A ball with a mass of # 5 kg# is rolling at #8 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 5    kg# is rolling at #8  m/s# and elastically collides with a resting ball with a mass of #4  kg#. What are the post-collision velocities of the balls?

Olivia Askew · Accepted Answer

#v_1^'=8/9 " "m/s#

#v_2^'=80/9" "m/s# #m_1=5" "kg# #v_1=8" "m/s# #P_1=m_1*v_1=5*8=40" " kg*m/s# #m_2=4" "kg# #v_2=0# #P_2=m_2*v_2=4*0=0# #Sigma P_b=P_1+P_2" total momentum"# #Sigma P_b=40+0=40" "kg*m/s # #"post collision velocities :"# #v_1^'=(2*Sigma P_b)/(m_1+m_2)-v_1# #v_1^'=(2*40)/(5+4)-8# #v_1^'=80/9-8# #v_1^'=(80-72)/9# #v_1^'=8/9 " "m/s# #v_2^'=(2*Sigma P_b)/(m_1+m_2)-v_2# #v_2^'=(2*40)/(5+4)-0# #v_2^'=80/9" "m/s#

Axel Acevedo · Answer

undefined To find the post-collision velocities, you can use the conservation of momentum and kinetic energy.

Let $ m_1 = 5 \, \text{kg} $ be the mass of the first ball rolling at $ v_1 = 8 \, \text{m/s} $, and $ m_2 = 4 \, \text{kg} $ be the mass of the second ball initially at rest.

By conservation of momentum:
$$ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 $$
$$ 5(8) + 4(0) = 5(u_1) + 4(u_2) $$
$$ 40 = 5u_1 + 4u_2 $$

By conservation of kinetic energy in an elastic collision:
$$ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 $$
$$ \frac{1}{2}(5)(8)^2 + \frac{1}{2}(4)(0)^2 = \frac{1}{2}(5)(u_1)^2 + \frac{1}{2}(4)(u_2)^2 $$
$$ 160 = \frac{5}{2}(u_1)^2 + \frac{4}{2}(u_2)^2 $$

Solving the system of equations:
$$ 40 = 5u_1 + 4u_2 $$
$$ 160 = \frac{5}{2}(u_1)^2 + \frac{4}{2}(u_2)^2 $$

We find $ u_1 = 2 \, \text{m/s} $ and $ u_2 = 10 \, \text{m/s} $. Therefore, the post-collision velocities of the balls are $ u_1 = 2 \, \text{m/s} $ and $ u_2 = 10 \, \text{m/s} $, respectively.