Elastic collision

Apr 2008
28
0
Pangkor Island, Perak, Malaysia.
\(\displaystyle m_1=3kg\)
\(\displaystyle u_1=10ms^{-1}\)
\(\displaystyle m_2=2kg\)
\(\displaystyle u_2=0\)
The collision is elastic.
Find:
\(\displaystyle v_1\) and \(\displaystyle v_2\)

By using total kinetic energy is conserved, I get 2 sets of answer.
\(\displaystyle v_1=10\) , \(\displaystyle v_2=0\)
\(\displaystyle v_1=2\) , \(\displaystyle v_2=12\)

Which is the correct answer?Or both are acceptable?
 
May 2008
29
9
There are two equations you can use:

Conservation of momentum:
\(\displaystyle m_1u_1+m_2u_2 = m_1v_1+m_2v_2\)

Difference in departing speeds = -(Difference in starting speeds)
\(\displaystyle u_1-u_2 = v_2-v_1\)
(Derived from conservation of kinetic energy)

Solving simultaneously, the only solution is \(\displaystyle v_1 = 2\), \(\displaystyle v_2 = 12\)

If you think about it practically, how could it be possible that the speeds are the same afterwards?
 
  • Like
Reactions: SengNee

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
\(\displaystyle m_1=3kg\)
\(\displaystyle u_1=10ms^{-1}\)
\(\displaystyle m_2=2kg\)
\(\displaystyle u_2=0\)

By using total kinetic energy is conserved, I get 2 sets of answer.
\(\displaystyle v_1=10\) , \(\displaystyle v_2=0\)

Which is the correct answer?Or both are acceptable?
DivideBy0 is absolutely correct, but I would like to mention the more general circumstance to the set of answers to any such problem. Typically you will get two "possible" answers, one of which will be unphysical. The problem isn't so much that the initial and final velocities of the particles are the same, the problem is that in order for this to be true particle 1 would have to pass through particle 2! This is typically the kind of thing that happens: one of the solutions has one particle passing through the other which is, of course, impossible.

-Dan
 
  • Like
Reactions: SengNee