Question

A 43.0 g ball is fired horizontally with initial speed v0toward a 107 g ball that is hanging motionless from a 1.2 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 107 g ball swings out to a maximum angle θmax = 50.0 ∘.

What was v0?

Answer 1

v0: Velocity of the smaller ball before colliding
v0′: Velocity of the smaller ball after colliding
v1: Velocity of the larger ball before colliding
v1′: Velocity of the larger ball after colliding

The larger ball was initially stationary:
►v1 = 0


So we are left with three unknowns.


The height of that heavier ball when swings up to is:

h = ℓ( 1 – cosθ )
h = ( 1.2 m )( 1 – cos 50º )
h = 0.42865 meters <=== maximum height of larger ball


GPE_f = KE_i
mgh = ½m( v1′ )²

...dividing both sides by m:
gh = ½( v1′ )²

...and solving for v1′

v1′ = √[ 2gh ]
v1′ = √[ 2( 9.8 m/s² )( 0.42865 meters) ]
v1′ = √[ 8.401 m²/s² ]

►v1′ = 2.90 m/s


We are now down to two unknowns, v0 and v0′.

We'll solve for these two unknowns with two equations, one using the conservation of momentum and the other using the definition of elasticity.

Conservation of momentum:
( m0 )( v0 ) + ( m1 )( v1 ) = ( m0 )( v0′ ) + ( m1 )( v1′ )
( 43g )( v0 ) + ( 107g )( 0 ) = ( 43g )( v0′ ) + ( 107g )( 2.9 m/s )
( 43g )( v0 ) = ( 43g )( v0′ ) + ( 310.3g·m/s )
dividing both sides by 43g:
v0 = v0′ + 7.216 m/s

v0′ = v0 - 7.216 m/s <=== Equation A



Definition of elasticity:
( v0 - v1 ) = e( v1′ - v0′ )
( v0 - 0 ) = 1( 2.9 m/s - v0′ )

v0 = 2.9 m/s - v0′

Now substitute Equation A into the above equation:
v0 = 2.9 m/s - ( v0 - 7.216 m/s )
2v0 = 23.428 m/s

►v0 = 5.058 m/s <=== Final answer


Since kinetic energy is conserved in an elastic collision, we can check the validity of our answer. Using Equation A:
v0′ = v0 - 7.216 m/s
v0′ = 5.058 m/s - 7.216 m/s
►v0′ = -2.158 m/s