A ball with a mass of #4 kg # and velocity of #9 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 5 m/s#. If #80%# of the kinetic energy is lost, what are the final velocities of the balls?

1 Answer
Apr 29, 2018

This is a relatively complicated collision problem. You won't see many of these unless you're majoring in physics or are an engineer.

By definition, energy is not conserved in this collision, due to the question's statement about kinetic energy being lost.

Therefore, the collision is inelastic and the conservation of momentum is observed, still.

However, given your data we can make a handy approximation,

#1/2m_"A"nu_"A"^2+1/2m_"B"nu_"B"^2 = 1/2m_"A"nu_"A"^('2)+1/2m_"B"nu_"B"^('2)+Q#

Where,

#Q = 0.80"KE"#

Let's start by finding the initial kinetic energy, and recall,

#"KE" = 1/2mnu^2#

#Sigma"KE" approx 250"J"#

So,

#Q = 0.80"KE" approx 200"J"#, and by extension

#"KE"^' = "KE" - Q = 50"J"#

Intuitively, the larger ball will move slower than the faster ball after the collision, even if it is inelastic.

We have two variables, so we need two equations!

#1/2m_"A"nu_"A"^('2)+1/2m_"B"nu_"B"^('2) = 50"J" " "(1)#

Now, remember what we said about conservation of momentum, the momentum after the collision will be the same as before, regardless of the energy lost.

#m_"A"nu_"A" + m_"B"nu_"B" = 1"N"*"m" = m_"A"nu_"A"^' + m_"B"nu_"B"^'#

#therefore m_"A"nu_"A"^' + m_"B"nu_"B"^' = 1"N"*"m" " " (2)#

From here on we will use some substitution by defining #nu_"B"^'# using #(2)#,

#nu_"B" = (1"N" * "m" - 4"kg" * nu_"A")/(7"kg")#

Now, let's solve for #nu_"A"#, and from here on it's smooth sailing,

#1/2 * 4"kg" * nu_"A"^('2) + 1/2 * 7"kg" * ((1"N" * "m" - 4"kg" * nu_"A")/(7"kg"))^2 =50"J"#

#therefore nu_"A"^' approx (4.08"m")/"s"#, and by extension,

#m_"A"nu_"A"^' + m_"B"nu_"B"^' = 1"N"*"m"#

#therefore nu_"B"^' approx (-2.19"m")/"s"#

Thanks to this answer for providing a push in the right direction for me!