Collision Problem?

Question

A uniform rod of length L1= 1.8m and M=2.4kg is supported by a hinge at the right end and is free to rotate in the counter-clockwise direction. The rod is released from a horizontal position and hits a particle of mass m (which is supported by a thin string of length L2=1.4m from the hinge. The particle sticks to the rod on contact. After the collision, theta max =35degrees

a) Find mass of particle, m

b) How much energy is dissipated during the collision

The particle is hanging from the hinge vertically and the rod is dropped from the horizontal position

plugged in 6.76kg and still got it wrong

Answer 1

We place the weight of the rod at the center of its length or 0.90m from one end. Its Potential Energy, PE at the horizontal position with respect to the particle, is 2.4g*1.40. This energy is converted to a lesser PE at theta=35 degrees. PE of the rod at this point is 2.4g*(1.40-0.90cos35). The particle had initial PE equal to zero, but at 35 degrees from its original position, it gained a PE equal to mg(1.40-1.40cos35). The PE gained by the particle is equal to the dissipated energy during the collision.

Based on the principle of conservation of energy:

PE of system, initial=PE of system, final

a)Substitute the above values:

2.4g*1.4=2.4g(1.40-0.90cos35)
+mg(1.40-1.40cos35)

g cancels out.

2.4*1.4=2.4*0.67+0.26m
m=(3.36-1.60)/0.26
=6.76kg

b)Dissipated energy=6.76g*0.26
=17.2 Joules

Pls draw a diagram showing the position of the rod and the particle, before and after collision. You'll then find my explanation easier to understand.