A ball with a mass of #2 kg# is rolling at #7 m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of #2  kg# is rolling at #7  m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Olivia Askew · Accepted Answer

In a perfectly elastic collision, #v_1=7/3m/s# and #v_2=28/3m/s#

In a perfectly elastic collision, momentum and mechanical energy are conserved.

As two objects collide, all of the kinetic energy they possess is transferred into elastic potential energy in their molecular bonds, and then all of that energy which is stored in the bonds is transformed back into the post-collision kinetic energy of the object. Although energy is transformed into potential energy during the collision, the mechanical energy before and after the collision is purely kinetic energy.

We are given that #m_1=2kg#, #v_1=7m/s#, #m_2=1kg#, and #v_2=0#.

You can determine the final velocities in terms of the given quantities using these equations(1):

#v_1=((m_1-m_2)/(m_1+m_2))v_(1o)+((2m_2)/(m_1+m_2))v_(2o)#

#v_2=((2m_1)/(m_1+m_2))v_(1o)-((m_1-m_2)/(m_1+m_2))v_(2o)#

Because the second ball is initially at rest (#v_2=0#), these simplify to:

#v_1=((m_1-m_2)/(m_1+m_2))v_(1o)#

#v_2=((2m_1)/(m_1+m_2))v_(1o)#

Therefore, the post-collision velocity of the first ball is:

#v_1=((2kg-1kg)/(2kg+1kg))(7m/s)#

#v_1=7/3m/s#

and the post collision velocity of the second ball is:

#v_2=((2(2kg))/(2kg+1kg))7m/s#

#v_2=28/3m/s#

Julia Aguilar · Answer

undefined To find the post-collision velocities of the balls, we can use the principle of conservation of momentum and the law of conservation of kinetic energy. Let $ v_1 $ and $ v_2 $ represent the velocities of the 2 kg and 1 kg balls respectively after the collision.

By conservation of momentum: 
$$ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_1 + m_2 \cdot v_2 $$

Where:
$ m_1 = 2 $ kg (mass of the first ball)  
$ m_2 = 1 $ kg (mass of the second ball)  
$ v_{1i} = 7 $ m/s (initial velocity of the first ball)  
$ v_{2i} = 0 $ m/s (initial velocity of the second ball)

Substituting the given values, we have: 
$$ 2 \cdot 7 + 1 \cdot 0 = 2 \cdot v_1 + 1 \cdot v_2 $$

By conservation of kinetic energy (since it's an elastic collision):
$$ \frac{1}{2} m_1 \cdot v_{1i}^2 + \frac{1}{2} m_2 \cdot v_{2i}^2 = \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 $$

Substituting the given values, we have: 
$$ \frac{1}{2} \cdot 2 \cdot 7^2 + \frac{1}{2} \cdot 1 \cdot 0^2 = \frac{1}{2} \cdot 2 \cdot v_1^2 + \frac{1}{2} \cdot 1 \cdot v_2^2 $$

Solving these equations simultaneously will give us the post-collision velocities of the balls.