A ball with a mass of #4# #kg# and velocity of #5# #ms^-1# collides with a second ball with a mass of #6# #kg# and velocity of #-1# #ms^-1#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?
2 Answers
Momentum is conserved. Kinetic energy is not, but in this case we know that final
The final velocity of the
Explanation:
Let's call the
Before the collision:
Momentum:
Kinetic energy:
After the collision:
Momentum:
Kinetic energy:
We know that momentum is conserved, so the momentum after the collision will be
20% of the kinetic energy is lost to other forms of energy like heat and sound, so 80% will remain:
That gives us two equations in two unknowns:
Equation 1:
Equation 2:
Rearranging Equation 1 to make
Substituting this into Equation 2:
I'll leave solving this equation for you - it's just algebra. It does involve solving a quadratic equation.
The solution is that
Substituting this back into Equation 1 and solving we find that
To figure out which set of answers we want, let's imagine that left-to-right is the "+" direction and right-to-left is the "-" direction.
Initially, the
The other set of answers doesn't really make physical sense, since it requires the left-hand ball to be traveling to the right faster than the right-hand ball.
Explanation:
We notice that velocity of one ball is given as
Before the collision
Momentum, in
Kinetic energy
#=50+3=53J#
After the collision
Let
Momentum:
and
Since momentum is conserved, we have
#14=4v_(1x)+6v_(2x)# .....(1)
#0=4v_(1y)+6v_(2y)# .....(2)
Kinetic energy:
Given is that
This gives us
#42.4=2(v_(1x)^2+v_(1y)^2)+3(v_(2x)^2+v_(2y)^2)# .....(3)
Thus we have three equations in four unknowns. To find a solution lets assume that even after collision the balls move along the
Dropping the suffix
Substituting value of
Solving the quadratic
From (6) corresponding values of