave you in lab. (Hint: Even though the projectile moves on a straight line, at the point of impact you can interpret its velocity as tangential and assign it an ωi. Also, assume that the mass of the pendulum is very concentrated at the distance l from the axis of rotation, which is also where the projectile hits.)

(d) Show that the translational kinetic energy K = 1 2 (m + M)v 2 f of the projectile–pendulum object can also be interpreted as a rotational kinetic energy K = 1 2 Iω2 . (These are the same quantity of energy calculated two different ways; they don’t both contribute to the total energy, just one does.)

Answer 1