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

m_(1)v_(1i) + m_(2)v_(2i) = m_(1)v_(1f) + m_(2)v_(2f)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) + mv_(0) = mv_(1) + mv_(2)m(0)+mv0=mv1+mv2

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

v_(0) = v_1 + v_2v0=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

m_(1)v_(1i) + m_(2)v_(2i) = (m_(1)+m_2)v_(f)m1v1i+m2v2i=(m1+m2)vf

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

v_2 = 2v_fv2=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 vv (i.e. the velocities get exchanged.

For a perfectly inelastic collision both carts move with a shared velocity of v/2v2

Explanation:

Momentum conservation leads to

m_1 v_(1i)+m_2 v_(2i) = m_1 v_(1f)+m_2 v_(2f)m1v1i+m2v2i=m1v1f+m2v2f

Since, in this problem m_1 = m_2 = mm1=m2=m, v_(1i) = 0v1i=0 and v_(2i) = vv2i=v, we have

v = v_(1f)+v_(2f)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.

v_(2f)-v_(1f) = v_(1i)-v_(2i) =-vv2fv1f=v1iv2i=v

Thus v_(2f) = 0, v_(2i) = vv2f=0,v2i=v

**Perfectly inelastic collision#

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

v_(1f) = v_(2f) = 1/2 (v_(1f) + v_(2f)) = 1/2 vv1f=v2f=12(v1f+v2f)=12v