# Pucks colliding at right angles

1. Apr 21, 2015

### Soniteflash

1. The problem statement, all variables and given/known data
Two pucks are moving on a frictionless air table are about to collide. The 1.5 kg puck is moving directly east at 2.0 m/s. The 4.0 Kg puck is moving directly north at 1.0m/s.

What is the total kinetic energy of the two-puck system before the collision?
A. Square root of 13 J
B.5.0J
C.7.0 J
D.10J
E.11J

What is the magnitude of the total momentum of the two-puck system after the collision?
A. 1.0 kg*m/s
B. 3.5 kg*m/s
C. 5.0 kg*m/s
D. 7.0 kg*m/s
E. kg*m/s
2. Relevant equations
(1/2)Mv2

P = mv

3. The attempt at a solution
F
or the first question I calculated the kinetic energy of the two pucks and then used the Pythagorean Theorem since the two of them are at right angles which gave me 5 Joules.

For the second question I used the Momentum formula for both pucks. I used the Pythagorean Theorem again and got 5 kg*m/s. I am unsure about this though.

2. Apr 21, 2015

### Suraj M

Is it an elastic or nonelastic collision?
I do not understand why you did this?
If the collision is perfectly non.elastic then you're right.

3. Apr 21, 2015

### AlephNumbers

Momentum would be conserved regardless of whether it was elastic or inelastic. That is why it is not necessary to specify whether or not the collision was elastic/inelastic in the problem statement.

4. Apr 21, 2015

### Suraj M

If the collision was elastic then the bodies wouldn't stick together, but then, well yeah you're right, question only asks about momentum of the system, not each body, sorry..

5. Apr 21, 2015

### Soniteflash

Ah, I see my error i think. Kinetic Energy is a scalar and not a vector quantity.

6. Apr 21, 2015

### haruspex

Ok, but I'm a bit puzzled. If you did erroneously use Pythagoras, why didn't you get sqrt(13)?

7. Apr 21, 2015

### Soniteflash

Oh boy......I added instead of doing sqrt(22+32).......Can't even do wrong stuff correctly......