Could someone help me with this momentum question?

Question

I don't get it. How come I will have two answer for each problem?

A 5 kg body moving in the +X directionat 5.5 m/s collides head-on with a 3.4 kg body moving in the -X direction at 4 m/s. Find the final velocity of each mass for each of the following situations. (Take the positive direction to be +X.)

(a) The bodies stick together.
m/s (5 kg mass)
m/s (3.4 kg mass)


(B) the collision is elastic
m/s (5 kg mass)
m/s (3.4 kg mass)

(c) the 5.0 kg body is at rest after the collision
m/s (5 kg mass)
m/s (3.4 kg mass)

(d) The 3.4 kg body is at rest after the collision.
m/s (5 kg mass)
m/s (3.4 kg mass)

(e) The 5 kg body has a velocity of 4 m/s in the -X direction after the collision.
m/s (5 kg mass)
m/s (3.4 kg mass)

can someone show me an example?

Answer 1

Two answers because generally the two bodies will have different post-collision velocities. An exception is when they stick together as in (a).
As the 1st answer says, we know momentum is conserved, so the total momentum is the same before and after the collision. Before, it's 5 kg * 5.5 m/s - 3.4 kg * 4 m/s. After, the total momentum is shared between the two bodies.
When both bodies stick together as in (a), the velocity of each is the total momentum divided by the total mass. This velocity (call it the "system velocity") is also useful to calculate if you're dealing with an elastic collision.
In an elastic collision (b), the bodies bounce apart with a velocity relative to the system velocity that is the same magnitude but opposite in sign to the relative velocity they had before the collision. Algebraically, for body n, Vafter(n) = Vsystem - (Vbefore(n) - VSystem).
In a collision in which one body has a given velocity and thus a calculatable momentum as in (c) and (d) where it is 0 and in (e) where it is not, the other body (n) has the remainder of the momentum, so divide the remaining momentum by body n's mass to get its velocity. Algebraically, if the body with the defined velocity is body m and the other is body n, Vafter(n) = (TotalMomentum - Mass(m) * Vafter(m)) / Mass(n).
Be careful to keep the correct signs of all velocities and momenta in doing these calculations.

Answer 2

The important thing in all these problems is conservation of momentum. The sum of the masses times velocities before equals the sum of masses times velocities afterward.

You know the masses and velocities before. You know the masses afterward. You just need to find the two velocities after.

So you have two unknowns--you need two equations to find them.

Conservation of momentum gives you one.

Each letter (a-e) gives you the second.

For part a, the two velocities are the same.
For part b, kinetic energy is conserved. The sum of the 1/2 mv^2 's is constant (this is the toughest).

For part c, they give you a velocity.

Etc.

You just have to do some algebra.

Make sense?

Answer 3

Each answer is the velocity of the mass involved. Actually, 'both' velocities taken together are the 'answer' to each question.

And the thing to remember is that momentum is *always* conserved, while kinetic energy may, or may not, be conserved.

HTH ☺

Doug