A ball with a mass of #1 kg # and velocity of #2 m/s# collides with a second ball with a mass of #3 kg# and velocity of #- 4 m/s#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Aug 13, 2018

Answer:

The solution is #=((-1.03, -3.68))ms^-1#

Explanation:

There is conservation of momentum

#m_1u_1+m_2u_2=m_1x+m_2y#

Plugging in the above values

#1*2+3*-4=1x+3y#

#x+3y=2-12=-10#

#x+3y=-10#.................................#(1)#

#20%# of kinetic energy is lost

#(1/2m_1u_1^2+1/2m_2u_2^2)*8/10=1/2m_1x^2+1/2m_2y^2#

Plugging the data

#(1*2^2+3*(-4)^2)*8/10=1x^2+3y^2#

#x^2+3y^2=41.6#

#x^2+3y^2=41.6#......................................#(2)#

Solving equations #(1)# and #(2)# graphically

graph{(x+3y+10)(x^2+3y^2-41.6)=0 [-18.02, 18.03, -9.01, 9.01]}

The solutions are #=(-1.03, -3.68)# or #=(-6.03, -1.32)#

The second solution is discarded