A ball with a mass of #3 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?

1 Answer
Jan 2, 2016

Answer:

Equations of conservation of energy and momentum.

#u_1'=1.5m/s#

#u_2'=4.5m/s#

Explanation:

As wikipedia suggests:

#u_1'=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2=#

#=(3-1)/(3+1)*3+(2*1)/(3+1)*0=#

#=2/4*3=1.5m/s#

#u_2'=(m_2-m_1)/(m_1+m_2)*u_2+(2m_1)/(m_1+m_2)*u_1=#

#=(1-3)/(3+1)*0+(2*3)/(3+1)*3=#

#=-2/4*0+6/4*3=4.5m/s#

[Equations' source]

Derivation

Conservation of momentum and energy state:

Momentum

#P_1+P_2=P_1'+P_2'#

Since momentum is equal to #P=m*u#

#m_1*u_1+m_2*u_2=m_1*u_1'+m_2*u_2'# - - - #(1)#

Energy

#E_1+E_2=E_1'+E_2'#

Since kinetic energy is equal to #E=1/2*m*u^2#

#1/2*m_1*u_1^2+1/2*m_2*u_2^2=1/2*m_1*u_1^2'+1/2*m_2*u_2^2'# - - - #(2)#

You can use #(1)# and #(2)# to prove the equations mentioned above. (I tried but kept getting two solutions, which is not right)