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?

Question

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?

Lily Adams · Accepted Answer

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

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

Olivia Askew · Answer

undefined To find the final velocities of the balls after the collision, we can use the principle of conservation of momentum and the equation for kinetic energy.

1. Conservation of momentum: $m_1v_1 + m_2v_2 = m_1v_{1f} + m_2v_{2f}$

2. Equation for kinetic energy: $KE = \frac{1}{2}mv^2$

Using these equations, we can solve for the final velocities $v_{1f}$ and $v_{2f}$:

For the first ball: 
Initial momentum: $4 \times 3 + 5 \times (-2) = 12 - 10 = 2$ kg m/s
Initial kinetic energy: $\frac{1}{2} \times 4 \times 3^2 = 18$ J
Final kinetic energy: $0.8 \times 18 = 14.4$ J
Final velocity: $\sqrt{\frac{2 \times 14.4}{4}} = 1.2$ m/s

For the second ball:
Initial momentum: $4 \times 3 + 5 \times (-2) = 12 - 10 = 2$ kg m/s
Initial kinetic energy: $\frac{1}{2} \times 5 \times (-2)^2 = 10$ J
Final kinetic energy: $0.8 \times 10 = 8$ J
Final velocity: $\sqrt{\frac{2 \times 8}{5}} \approx 0.8$ m/s

So, the final velocities of the balls are approximately $1.2 \, \text{m/s}$ and $0.8 \, \text{m/s}$ respectively.