10/12/09 - Momentum, Explosions and Collisions

			"Explosions"

If we have two carts of equal mass m at rest with a compressed spring between them

when we release the spring, they go flying apart with equal speeds but in opposite directions.

If we repeat the experiment but double the mass of the cart on the right

we find that the cart with the doubled mass moves with half the speed.

This suggests that the product of mass and velocity is a special quantity, and we give it the name momentum. Note that direction is important, meaning that momentum is a vector.

If we look at the initial momentum of each cart before the "explosion", we find that each cart has zero momentum because neither cart is moving (v = 0).

If we look at the initial momentum of each cart after the "explosion", we find that the left cart has momentum -mv and the right cart has momentum mv (mv or (2m)(v/2) cases) and their sum is zero. This suggests that the total momentum before and after the "explosion" is unchanged. This is called the law of conservation of momentum, and what it says is that if you add the momentum vectors of each particle in a system to get the total momentum, the total momentum will not change unless a force from outside the system acts on the system for some amount of time.

This is summarized by the statement:

F represents an external force and ∆t is the length of time the force acts on the system. The quantity F∆t is called the impulse and it tells us how the momentum changes. If the external force F = 0, then m∆v = 0, which means the velocity (and hence the momentum) is unchanged. If the external force F ≠ 0, then the momentum of the system will change by the amount m∆v = F∆t. Note that F∆t = m∆v can be rearranged to F = m∆v/∆t = ma, which is Newton's second law.

			Collisions

			The law of conservation of momentum applies whenever masses in a system interact without external forces. For example, when two masses collide.

			Inelastic collision with equal masses

Assume the incident mass is moving to the right with velocity v and the right mass is at rest. The initial momentum is mv to the right. If the two masses stick together and move off with velocity u to the right, conservation of momentum tells us:

mv = (2m)u

So u = v/2.

If we look at the kinetic energy before the collision:  KEinitial = (1/2)mv2

If we look at the kinetic energy after the collision, we get:  KEfinal = (1/2)(2m)(v/2)2 = (1/4)mv2

So the change in kinetic energy is ∆KE = KEfinal - KEinitial = (1/4)mv2 - (1/2)mv2 = -(1/4)mv2, which means we lost half the kinetic energy in the collision.  When kinetic energy is lost in a collision, the collision is said to be "inelastic".

			Elastic collision with equal masses

Assume the incident mass is moving with velocity v to the right, and the right mass is at rest.

This time assume that after the collision, the incident mass moves to the right with velocity u1 and the rest mass moves to the right with velocity u2:

To conserve momentum we must have:

mv = mu1 + mu2

To conserve kinetic energy we must have:

(1/2)mv2 = (1/2)mu12 + (1/2)mu22

From the momentum equation, we can divide out the m's and get:

v = u1 + u2

Substituting this result for v into the left side of the kinetic energy equation, we get:

(1/2)m(u1 + u2)2 = (1/2)m(u12 + 2u1u2 + u22) = (1/2)mu12 + (1/2)mu22

Simplification results in mu1u2 = 0.

This tells us that either u1 = 0, in which case u2 = v (that is, the incident cart stops, and the cart initially at rest moves off with the same velocity the incident cart had initially):

or u2 = 0 and u1 = v, which means that the incident cart never hit the rest cart.

			Elastic collision with incident mass larger than rest mass

If we assume that, after the collision, the mass initially at rest moves to the right with velocity u2 and the incident mass continues to the right with velocity u1:

Conservation of momentum tells us that:

Conservation of kinetic energy tells us that:

Solving the momentum equation for u2 gives us:

Substituting this into the kinetic equation for u2 give us:

After much simplification we get the following quadratic equation:

which we can solve with the quadratic formula to get:

which simplifies to

Substituting this into the previous equation for u2:

To see what this means, assume M = 5m. Then

So the incident mass slows down to 2/3 its original velocity and the mass initially at rest goes off with 5/3 the original velocity of the incident mass.

			Elastic collision with incident mass smaller than rest mass

If we assume that, after the collision, the rest mass moves to the right with velocity u2 and the incident mass continues to the right with velocity u1: