A ball with a mass of #2 kg # and velocity of #8 m/s# collides with a second ball with a mass of #4 kg# and velocity of #- 1 m/s#. If #15%# of the kinetic energy is lost, what are the final velocities of the balls?
1 Answer
Explanation:
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(blue)(m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f))#
We are given the following information:
#|->"m"_1=2"kg"# #|->"v"_(1i)=8"m"//"s"# #|->"m"_2=4"kg"# #|->"v"_(2i)=-1"m"//"s"# #"kinetic energy lost:" 15%#
We can begin by calculating the momentum before the collision:
#P_i=m_1v_(1i)+m_2v_(2i)#
#=(2kg)(8m/s)+(4kg)(-1m/s)#
#=color(blue)(12" kgm"//"s")#
Note that momentum is a vector quantity and the positive value indicates direction. Let's set to the right is the positive direction.
By momentum conservation, the total momentum after the collision should also be
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((2"kg")(8" m"//"s")^2+(4"kg")(-1" m"//"s")^2)#
#K_i=1/2(132("kgm"^2)/"s"^2)#
#K_i=66("kgm"^2)/"s"^2#
#=color(darkblue)(66"J")#
We are given that
#15%=0.15#
#=>0.15*66"J"#
#=9.9"J"#
#=>66"J"-9.9"J"=56.1"J"#
This tells us that:
#color(darkblue)(K_f=1/2m_1(v_(1f))^2+1/2m_2(v_(2f))^2=56.1"J")#
We now have two equations expressing final momentum and kinetic energy:
#color(darkblue)(m_1v_(1f)+m_2v_(f2)=12" kgm"//"s")#
#color(darkblue)(1/2(m_1(v_(1f))^2+m_2(v_(2f))^2)=112.2"J")#
As the masses of the objects do not change, we can fill them in and simplify:
#color(darkblue)(2v_(1f)+4v_(f2)=12" kgm"//"s")#
#color(darkblue)(2(v_(1f))^2+4(v_(2f))^2=112.2)#
We now have two equations and two unknowns. We can now manipulate one of these equations to solve for either
#2v_(1f)+4v_(2f)=12#
#=>2v_(1f)=12-4v_(2f)#
#=>color(darkblue)(v_(1f)=1/2(12-4v_(2f)))#
Substituting into the second equation, we now have everything in terms of
#2(1/2(12-4v_(2f)))^2+4(v_(2f))^2=112.2#
Simplifying:
#1/2(144+16(v_(2f))^2-96v_(2f))+4(v_(2f))^2=112.2#
Multiplying through by a half to eliminate the fraction:
#144+16(v_(2f))^2-96v_(2f)+8(v_(2f))^2=224.4#
#color(blue)(24(v_(2f))^2-96v_(2f)-80.4=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(24)(-80.4)))/(2*24)#
#=>color(blue)((96+-sqrt(16934.4))/(48))#
We therefore have two answers for
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
Only one of the calculated velocities will work, and so
We can use the above answer to solve for the final velocity of the first ball,
#2v_(1f)+4v_(f2)=12#
#=>2v_(1f)+4(4.71)=12#
#=>2v_(1f)=-6.84#
#=>color(crimson)(v_(1f)=-3.42"m"//"s")#
Therefore:
#color(darkblue)(v_(1f)=-3.42"m"//"s")# #color(darkblue)(v_(2f)=4.71"m"//s")#
Note that these answers can be checked with momentum conservation. Simply calculate the final momentum using the velocities determined and confirm it is
