# FORCE, MASS and ACCELERATION CONCEPTUAL QUESTION

1. Sep 29, 2014

### kirsten_2009

1. The problem statement, all variables and given/known data

A 30 kg girl standing on slippery ice catches a 0.5 kg ball thrown with a speed of 16 m/s. What then happens to the girl?

2. Relevant equations

F=m*a

3. The attempt at a solution

Assuming that the ice is completely smooth (zero friction) then…

F = m x a

F = 0.5 kg x 16 ms² = 8 kg*ms² (N)

A force of 8 N is acting on the girl, but assuming there is still the force of gravity acting on her then nothing would happen since this force is not sufficient to move the girl in any direction. However, if we assume that there is no force of gravity acting on her and that air resistance is negligible then the girl would begin to move slowly at a steady velocity and unless there were any forces to stop her she would continue moving at that velocity forever?

2. Sep 29, 2014

### nasu

The acceleration is not given. The 16 m/s is the speed. (m/s and not m/s^2).
Did you learn about momentum and conservation laws?

3. Sep 29, 2014

### CWatters

It says she is "standing on slippery ice" so you can ignore friction and assume she will move.

4. Sep 29, 2014

### kirsten_2009

Hi,

oh yes, I see what you're saying about the speed. Yes, we just touched on the concept of momentum and conservation of momentum in an isolated system. So, since there is no friction, why would she move? Wouldn't she need a lot more force to overcome the force of gravity? Or should I just assume that gravity is not acting on her?

5. Sep 30, 2014

### Orodruin

Staff Emeritus
Gravity is acting in a direction perpendicular to the ice. I believe the problem constructor intends the ball's velocity to be parallel to it and so gravity does not have any influence in the direction relevant to the problem.

6. Sep 30, 2014

### dean barry

Looks like a classic conservation of momentum prob.

7. Sep 30, 2014

### kirsten_2009

Thanks for your reply! So, if gravity isn't playing a role, the ice is slippery and air resistance and anything else that could prevent the girl from moving is negligible then would it be correct to say that the ball exerts a force of 8 N on the girl (and the girl exerts the same force on the ball but in opposite direction), momentum is conserved in this "inelastic collision" and both the ball and the girl are accelerated (ball is decelerated and girl is accelerated into motion)?

8. Sep 30, 2014

### Orodruin

Staff Emeritus
As nasu said, you cannot say anything about the actual force during the catch (you do not know the acceleration). What you can do is to use the law of momentum conservation. So if we are going to try that: What does the law of momentum conservation say?

9. Sep 30, 2014

### kirsten_2009

O.K I think I get it....the momentum of the system must be conserved.
p=m x v

Though 16 m/s is not a velocity, it's a speed, I'm going to assume that the direction is not important since it's a ball that's being caught so the ball is coming towards the girl and would "push" her back....so....

momentum of the ball = 0.5 kg x 16 m/s = 8kg/ms and then the mass for both the girl and ball = 30 kg + 0.5 kg = 30.5 kg * v (so I could solve for v)
v = 8 kg/ms / 30.5 kg = 0.26 m/s and that's how fast she would be moving after catching the ball....correct?

10. Sep 30, 2014

### Orodruin

Staff Emeritus
Correct. Extra credit for not doing the very common mistake of forgetting the mass of the ball in the momentum after the catch.

Regarding the speed/velocity: You can simply define your coordinate system such that the ball is moving in the positive x-direction before the catch.

11. Sep 30, 2014

### kirsten_2009

Thanks for taking the time to help me!