A ball with a mass of #3 kg # and velocity of #1 m/s# collides with a second ball with a mass of #4 kg# and velocity of #- 2 m/s#. If #80%# 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 #1  m/s# collides with a second ball with a mass of #4 kg# and velocity of #-  2 m/s#. If #80%# of the kinetic energy is lost, what are the final velocities of the balls?

Lily Adams · Accepted Answer

Ball 1 (the 3 kg ball) reverses direction, moving at -0.92 m/s, and ball 2 (the 4 kg ball) still moving to the left at -0.56 m/s.

This is a lengthy problem that asks us to solve for two unknowns (the two final velocities). To do this, we must generate two equations involving these two unknowns, and solve them simultaneously. One equation will come from conservation of momentum, the other will come from the 80% kinetic energy lost condition. First, cons. of momentum: #m_1v_(1i) + m_2v_(2i) = m_1v_(1f) + m_2v_(2f)# Inserting the values we know: #3(1) + 4(-2) = 3v_(1f) + 4v_(2f)# #-5 = 3v_(1f) + 4v_(2f)# Now, the kinetic energy condition. The initial KE is: #1/2m_1v_(1i)^2 + 1/2m_2v_(2i)^2# #1/2(3)(1)^2 + 1/2(4)(-2)^2 = 1.5 + 8 = 9.5 J# The final KE is: #1/2m_1v_(1f)^2 + 1/2m_2v_(2f)^2# Since 80% of the KE is lost, the final KE equals 20% of the initial #1/2(3)v_(1f)^2 + 1/2(4)v_(2f)^2 = 9.5 xx 0.2=1.9J# With all that complete, our two equations are: #-5 = 3v_(1f) + 4v_(2f)# #3/2v_(1f)^2 + 4/2v_(2f)^2 = 1.9# Rewrite the first equation as #v_(1f)=(-5-4v_(2f))/3# Substitute this value into the second equation: #3/2((-5-4v_(2f))/3)^2 + 4/2v_(2f)^2 = 1.9# Now, we must solve this equation. It will be simpler if we multiply every term by 18, to eliminate the denominators: #3(-5-4v_(2f))^2 + 36v_(2f)^2 = 34.2# #3(25+40v_(2f)+16v_(2f)^2) + 36v_(2f)^2 = 34.2# #75+120v_(2f)+48v_(2f)^2 + 36v_(2f)^2 = 34.2# #84v_(2f)^2+120v_(2f)+40.8=0# Use the quadratic formula to solve for #v_(2f)# #v_(2f)= (-120+- sqrt((120)^2 - 4(84)(40.8)))/(2(84))# #= (-120+-26.3)/168# There are two answers: #v_(2f)= (-120+26.3)/168=-0.56m/s# and #v_(2f)= (-120-26.3)/168=--0.87m/s# Substituting this answer back into #v_(1f)=(-5-4v_(2f))/3# we get #v_(1f)=(-5-4(-0.56))/3=-0.92m/s# and #v_(1f)=(-5-4(-0.87))/3=-0.51m/s# We must reject the second answer in each case, as this would result in ball 2 (which was originally travelling to the left and was in a position to the right of ball 1) now travelling faster than ball 1, with both balls moving to the left. This would mean that it had somehow managed to go through ball 1, and was now to the left of it! So, the only acceptable answers are #v_(1f)=-0.92m/s# #v_(2f)=-0.56m/s#

Adam Arnold · Answer

undefined The final velocities of the balls are 0.2 m/s for the first ball and -0.6 m/s for the second ball.