A ball with a mass of #1 kg # and velocity of #7 m/s# collides with a second ball with a mass of #6 kg# and velocity of #- 4 m/s#. If #40%# of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Feb 21, 2018

#v_1 = -4.62 m/s and v_2 = -2.06 m/s#

Explanation:

KE initial= #1/2m_1*u_1^2+1/2m_2*u_2^2#
KE initial= #1/2(1kg)(7m/s)^2+1/2(6kg)(-4m/s)^2#
KE initial= 72.5 J

If 60% KE is conserved then 72.5*.6= 43.5 J is KE final

Another consideration: momentum, p, must be conserved.

Momentum before = momentum after

#p = m_1*u_1 + m_2*u_2 = m_1*v_1 + m_2*v_2#

#(1kg)(7m/s)+(6kg)(-4m/s) = 1 kg*v_1 + 6 kg*v_2#

#7 (kg*m)/s - 24 (kg*m)/s = 1 kg*v_1 + 6 kg*v_2#

#-17 (kg*m)/s = 1 kg*v_1 + 6 kg*v_2#

Solving for #v_1#,

#v_1 = (-17 (kg*m)/s - 6 kg*v_2)/(1 kg) = -17 m/s - 6*v_2#

Going back to the kinetic energy

#1/2m_1v_1^2+1/2m_2v_2^2= # KE final which, from above, is 43.5 J.

#1/2(1kg)v_1^2+1/2(6kg)v_2^2= 43.5 J#

Plugging in the expression for #v_1# from the momentum analysis,

#1/2(1kg)(-17 m/s - 6*v_2)^2+1/2(6kg)v_2^2= 43.5 J#

#1/2(1kg)(17^2 m^2/s^2 + 2*17 m/s*6*v_2 + 6^2*v_2^2)+1/2(6kg)v_2^2= 43.5 J#

#1/2(1kg)(289 m^2/s^2 + 204 m/s*v_2 + 36*v_2^2)+1/2(6kg)v_2^2= 43.5 J#

Now I will multiply thru by 2, multiply thru the first set of parentheses by that 1 kg, and substitute #J# for the combination of units #(kg*m^2)/s^2#,

#289 J + 204 (kg*m)/s*v_2 + 36 kg*v_2^2+6kg*v_2^2= 87 J#

Now, notice the units in that equation. It is clear that all terms are energy and that #v_2# will be in units of m/s when the quadratic equation is solved. Therefore I will drop the units for simplicity.

#289 + 204 *v_2 + 36 *v_2^2+6kg*v_2^2= 87#

# 42 *v_2^2 + 204 *v_2+ 202 = 0#

My working of the quadratic equation yields 2 values and neither is obviously invalid.

#v_1 = -4.62 m/s, -0.24 m/s#

Let's see what value for #v_2# each gives. Using the last equation from the momentum study again,

#-4.62 m/s = -17 m/s - 6*v_2#

# - 6*v_2 = -4.62 m/s + 17 m/s#

# v_2 = (-4.62 m/s + 17 m/s)/-6 = -2.06 m/s#

Repeating with the other result from the quadratic equation work,

#-0.24 m/s = -17 m/s - 6*v_2#

# - 6*v_2 = -0.24 m/s + 17 m/s#

# v_2 = (-0.24 m/s + 17 m/s)/-6 = -2.49 m/s#

I expected a clear-cut way to eliminate one of those results. I do notice that both balls are going the direction that the second ball was going before the collision. Therefore the solution that has #v_1 < v_2# (smaller one running away from the big one) looks better.

I will post this now but review my work and ask for a double check.

I hope this helps,
Steve