A ball with a mass of #7 kg # and velocity of #2 m/s# collides with a second ball with a mass of #6 kg# and velocity of #- 3 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

A ball with a mass of #7 kg # and velocity of #2  m/s# collides with a second ball with a mass of #6 kg# and velocity of #-  3 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Leo Abeyta · Accepted Answer

#v_(1f)=-1.435" m"//"s"#

# v_(2f)=1.007" m"//s"# There may be a simpler way to solve this, but here's what I came up with: Momentum is conserved in all collisions. In an inelastic collision, momentum is conserved as always, but energy is not; part of the kinetic energy is transformed into some other form of energy. Therefore, we have an inelastic collision. #vecp=mvecv# The equation for momentum. We can use momentum conversation and kinetic energy to find the final velocities of the balls. Momentum conservation: #DeltavecP=0# #=>vecp_f=vecp_i# For multiple objects, we use superposition as with forces: #vecP=vecp_(t o t)=sumvecp=vecp_1+vecp_2+...+vecp_n# So we have: #=>color(darkblue)(m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f))# We are given the following information: We can begin by calculating the momentum before the collision: #P_i=m_1v_(1i)+m_2v_(2i)# #=(7kg)(2m/s)+(6kg)(-3m/s)# #=color(blue)(-4" kgm"//"s")# Note that momentum is a vector quantity and the negative sign indicates direction. Let's set left is the negative direction. Therefore, by momentum conservation, the total momentum after the collision should also be #4"Ns"# (directed toward the left) #=>color(blue)(P_f=m_1v_(1f)+m_2v_(f2)=-4" kgm"//"s")# We can also calculate the initial kinetic energy before the collision: #K=1/2mv^2# #K_i=1/2m_1(v_(1i))^2+1/2m_2(v_(2i))^2# #K_i=1/2((7"kg")(2" m"//"s")^2+(6"kg")(-3" m"//"s")^2)# #K_i=1/2(82("kgm"^2)/"s"^2)# #K_i=41("kgm"^2)/"s"^2# #=color(darkblue)(41"J")# We are given that #75%# of the kinetic energy is lost in the collision, so we can calculate what the final kinetic energy of the system should be. #75%=0.75# #=>0.75*41"J"# #=30.75"J"# #:.30.75# joules of kinetic energy are lost during the collision. #=>41"J"-30.75"J"=10.25"J"# #:.10.25# joules of kinetic energy remain after the collision. This tells us that: #color(darkblue)(K_f=1/2m_1(v_(1f))^2+1/2m_2(v_(2f))^2=10.25"J")# We now have two equations expressing final momentum and kinetic energy: #color(blue)(m_1v_(1f)+m_2v_(f2)=-4" kgm"//"s")# #color(blue)(1/2(m_1(v_(1f))^2+m_2(v_(2f))^2)=10.25"J")# As the masses of the objects do not change, we can fill them in and simplify: #color(blue)(7v_(1f)+6v_(f2)=-4" kgm"//"s")# #color(blue)(7(v_(1f))^2+6(v_(2f))^2=20.5)# We now have two equations and two unknowns. We can now manipulate one of these equations to solve for either #v_(1f)# or #v_(2f)# and substitute this into the second equation. Neither looks particularly friendly, so I will solve the first equation for #v_(1f)#. #7v_(1f)+6v_(2f)=-4# #=>7v_(1f)=-4-6v_(2f)# #=>color(darkblue)(v_(1f)=1/7(-4-6v_(2f)))# Substituting into the second equation, we now have everything in terms of #v_(2f)#: #7(1/7(-4-6v_(2f)))^2+6(v_(2f))^2=20.5# Simplifying: #1/7(16+36(v_(2f))^2+48v_(2f))+6(v_(2f))^2=20.5# Im going to multiply both sides by 7 to get rid of the fractions: #(16+36(v_(2f))^2+48v_(2f))+42(v_(2f))^2=143.5# #=>78(v_(2f))^2+48v_(2f)+16=143.5# Multiplying by two to get rid of the decimal on the right: #=>156(v_(2f))^2+96v_(2f)+32=287# #=>color(darkblue)(156(v_(2f))^2+96v_(2f)-255=0)# We have a quadratic equation which we can solve using the quadratic formula: For an equation of the form #ax+by+c=0# #x=(-b+-sqrt(b^2-4ac))/(2a)# Therefore: #v_(2f)=(-96+-sqrt(96^2-4(156)(-255)))/(2*156)# #=>(-96+-sqrt(9216+159120))/(312)# #=>color(darkblue)((-96+-sqrt(168336))/(312))# We therefore have two possible answers for #v_(f2)# and will also have two possible answers for #v_(f1)#. However both answers may not be possible. Since this is an elastic collision and we expect the balls to move off in the opposite direction that they approached each other from, we expect a negative velocity for #"m"_1# and a positive velocity for #"m"_2#. Only one of the calculated velocities will work, and #v_(2f)=1.007"m"//s"# We can use the above answer to solve for the final velocity of the first ball, #v_(1f)#: #7v_(1f)+6v_(f2)=-4# #=>7v_(1f)+6(1.007)=-4# #=>7v_(1f)=-10.042# #=>color(indigo)(v_(1f)~~-1.435"m"//"s")# Therefore: Note that these answers can be checked with momentum conservation. Simply calculate the final momentum using the velocities determined and confirm that it is #-4" kgm"//"s"#.

Nora Arzate · Answer

undefined To find the final velocities of the balls after the collision, we can use the principle of conservation of momentum and the fact that 75% of the kinetic energy is lost during the collision.

According to the conservation of momentum:
Initial momentum = Final momentum

The initial momentum of the system is the sum of the momenta of the two balls before the collision, and the final momentum is the sum of the momenta of the two balls after the collision.

Initial momentum:
$m_1 \times v_{1i} + m_2 \times v_{2i}$

Final momentum:
$m_1 \times v_{1f} + m_2 \times v_{2f}$

Given that 75% of the kinetic energy is lost, the kinetic energy after the collision is 25% of the initial kinetic energy.

We know that kinetic energy is given by:
$KE = \frac{1}{2}mv^2$

Substituting the given values, we can find the initial kinetic energy of the system. Then, we can calculate the final kinetic energy by multiplying the initial kinetic energy by 0.25.

Using the fact that kinetic energy is also equal to the change in kinetic energy during the collision:
$KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}mv^2_{\text{initial}} - \frac{1}{2}mv^2_{\text{final}}$

After solving for the final velocities using the conservation of momentum equation and the equation for kinetic energy loss, we can find the final velocities of the balls.

Substituting the given values and solving the equations will give the final velocities of the balls.