Question

In elastic collisions,total kinetic energy and momentum are both conserved.

On a frictionless track,cart 1 is moving with a constant,rightward(+) velocity of 1.0m/s.Cart 2 is also moving rightward with constant velocity of 5.0m/s.  After a while,cart 2 collides cart 1 from behind.(It is an elactic collision.)If the final velocity of cart 2 becomes 3.0m/s, what is the final velocity of cart 1?

Answer 1

Let m1 and m2 be the mass of the first and second body,.

Before collision we have: m2, m1 with velocity 5m/s , 1 m/s respectively
After collision we have : m2, m1 with velocity 3m/s, v respectively.

Momentum is constant: 5*m2 + 1*m1 = 3*m2 + v*m1 ..........................(1) ,

simplifying we get
2*m2 + m1 = v*m1
(v-1)*m1 = 2*m2

(v-1)/2 = m2/m1

Kinetic enrgy is constant: m2*3²/2 + m1*v²/2 = m2*5²/2 + m1*1²/2.........(2) , simplify

9m2 + v²m1 = 25m2 + m1 ............(2) , simplify
v²*m1 = 16*m2 + m1 , divide by m1 =>
v² = 16*(m2/m1) + 1

v² = 16*(v - 1)/2 + 1
v² = 8v - 7
v² - 8v + 7 = 0
(v - 7)(v - 1) = 0 , (v = 1=> m2/m1 = 0)

v = 7
The final velocity of cart1 is 7 m/s