Ball launching off of a tray connected to a spring

• Elias Waranoi
In summary: then according to my calculations, the equilibrium height of the system is at -0.15 cm below the equilibrium point.
Elias Waranoi

Homework Statement

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point A, which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray?

mt =1.5 kg
mb = 0.275 kg
k = 185 N/m
A = -0.15 m

F = -kx
F = ma

The Attempt at a Solution

My attempt:
The maximum downward acceleration of the ball is -g = -9.8 so the ball should leave the tray when the acceleration of the tray is also -g.
F = -kx = mta = -mtg
x = mtg/k = 0.0795 m
Δx = x - A = 0.229 m

This is wrong and I would have gotten the right answer if I used mt + mb instead of just mt in the above equations but I can't understand why. When the ball leaves the tray the weight of the ball on the tray should be zero (because they're not in contact anymore) combined with the acceleration of the tray being -g should give the equation x = mtg/k. But apparently this logic is flawed since the right answer is given by x = (mt + mb)g/k

Please tell me what's wrong with my logic.

Can you clarify for me what is the condition of the spring when the acceleration of the tray is -g?

Edit: You said that the weight of the ball on the tray is 0 and the acceleration of the tray is -g right at the point where separation occurs. Here is another question: What is it that actually causes the separation - a gap to appear between the ball and the tray. Since the acceleration of the ball at the point of, and after separation, is -g, what could cause a separation - an actual gap to appear between the ball and tray? What has to be true about the acceleration of the tray for an actual gap to appear between the ball and tray?

Last edited:
Elias Waranoi
Elias Waranoi said:

Homework Statement

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point A, which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray?

mt =1.5 kg
mb = 0.275 kg
k = 185 N/m
A = -0.15 m

F = -kx
F = ma

The Attempt at a Solution

My attempt:
The maximum downward acceleration of the ball is -g = -9.8 so the ball should leave the tray when the acceleration of the tray is also -g.
F = -kx = mta = -mtg
x = mtg/k = 0.0795 m
Δx = x - A = 0.229 m

This is wrong and I would have gotten the right answer if I used mt + mb instead of just mt in the above equations but I can't understand why. When the ball leaves the tray the weight of the ball on the tray should be zero (because they're not in contact anymore) combined with the acceleration of the tray being -g should give the equation x = mtg/k. But apparently this logic is flawed since the right answer is given by x = (mt + mb)g/k

Please tell me what's wrong with my logic.
Ask the question in the way "where is the tray just before the ball leaves?".

Elias Waranoi
Oh, now I get it. According to my calculations the tray is 0.0795 m above the equilibrium height of the spring and tray system. While A = -0.15 is how far below the equilibrium height it is for the spring, tray and ball system. I get the correct answer now, thanks!

Well done!

I don't if this is still active, but can someone explain why the equilibrium condition holds 9.24 cm above equilibrium position?! I mean, shouldn't equilibrium happen at the equilibrium position?

BigBrains said:
I don't if this is still active, but can someone explain why the equilibrium condition holds 9.24 cm above equilibrium position?! I mean, shouldn't equilibrium happen at the equilibrium position?
I don't know how you calculated 9.24cm, so maybe I misunderstand your question, but the thread discusses two different equilibrium positions, one for tray+ball+spring and a higher one for tray+spring.

Let me rephrase this, when an object is in equilibrium shouldn't it be at the equilibrium position ? If this is true, then how come the equilibrium position in the question is above equilibrium point? Shouldn't they be the same thing?

If we consider the spring, ball and tray system; their equilibrium point is below the spring tray system's, which is below the spring's. Then how come equilibrium be above? It should be below.

BigBrains said:
Let me rephrase this, when an object is in equilibrium shouldn't it be at the equilibrium position ? If this is true, then how come the equilibrium position in the question is above equilibrium point? Shouldn't they be the same thing?
The equilibrium point mentioned in the question is for the tray+ball+spring system. Where in the thread is mention of this other position you are referring to?

1. How does the spring affect the ball's launch?

The spring acts as a source of potential energy that is transferred to the ball, causing it to launch off of the tray. As the spring is compressed, it stores potential energy, which is then released as the spring expands, propelling the ball forward.

2. What factors affect the distance the ball travels?

The distance the ball travels is affected by the initial compression of the spring, the mass of the ball, the angle of the launch, and air resistance. The greater the initial compression and the lower the angle of launch, the farther the ball will travel.

3. How does air resistance impact the ball's trajectory?

Air resistance, also known as drag, acts against the motion of the ball and can cause it to slow down and deviate from its intended path. The shape and velocity of the ball can also affect the amount of air resistance it experiences.

4. Can the ball's launch height be controlled?

Yes, the launch height of the ball can be controlled by adjusting the angle of the tray or by using a spring with a different stiffness. A higher angle or a stiffer spring will result in a higher launch height.

5. What is the relationship between the spring constant and the ball's launch velocity?

The spring constant, which represents the stiffness of the spring, is directly proportional to the launch velocity of the ball. This means that as the spring constant increases, the launch velocity of the ball also increases.

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