A ball with a mass of # 5 kg# is rolling at #1 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

1 Answer
Jun 28, 2018

Answer:

The post collisions velocities are #=011ms^-1# and #=1.411ms^-1#

Explanation:

Since the collision is elastic, we have conservation of momentum

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

And conservation of kinetic energy

#1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2#

The mass the first ball is #m_1=5kg#

The velocity of the first ball before the collision is #u_1=1ms^-1#

The mass of the second ball is #m_2=4kg#

The velocity of the second ball before the collision is #u_2=0ms^-1#

The velocity of the first ball after the collision is #=v_1ms^-1#

The velocity of the second ball after the collision is #=v_2ms^-1#

Therefore,

#5*1+4*0=5*v_1+4*v_2#

#4v_2+5v_1=5#....................................#(1)#

#1/2*5*1^2+1/2*4*0=1/2*5*v_1^2+1/2*4*v_2^2#

#4v_2^2+5v_1^2=5#..............................#(2)#

Solving for #v_1# and #v_2# in equations #(1)# and #(2)#

From equation #(1)#, #v_2=(5(1-v_1))/4#

Substituting in equation #(2)#

#5v_1^2+4*((5(1-v_1))/4)^2=5#

#4v_1^2+5+5v_1^2-10v_1=4#

#9v_1^2-10v_1+1=0#

#v_1=(10+-sqrt(((-10)^2)-4(9)(1)))/(18)=(10+-8)/18#

Therefore,

#v_1=1ms^-1# or #v_1=1/9=0.11ms^-1#

#v_2=0ms^-1# or #v_2=10/9=1.11ms^-1#

The first solutions are the initial conditions.

The post collisions velocities are #=011ms^-1# and #=1.411ms^-1#