Question

A pendulum of length 2.0 m is raised so that it maeks an angle of 43^o with the vertical. The mass of the pendulum bob is 1.0 kg. It is released with an initial speed of 1.0 m/s and swings downward. At the bottom of its arc, it collides with the bob of a second pendulum of length 2.0 m and mass 3.0 kg which is initially at rest. Assume the collision is elastic.

Calculare the speed with which the first pendulum bob strikes the second.

Determine the speeds of the two masses immediately after the collision.

Determine the heights that each mass will reach after the collision.

Answer 1

Initial Speed = 1.0 m/s
Initial Height = (2.0 - 2.0 * cos(43))

Using Energy Conservation -
Initial Potential Energy + Initial Kinetic Energy = Final Kinetic Energy
m*g*h + 1/2 * mv^2 = 1/2 * mvf^2
9.8 *  (2.0 - 2.0* cos(43)) + 1/2 * 1^2 = 1/2 * vf^2
vf = 3.4 m/s
Speed with which the first pendulum bob strikes the second, vf = 3.4 m/s

Using Momentum Conservation -
m1*vf = m1*vf1 + m2*vf2
3.4 * 1 = 1*vf1 + 3 * vf2 -------1

Using Energy Conservation as the Collison is Elastic
1/2 * m1*vf^2 = 1/2 * m1*vf1^2 + 1/2 * m2*vf2^2
1 * 3.4^2 = 1 * vf1^2 + 3*vf2^2 -----2

Using eq 1 & 2 -
vf1 = - 1.7 m/s
vf2 = 1.7 m/s
Speed of two masses immediately after collison -
Speed of Mass 1, vf1 = - 1.7 m/s
Speed of Mass 2, vf2 = 1.7 m/s

(-ve sign just means that the direction is opposite to the original direction)

Now, Again Using Energy conservation-

m*g*h = 1/2 * m*v^2
h = (1/2 * v^2)/g
h = (1/2 * 1.7^2)/9.8 m
h = 0.147 m
Heights that each mass will reach after the collision. h =0.147 m