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Impulse of normal force

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Homework Statement




Stick of mass M resting on frictional surface with friction coefficients μks. From the center of the stick at time t=0 a ball of mass m thrown to the right with an angle θ above the horizontal and with speed v0. As a result of the throwing there is an impulse on the y axis (perpendicular to the horizontal stick) between the stick and the surface for short time Δt.

Calculate the impulse of the normal force between the surface and the stick. Assume the interaction time is short Δt→0?

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How am I formulate the equation- 1. When I calculate the change in momentum is pf=0 because the stick does not move vertically? 2. Is the acting force on the stick is F=N(t)−Mg because the normal on M by the ball is internal force?
 

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  • #2
haruspex
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Since we are considering a vanishingly short period of time, any steady forces make negligible contribution to the momentum. Just consider the vertical and horizontal changes in momentum of the ball. Clearly the stick does not move vertically, so the change in vertical momentum of the ball must be balanced by a vertical impulse from the stick onto the ground.
You say the stick does not move, but that is not stated in your quote of the problem. Also, the question only asks about vertical impulse (if I'm reading it correctly) so I don't understand why it tells you about frictional coefficients. Are there more parts to the question?
 
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...I don't understand why it tells you about frictional coefficients. Are there more parts to the question?
Perhaps that is their way of saying it doesn't move? But I agree I feel like there would need to be more to the question..
 
  • #4
vela
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Homework Statement




Stick of mass M resting on frictional surface with friction coefficients μks. From the center of the stick at time t=0 a ball of mass m thrown to the right with an angle θ above the horizontal and with speed v0. As a result of the throwing there is an impulse on the y axis (perpendicular to the horizontal stick) between the stick and the surface for short time Δt.

Calculate the impulse of the normal force between the surface and the stick. Assume the interaction time is short Δt→0?

-------------
How am I formulate the equation- 1. When I calculate the change in momentum is pf=0 because the stick does not move vertically?
Right. The stick doesn't move vertically, so ##\Delta p_y = 0##.

2. Is the acting force on the stick is F=N(t)−Mg because the normal on M by the ball is internal force?
What is F supposed to represent? It's the force due to what? You're only considering the stick, so forces like gravity and the normal force from the ground which are all caused by things outside the stick are external forces. The first thing you need to do it identify all the forces on the stick.

In this problem, there's apparently a force acting on the ball which causes it to fly off with speed ##v_0## and angle ##\theta##. How is this force related to the forces on the stick?
 
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Are there more parts to the question?
Yes there are. Still I can't see how to calculate the impulse of the normal force between the stick and the surface...

I thought that the change in momentum is 0-(-mv0sinθ), and when I show the forces that act on the stick N will change because the N that come from the ball will stop acting when the ball will be thrown, so F=N(t)-Mg.
 

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haruspex
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I thought that the change in momentum is 0-(-mv0sinθ), and when I show the forces that act on the stick N will change because the N that come from the ball will stop acting when the ball will be thrown, so F=N(t)-Mg.
The question is about momentum, not forces. In time Δt, the momentum from the weight of the stick is MgΔt, but what happens to that as Δt → 0 ?
You have the right change in vertical momentum for the ball. The only other vertical impulse is between the stick and the ground, so the magnitude of that must be...?
 
  • #7
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1.Is the analysis at the time when the ball is landing on the stick (because then pfinal=0 pi=-v0sinθ )?
2.Why there is no momentum from the normal force between the ball and the stick?
3. Is the normal force between the stick and plane not constant beacause of the normal that acts between the stick and the ball is changing?
 
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haruspex
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1.Is the analysis at the time when the ball is landing on the stick (because then pfinal=0 pi=-v0sinθ )?
The ball does not land on the stick. The ball is thrown from the stick.
I can't answer questions about values of variables if you don't define them. What are these initial and final momenta momenta of?
2.Why there is no momentum from the normal force between the ball and the stick?
It depends what you mean by normal force here. There was presumably an ongoing normal force equal to mg before the ball was thrown. During the Δt of the throwing event, its contribution to momentum changes is mgΔt, which is negligible. The throw itself would have involved a much larger force acting for the short time Δt. We don't care how large exactly that force was, or whether it was constant. All we care about is that the integral of the force over the brief duration of the throw gave the ball a certain vertical momentum. Since the stick neither becomes airborne nor sinks into the ground, the same impulse must have operated between the stick and the ground.
 
  • #9
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For formulating the change in momentum of the ball (to be equating with the impulse of the net force): at first it is given an initial momentum p i=v0sinθ after Δt wouldn't pf still have the same value as at the initial momentum?
 

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