Question

Two carts, of inertias m1 and m2, collide head-on on a low-friction track. Before the collision, which is elastic, cart 1 is moving to the right at 14 m/s and cart 2 is at rest. After the collision, cart 1 is moving to the left at 6 m/s . If the positive direction is from left to right, what is the speed of cart 2 after the collision?

Answer 1

Answer;

v2 = sqrt(160 m1/m2)

Solution and explanation;

The collision is elastic, which means the total kinetic energy is conserved. Let's call situation 1 before the collision and situation 2 after the collision.

So, initial kinetic energy 1 = 1/2 m1 v1^2 + 1/2 m2 v2^2

= 1/2 m1 × 14^2

= 98 m1

Final kinetic energy = 1/2 m1 v1^2 + 1/2 m2 v2^2

= 1/2 m1 × 6^2 + 1/2 m2 v2^2

= 18 m1 + 1/2 m2 v2^2

Conservation of kinetic energy: Initial kinetic energy = Final kinetic energy

98 m1 = 18 m1 +1/2 m2 v2^2

1/2 m2 v2^2 = 80 m1

v2^2 = 160 m1/m2

v2 = sqrt(160 m1/m2)

Answer 2

`v_(2f) = 8 m/s`

Step-by-step explanation:

As we know that momentum is always conserved in this type of collision

so we will have

`m_1v_(1i) + m_2v_(2i) = m_1v_(1f) + m_2v_(2f)`

so here we will have

`m_1(14) + 0 = m_1(-6) + m_2v_(2f)`

`20 m_1 = m_2v_(2f)`

also by the coefficient of restitution we know

`v_(2f) - v_(1f) = v_(1i) - v_(2i)`

so we have

`v_(2f) - (-6) = 14 - 0`

`v_(2f) = 8 m/s`