FORCE, MASS and ACCELERATION CONCEPTUAL QUESTION

AI Thread Summary
In the scenario, a 30 kg girl standing on slippery ice catches a 0.5 kg ball moving at 16 m/s, which leads to a discussion about momentum conservation. The force exerted by the ball on the girl is 8 N, but gravity does not influence her horizontal motion due to the lack of friction. The conservation of momentum principle indicates that the total momentum before and after the catch remains constant. After calculating, it is determined that the girl would move at approximately 0.26 m/s after catching the ball. The discussion emphasizes the importance of understanding momentum and the effects of external forces in this context.
kirsten_2009
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Homework Statement


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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?

Homework Equations



F=m*a

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?


 
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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?
 
It says she is "standing on slippery ice" so you can ignore friction and assume she will move.
 
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?
 
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.
 
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Looks like a classic conservation of momentum prob.
 
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Orodruin said:
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.

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)?
 
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?
 
Orodruin said:
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?

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?
 
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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
Orodruin said:
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.
Thanks for taking the time to help me!
 
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