A ball with a mass of #2 kg# is rolling at #3 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 #3   m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Charles Aeschliman · Accepted Answer

#v_1^'=1" m/s"#
#v_2^'=4 " "m/s# #Sigma vec P_b=m_1*vec v_1+m_2*vec v_2# #Sigma vec P_b=2*3+1*0# #Sigma vec P_b=6 kg*m/s" Total momentum before collision"# #v_1^'=(2*Sigma P_b)/(m_1+m_2)-v_1# #v_1^'=(2*6)/(2+1)-3# #v_1^'=12/3-3# #v_1^'=4-3=1 " "m/s " velocity of "m_1 " after collision"# #v_2^'=(2*Sigma P_b)/(m_1+m_2)-v_2# #v_2^'=(2*6)/(2+1)-0# #v_2^'=4 " "m/s" velocity of " m_2 " after collision"#

Layla Ahmed · Answer

undefined To find the post-collision velocities of the balls, we can use the principle of conservation of momentum and kinetic energy. Let $ v_1 $ and $ v_2 $ be the velocities of the 2 kg and 1 kg balls respectively after the collision. We have the following equations:

1. Conservation of momentum: 
   $$ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 $$
2. Conservation of kinetic energy (since it's an elastic collision): 
   $$ \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 $$

Substituting the given values:
$$ 2 \times 3 + 1 \times 0 = 2 \times u_1 + 1 \times 0 $$
$$ \frac{1}{2} \times 2 \times 3^2 + \frac{1}{2} \times 1 \times 0^2 = \frac{1}{2} \times 2 \times v_1^2 + \frac{1}{2} \times 1 \times v_2^2 $$

Solving these equations simultaneously, we find:
$$ u_1 = 1 \, \text{m/s} $$
$$ v_1 = 2 \, \text{m/s} $$
$$ v_2 = 4 \, \text{m/s} $$