A ball with a mass of # 5 kg# is rolling at #18 m/s# and elastically collides with a resting ball with a mass of #2 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 5    kg# is rolling at #18    m/s# and elastically collides with a resting ball with a mass of #2  kg#. What are the post-collision velocities of the balls?

Layla Ahmed · Accepted Answer

After collision, the 5-kg ball moves in the same direction at 7.71m/s while the 2-kg ball also moves in this direction at 25.7 m/s.

This is a lengthy problem that asks us to solve for two unknowns (the two final velocities). To do this, we must generate two equations involving these two unknowns, and solve them simultaneously. One equation will come from conservation of momentum, the other will come from kinetic energy conservation. First, cons. of momentum: #m_1v_(1i) + m_2v_(2i) = m_1v_(1f) + m_2v_(2f)# Inserting the values we know: #5(18) + 2(0) = 5v_(1f) + 2v_(2f)# #90 = 5v_(1f) + 2v_(2f)# Now, kinetic energy conservation. #1/2m_1v_(1i)^2 + 1/2m_2v_(2i)^2=1/2m_1v_(1f)^2 + 1/2m_2v_(2f)^2# #1/2(5)(18)^2 + 1/2(2)(0)^2 = 1/2(5)v_(1f)^2 + 1/2(2)v_(2f)^2# #810=2.5v_(1f)^2 + (1)v_(2f)^2# With all that complete, our two equations are: #90 = 5v_(1f) + 2v_(2f)# #810=2.5v_(1f)^2 + (1)v_(2f)^2# Rewrite the first equation as #v_(2f)=(90-5v_(1f))/2# Substitute this value into the second equation: #2.5v_(1f)^2 + ((90-5v_(1f))/2)^2= 810# Now, we must solve this equation. It will be simpler if we multiply every term by 4, to eliminate the denominators: #10v_(1f)^2 + (90-5v_(1f))^2= 3240# #10v_(1f)^2 + (8100-900v_(1f)+25v_(1f)^2)= 3240# #35v_(1f)^2 -900v_(1f)+4860= 0# Use the quadratic formula to solve for v_(1f) #v_(1f)= (900+- sqrt((-900)^2 - 4(35)(4860)))/(2(35))# #= (900+-360)/70# There are two answers: #v_(1f)= (900+360)/70=18m/s# and #v_(1f)= (900-360)/70=7.71m/s# Substituting each answer back into #v_(2f)=(90-5v_(1f))/2# we get #v_(2f)=(90-5(18))/2=0m/s# and #v_(2f)=(90-5(7.71))/2=25.7m/s# We must reject the first answer in each case, as this would result in both balls continuing in their original directions. This could only happen if there was no collision. So, the only acceptable answers are #v_(1f)=-7.71m/s# #v_(2f)=25.7m/s#

Leo Abeyta · Answer

undefined To find the post-collision velocities of the balls, we can use the conservation of momentum and kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved.

The conservation of momentum equation is:
$$ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} $$
Where:
- $ m_1 $ and $ m_2 $ are the masses of the two objects,
- $ v_{1i} $ and $ v_{2i} $ are the initial velocities of the two objects, and
- $ v_{1f} $ and $ v_{2f} $ are the final velocities of the two objects.

The conservation of kinetic energy equation is:
$$ \frac{1}{2} m_1 \cdot v_{1i}^2 + \frac{1}{2} m_2 \cdot v_{2i}^2 = \frac{1}{2} m_1 \cdot v_{1f}^2 + \frac{1}{2} m_2 \cdot v_{2f}^2 $$

Given:
- $ m_1 = 5 $ kg,
- $ v_{1i} = 18 $ m/s,
- $ m_2 = 2 $ kg,
- $ v_{2i} = 0 $ m/s (initially at rest).

We need to solve these two equations simultaneously to find the final velocities $ v_{1f} $ and $ v_{2f} $.

First, apply conservation of momentum to find $ v_{1f} $:
$$ 5 \cdot 18 + 2 \cdot 0 = 5 \cdot v_{1f} + 2 \cdot v_{2f} $$

Second, apply conservation of kinetic energy to find $ v_{2f} $:
$$ \frac{1}{2} \cdot 5 \cdot 18^2 = \frac{1}{2} \cdot 5 \cdot v_{1f}^2 + \frac{1}{2} \cdot 2 \cdot v_{2f}^2 $$

Solve these two equations to find $ v_{1f} $ and $ v_{2f} $.