# Momentum and collision physics Question

1. Dec 9, 2004

### Soaring Crane

1. Two objects of mass 0.2 kg and 0.1 kg, respectively move parallel to the x-axis. The 0.2 kg object overtakes and collides with the 0.1 kg object. Immediately after the collision, the y-component of the velocity of the 0.1 kg is 1 m/s upward. What is the y-component of the velociy of the 0.1 kg object immediately after the collision?

a. 2 m/s
b. .5 m/s downward
c. 0 m/s
d. .5 m/s upward
e. 2 m/s upward

0.2 kg O-------->
---------0.1 kg 0---->

This is probably an easy one; the answer is not 0 m/s. Can anyone explain to me how to go about getting the answer and why 0 m/s is not the answer?

2. 2 people of unequal mass are initially standing still on ice with negligible friction. They then simultaneously push each other horizontally. Afterward, which is true?

a. The kinetic energies of the 2 people are =.
b. Speeds of 2 people are =.
c. Momenta of 2 people are of = magnitude.
d. Center of mass of 2-person system moves in the direction of less massive person.
e. Less massive person has a smaller initial acceleration than more massive person.

A and B are incorrect, but I have no idea why. Could anyone explain to me why the correct answer answer is correct and the incorrect ones are incorrect?

Thanks for clearing these up.

Last edited: Dec 9, 2004
2. Dec 10, 2004

### cepheid

Staff Emeritus
In part 1...the answer is *not* 0 m/s because (in the absence of an external force) momentum has to be conserved for the system of two masses. It is very important that you understand this point. Initially (before the collision), there was no momentum in the y-direction. So after the collision, there should still be no momentum in the y-direction. (Remember, momentum is a vector quantity, so if it changes direction, then it has changed, in violation of the law of cons. of momentum). Yet one of the objects now has a vertically upward component of velocity! Is this possible? Only if the vertical momentum of the first object is equal and opposite to the vertical momentum of the second...that way the vector sum of the two vertical momenta will be zero, and so will the overall momentum of the system in the vertical direction. Knowing that this must be so, you can deduce the required downward velocity of the second object. (Remember that the two objects do not have the same mass, so just because the vertical components of their momenta are equal and opposite does not mean that the vertical components of their velocities are).

3. Dec 10, 2004

### cepheid

Staff Emeritus
For the second problem...If we know our physics...we should arrive at the right answer without looking at the choices. Still, it never hurts to look at the answers and eliminate any that are definitely wrong.

I'm sure you figured out that e is wrong. The lighter guy accelerates less?! No way. That violates Newton's 2nd Law. The lighter guy has a greater acceleration, given that the two guys are subjected to the same equal and opposite force (by Newton's 3rd law). That statement is the key to solving this problem.

I'm not sure what's going on with d...I'm too lazy to work it out. :grin: I have reason to believe it's not the right answer.

Leaves us with a, b and c. So you want to know why a and b are incorrect, and c is the right answer? Well, b cannot be true. We established in eliminating e that the lighter guy has a larger acceleration. But the force causing it acts for the same amount of time on both guys. So both guys accelerate for the same duration, one guy with a larger acceleration than the other...obviously at the end of that interval of acceleration, the guy with the larger acceleration will end up with a higher final velocity than the other guy.

c has to be true. From Newton's third, the forces are equal. For convenience, let's assume that the one guy is 'k' times more massive than the other guy (k>1). Then his acceleration is a = F/km ...the lighter guy's acceleration is a = F/m. So the lighter guy has an accleration 'k' times greater. Since the accleration is constant, that means that in the same time interval, the lighter guy achieves a final velocity of k times the final velocity of the heavy guy. So what are their two final momenta?

heavy guy: p = (km)(v) = kmv

light guy: p = (m)(kv) = kmv

Their momenta are equal and opposite.

Have you covered the impulse-momentum theorem? If so, can you see that c must be true in this case without doing a calculation?

What does that leave? Oh yeah...prove that their kinetic energies aren't equal. It should be obvious now...since KE does not vary linearly with velocity:

heavy guy:
K = 1/2kmv2

light guy: K = 1/2m(kv)2 = 1/2mk2v2

They aren't equal.

4. Dec 10, 2004

### Soaring Crane

For part 2 wouldn't the center of mass be closer to the more massive peron. Do I show this mathematically by using (m1x1 +m2x2)/(m1 + m2)?

5. Dec 10, 2004

### HallsofIvy

Problem 2:
As far as "d" is concerned: since there are no external forces, the center of mass does not move. d is incorrect.

"c" is correct: Since there are no external forces, the total momentum remains 0. The momenta of the two individuals must be equal in magnitude and opposite in direction.

Since momentum equals "mass times velocity" so that its "magnitude" is "mass times speed", and the two momenta are the same, the smaller person must have the larger speed (and, so, larger initial acceleration). That tells us that a, b, and e are wrong.