A ball with a mass of #4 kg # and velocity of #3 m/s# collides with a second ball with a mass of #5 kg# and velocity of #- 2 m/s#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Mar 3, 2018

Ball 1: #\vec(v_(1_f)) = -2.11# m/s
Ball 2: #\vec(v_(2_f)) = 2.21# m/s

Explanation:

Collisions can be represented using momentum relations.

Momentum is defined as: #\vecp=m\vecv#

Using the values provided, we see the initial momenta are:
#\vec(p_(1_i)) = (4)(3) = 12# kg#*#m/s
#\vec(p_(2_i)) = (5)(-2) = -10# kg#*#m/s

The total net momentum of the system initially is:
#\vec(p_(Total_i)) = \vec(p_(1_i))+\vec(p_(2_i)) = 12 -10=2# kg#*#m/s

Since momentum is conserved, we know the magnitude of the momentum after the collision is also #2# kg#*#m/s.

Energy (kinetic) is defined as: #E = 1/2 m v^2#

Using the values provided, we see the initial energies are:
#E_(1_i) = 1/2(4)(3)^2 = 18# J
#E_(2_i) = 1/2 (5)(-2)^2 = 10# J

The total energy of the system initially is:
#E_(Total_i) = E_1 + E_2 = 18+10=28# J

Since we are told 20% of the energy is lost (80% remains), the energy is NOT conserved after the collision. The energy after the collision is:
#E_(Total_f) = 0.8E_(Total_i) = (0.8)(28) = 22.4# J

Now we can form some equations describing the post-collision system, which we can use to find the velocities of the balls.

They are magnitude of total momentum:
#4v_(1_f) + 5v_(2_f) = 2# kg#*#m/s

and total energy:
#2v_(1_f)^2 + 5/2 v_(2_f)^2 = 22.4# J

We can now solve this system of equations
(1) #4v_(1_f) + 5v_(2_f) = 2#
(2) #2v_(1_f)^2 + 2.5 v_(2_f)^2 = 22.4#

We can divide equation (1) by 4 and solve for #v_(1_f)#:
(3) #v_(1_f) = 0.5 - 1.25v_(2_f)#

Substituting (3) into (2) gives:
#2(0.5 - 1.25v_(2_f))^2 + 2.5 v_(2_f)^2 = 22.4#
#2(0.25-1.25v_(2_f)+1.5625v_(2_f)^2) + 2.5 v_(2_f)^2 = 22.4#
#0.5-2.5v_(2_f)+3.125v_(2_f)^2 + 2.5 v_(2_f)^2 = 22.4#
#5.625v_(2_f)^2-2.5v_(2_f)-21.9 = 0#

This quadratic can now be solved using the quadratic equation:
#v_(2_f)=(-b+-sqrt(b^2-4ac))/(2a)#
where #a=5.625#, #b=-2.5#, and #c=-21.9#

This gives: #v_(2_f) = 2.21# m/s and #v_(2_f) = -1.76# m/s

Since ball 2 originally travels in the negative direction, after the collision it will go in the positive direction. So we accept the root that is positive. Hence:
#v_(2_f) = 2.21# m/s

Now we can use this value in equation (1) to find the value of #v_(1_f)#:
#4v_(1_f) + 5(2.21) = 2#
#v_(1_f) = -2.11# m/s

It is good that this value is negative, as ball 1 was originally traveling in the positive direction. Post-collision, it should go the opposite direction.

Hence we have found the final velocities of the balls:

Ball 1: #\vec(v_(1_f)) = -2.11# m/s
Ball 2: #\vec(v_(2_f)) = 2.21# m/s