Question

Part A

Two identical steel balls, each of mass 2.40 kg, are suspended from strings of length 33.0 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle θ = 60.0° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

Part B

Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?

Answer 1

a)

L = length of string

h = height gained by ball as it is pulled by angle 60 = L - L cos60 = L/2 = 0.33/2 = 0.165 m

v = velocity of ball just before hitting the other ball

using conservation of energy

PE at Top = kinetic energy at bottom

(0.5) m v² = mgh

v = sqrt(2gh) = sqrt(2g(L/2)) = sqrt(gL) = sqrt(9.8 x 0.33) = 1.8 m/s

let v₁ and v₂ be the velocities after the collision

m = mass of each ball

using conservation of momentum

mv = m v₁ + m v₂

v₂ = v₁ - 1.8                          eq-1

using conservation of kinetic energy

(0.5) m v² = (0.5) m v₁² + (0.5) m v₂²

v² = v₁² + v₂²

1.8² = v₁² + (v₁ - 1.8)²

v₁ = 0 m/s

v₂ = 1.8 m/s

using conservation of energy again

mgh = (0.5) m v₂²

h = v₂² /2g = 1.8²/(2 x 9.8) = 0.165 m

part B)

v = 1.8 m/s

V = velocity after collision

using conservation of momentum

mv = (m + m) V

V = 1.8/2 = 0.9 m/s

using conservation of energy again

(2m)gh = (0.5) (2m) V²

h = V² /2g = 0.9²/(2 x 9.8) = 0.041 m