Question

Two pendulums of equal length l = 2.00 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 0.25 kg and the second has mass 0.85 kg, how high will the second bob rise above its initial position when struck elastically by the first bob after it is released?

Answer 1

the velocity of the first bob when hitting the second bob is obtained by converting potential energy to kinetic energy:

1/2 mv^2 = mgh

v^2 = 2gh with h = 2(1 - cos35)

v^2 = 2*9.81*2(1 - cos35)

v = 2.66 m/s

the initial velocity of the second bob is obtained by

v2' = 2(v1m1 + v2m2)/(m1+m2) - v2

with

v1= 2.66 m/s

v2= 0

v2' = 2(0.25*2.66)/(0.85+0.25)

v2' = 1.209 m/s

the hight of bob 2 is again obtained by converting kinetic energy to potential energy:

1/2*mv^2 = mgh

h = v^2/(2g)

h = 1.209 ^2/(2*9.81)

h = 0.0745 m

or h = 7.45 cm