Question

Two pendulums of equal length l = 0.45 m are suspended from the same point. The pendulum bobs are steel spheres. The first bob is drawn back to make a 35° angle with the vertical. If the first bob has mass 0.25kg and the second has mass .54 how high will the second bob rise above its initial position when struck elastically by the first bob after it is released?

Answer 1

Solution:

Let us go to the basics first.

First bob has height
h = L(1 - cosΘ) = 0.45m * (1 - cos35º) = 0.0814 m
and so arrives at the impact point with velocity
V = √(2gh) = √(2 * 9.8m/s² * 0.0814m) = 1.263 m/s

For an elastic, head-on collision, we know (from CoE) that
the relative velocity of approach = relative velocity of separation, or
V = 1.263m/s = v - u → for u, v the post-collision velocities of 0.25kg, 0.54kg bob
So u = v - 1.263m/s

Now conserve momentum: initial mv = Σ final mv
0.25kg * V = 0.25kg * u + 0.54kg * v
0.25 * 1.263 = 0.25(v - 1.263) + 0.54*v = 0.79v - 0.316
v = 0.631 / 0.79 = 0.798 m/s

Then h' = v² / 2g = (0.798m/s)² / 19.6m/s² = 0.0325 m = 3.25 cm (Answer)

Thanks!!!