# Homework Help: To what angle does each ball rebound?

1. Dec 1, 2004

### viendong

need help ASAP [solved] :)

I've been trying for hours..... and couldn't get it right though

A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
Question: To what angle does each ball rebound.?
Answer: m =100 g >>> 79.3 degree
m = 200 g>>>>>14.7 degree
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1) find (y_0)A when the 100 ball is pulled to the left w/ 45 degree angle
(y_1)A = L(1-cos(theta))
=1(1-cos(45)
= 0.293
2) From that I find (V_1)A = sqr(2*g*(y_0)A)
(V_1)A = 2.40 m/s

3) Now I find the (V_2)A = (m1-m2)/(m1+m2)*(V_1)A
V_2)A = - 0.8 m/s

4) Now I find (Y_2)A = (V_2)A ^2/2*g

4) Now I need to find the angle that rebound for Ma=100 g

(theta) = arcosin(1-(Y_2)A /L) = 37 degree....

Last edited: Dec 1, 2004
2. Dec 1, 2004

### Andrew Mason

Think of this in terms of an elastic collision between the two balls at the bottom. You can determine their speeds at the bottom using:
mgh = mv2/2

Since the collision speed is independent of m, v1 = -v2
Using conservation of momentum you can determine their respective velocities after collision. From that, determine their respective kinetic energies immediately after the collision. Using mgh = mv2/2 you can determine the maximum height the respective masses will reach.

AM