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Projectile Motion and Spring Question


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These are two questions from an assignment which I just don't get. I'm in grade 12. I just want to be pointed in the direction, not given the answer. I wanna make sure I know this for the test, my teachers hard :(

 

1) A 1.00kg mass and a 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0N/m. The 2.00kg mass is suddenly removed. How high above this starting position does the 1.00kg reach?

 

I was thinking of determining the energy at equilibrium, because Ee = 0 and Ek = max at that point...but it would become zero, would it not? or, could i equal Ek1 + Ee1 = Ee2 + Ek2? I dunno...

 

2) A ball of mass 0.25kg is thrown vertically upward from the roof of a building 18m high with a speed of 16m/s, and just misses the building on the way down

a) To what vertical height above Earth does the ball rise?

b) With what vertical velocity does the ball hit the ground?

 

dy = v1t + -.5at^2? But I don't know time, or the vertical distance...I dunno where to begin, I used to be able to do this...but slipped my mind completly.

 

Thanks for any help :D

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These are two questions from an assignment which I just don't get. I'm in grade 12. I just want to be pointed in the direction' date=' not given the answer. I wanna make sure I know this for the test, my teachers hard :(

 

1) A 1.00kg mass and a 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0N/m. The 2.00kg mass is suddenly removed. How high above this starting position does the 1.00kg reach?

 

I was thinking of determining the energy at equilibrium, because Ee = 0 and Ek = max at that point...but it would become zero, would it not? or, could i equal Ek1 + Ee1 = Ee2 + Ek2? I dunno...

 

2) A ball of mass 0.25kg is thrown vertically upward from the roof of a building 18m high with a speed of 16m/s, and just misses the building on the way down

a) To what vertical height above Earth does the ball rise?

b) With what vertical velocity does the ball hit the ground?

 

dy = v1t + -.5at^2? But I don't know time, or the vertical distance...I dunno where to begin, I used to be able to do this...but slipped my mind completly.

 

Thanks for any help :D[/quote']

 

1. The energy won't become zero - where would the energy go? How much energy is stored in the spring?

 

2. Use equations that done include t. Conservation of mechanical energy, or vf2=vo2 + 2a(dot)s

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2) A ball of mass 0.25kg is thrown vertically upward from the roof of a building 18m high with a speed of 16m/s' date=' and just misses the building on the way down

a) To what vertical height above Earth does the ball rise?

b) With what vertical velocity does the ball hit the ground?[/quote']

 

What you have to use for these two questions is the formula v^2 = u^2 +2as. One thing you may want to keep in mind is that all of the values you substitute into the formula are vectors, viz. they have magnitude AND direction.

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1) A 1.00kg mass and a 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0N/m. The 2.00kg mass is suddenly removed. How high above this starting position does the 1.00kg reach?

 

You can use F=ke here? First find total force which is 20N and find the extension for this using F=ke. Then using 10N alone find extension for 1kg. Deduct the two values and you will find the height 1 kg will reach. Do confirm this with your teacher, though.

 

2) A ball of mass 0.25kg is thrown vertically upward from the roof of a building 18m high with a speed of 16m/s, and just misses the building on the way down

a) To what vertical height above Earth does the ball rise?

b) With what vertical velocity does the ball hit the ground?

 

since the mass is given, use 1/2mv^2 = mgh for both a and b. For a, you have the v... so use plug in the values to get h(max). For b, use h(max) + 18 to obtain the value for v. The answers will be the same as TokenMonkey's though.

 

-mak10

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It's HW help. The point is to help the poster with the HW question, not to do it for them. The amount you learn by watching someone else work a problem is extremely limited.

 

Ah, I see. Yes, you're absolutely right. Well, it doesn't seem as if the guy who posted has seen it yet, so I'll just go ahead and edit my post...

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