If one cart was at rest, and was struck by another cart of equal mass, what would the final velocities be for a perfectly elastic collision? For a perfectly inelastic collision?

2 Answers
Apr 4, 2018

For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart.

For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.

Explanation:

For an elastic collision, we use the formula

m1v1i+m2v2i=m1v1f+m2v2f

In this scenario, momentum in conserved between the two objects.
In the case where both objects have equal mass, our equation becomes

m(0)+mv0=mv1+mv2

We can cancel out m on both sides of the equation to find

v0=v1+v2

For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart.

For inelastic collisions, we use the formula

m1v1i+m2v2i=(m1+m2)vf

By distributing out the vf, and then cancelling out m, we find

v2=2vf

This shows us that the final velocity of the two cart system is 1/2 the velocity of the initial moving cart.

Apr 4, 2018

For a perfectly elastic collision, the cart that was initially moving comes to a halt, while the other cart moves with velocity v (i.e. the velocities get exchanged.

For a perfectly inelastic collision both carts move with a shared velocity of v2

Explanation:

Momentum conservation leads to

m1v1i+m2v2i=m1v1f+m2v2f

Since, in this problem m1=m2=m, v1i=0 and v2i=v, we have

v=v1f+v2f

This holds for both elastic and inelastic collision.

Perfectly elastic collision

In a perfectly elastic collision, the relative velocity of separation is the same as that of approach (with a negative sign)
So.

v2fv1f=v1iv2i=v

Thus v2f=0,v2i=v

**Perfectly inelastic collision#

For a perfectly inelastic collision, the two bodies stick together, so that

v1f=v2f=12(v1f+v2f)=12v