A ball with a mass of #3 kg # and velocity of #5 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 2 m/s#. If #25%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

A ball with a mass of #3 kg # and velocity of #5  m/s# collides with a second ball with a mass of #7 kg# and velocity of #-  2 m/s#. If #25%# of the kinetic energy is lost, what are the final velocities of the balls?

William Audy · Accepted Answer

The final velocities are #=-2.78ms^-1# and #=1.33ms^-1#

We have conservation of momentum

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The kinetic energy is

#k(1/2m_1u_1^2+1/2m_2u_2^2)=1/2m_1v_1^2+1/2m_2v_2^2#

Therefore,

#3xx5+7xx(-2)=3v_1+7v_2#

#3v_1+7v_2=1#

#7v_2=1-3v_1#

#v_2=((1-3v_1))/7#........................#(1)#

and

#0.75(1/2xx3xx5^2+1/2xx7xx(-2)^2)=1/2xx3xxv_1^2+1/2xx7xxv_2^2#

#3v_1^2+7v_2^2=77.25#...................#(2)#

Solving for #v_1# and #v_2# in equation s #(1)# and #(2)#

#3v_1^2+7(((1-3v_1))/7)^2=77.25#

#21v_1^2+63v_1^2-42v_1+7-540.75=0#

#84v_1^2-42v_1-533.75=0#

Solving this quadratic equation in #v_1#

#v_1=(-42+-sqrt(42^2-4xx84xx(-533.75)))/(2*84)#

#v_1=(-42+-sqrt(181104))/(168)#

#v_1=(-42+-425.56)/(168)#

#v_1=-2.78ms^-1# or #v_1=2.28ms^-1#

#v_2=1.33ms^-1# or #v_2=-0.83ms^-1#

Nicholas Agrusa · Answer

undefined To find the final velocities of the balls, we can use the principle of conservation of momentum and the equation for kinetic energy. First, calculate the total momentum before the collision. Then, use the given information about the percentage of kinetic energy lost to find the total kinetic energy after the collision. Finally, use the equations involving momentum and kinetic energy to solve for the final velocities of the balls.