Hopefully easy physics?

Question

A 25 g ball is fired with an initial speed v1 toward a 120 g ball that is hanging motionless from a 1.05 m string. The balls have a perfectly elastic collision. As a result, the 120 g ball swings out until the string makes an angle of 37° with the vertical. What was v1?

______m/s

Answer 1

Elastic collisions conserve kinetic energy.

In all collisions momentum is conserved.

We can assume that energy is conserved as the 120g ball swings up, so that all of its kinetic energy just after the collision is converted to potential energy at the top of the swing.

First calculate the potential energy of the 120 g ball.

PE = mgh (use m=0.120 kg so that units are SI)

PE = 0.2489 joules

Therefore the kinetic energy of the 120 g ball immediately after the collision is = 0.2489 joules

KE = 1/2 m v^2

so its velocity = 2.037 m/s

and its momentum = +0.2444 kg-m/s

The initial kinetic energy of the 25 g ball = 0.0125 v1^2

Its initial momentum = 0.025 v1

Its final kinetic energy = 0.0125 v2^2

Its final momenutm = 0.025 v2

Kinetic energy is conserved:

0.0125 v1^2 = 0.0125 v2^2 + 0.2489

Momentum is conserved:

0.025 v1 = 0.025 v2 + 0.2444

You have two equations and two unknowns.

solving for v1:

v1 = 5.906 m/s

(Note that v2 -- the final velocity of the smaller ball -- is negative)