A 250 g steel ball and a 500 g steel ball each hang from 1.0-m-long strings. At

Question

A 250 g steel ball and a 500 g steel ball each hang from 1.0-m-long strings. At rest, the balls hang side by side, barely touching. The 250 g ball is pulled to the left until the angle between its string and vertical is 28 o. The 500 g ball is pulled to a 28 o angle on the right. The balls are released so as to collide at the very bottom of their swings. To what angle does 250 g ball rebound? Express your answers using two significant figures. To what angle does 500 g ball rebound? Express your answers using two significant figures.

Explanation / Answer

m1 = 250 g = 0.25 Kg
m2 = 500 g = 0.5 Kg
L = 1.0 m

Initial Potential Energy = m*g* L(1 - cos(28))

This Potential Energy Gets Converted to Kinetic Energy Before Collision.

1/2 m*v^2 =  m*g* L(1 - cos(28))
v = sqrt(2*g*L(1 - cos(28))) m/s
v = sqrt(2*9.8*1(1 - cos(28))) m/s
v = 1.52 m/s

After Collision ,
Let Velocity of m1 = v1
Let Velocity of m2 = v2

Substituing Values -

Using Momentum Conservation -
Initial Momentum = Final Momentum
m1 v - m2 v = m2 v2 + m1 v1
(0.25 - 0.5 ) * 1.52 = 0.25 * v1 + 0.5 *v2 ------------1


Using Energy Conservation -
1/2 m1 v² + 1/2 m2 v² = 1/2 m1 v1² + 1/2 m2 v2²
(0.25 + 0.5 ) * 1.52^2 = 0.25 * v1^2 + 0.5 * v2^2 -----------2

Using 1 and 2 , solving for v1 & v2 -
v1 = - 2.53 m/s (-e sign indicates Ball moves towards left direction.)
v2 =  0.51 m/s

Now, Kinetic Energy After Collision Get's Converted to Potential Energy at Max Height.

Let m1 rebound to angle 1 , Therefore -
m*g* L(1 - cos(1)) = 1/2 m*v^2
9.8 *1(1 - cos(1)) = 0.5 * 2.53^2
1- cos(1) = 0.327
1 = cos^1 (1 - 0.327)
1 = 47.7o

250 g Ball Rebounds to angle, 1 = 47.7o

Let m2 rebound to angle 2 , Therefore -
m*g* L(1 - cos(2)) = 1/2 m*v^2
9.8 *1(1 - cos(2)) = 0.5 * 0.51^2
2 = cos^-1( 1 - (0.5 * 0.51^2)/9.8)
2 = 9.35o

500 g Ball Rebounds to angle, 2 = 9.35o

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