A 2.00 kg , horizontal, uniform tray is attached to a vertical ideal spring of f

Question

A 2.00 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 300 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 17.8 cm below its equilibrium point (call this point A) and released from rest.

Part A: How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (answer in cm)

Part B: How much time elapses between releasing the system at point A and the ball leaving the tray? (answer in s)

Part C: How fast is the ball moving just as it leaves the tray?

Thank you!

Explanation / Answer

Part A

The tray+ball system will oscillate with the mean/equilibrium position at point, say, B.

if initial compression of the spring is x , then

kx = (M+m)g

or x = [(2 kg + 0.3 kg)(9.8 m/s2)]/(185 N/m)

or x = 12.2 cm

Now, the tray is pushed 17.8 cm below its equilibrium position B. While the system is in motion the acceleration on it would be towards B through out the motion.

Let us suppose the block separates from the tray at a distance h above the point A.

The elongation in the spring will be x0 = h - (x + 17.8 cm)

At this position the acceleration of the system is a and it's downward.

So the equation of motion for the ball is,

mg - N = ma , where N is the normal contact force by the tray on the ball and m is the mass of the ball

Now, the ball at this point separates, so N = 0, thus we get,

mg = ma

or a = g, so the system will have an acceleration g downward.

Let us consider the equation of tray+ball system at this point, we have,

kx0 + (M+m)g = (M+m)a , but since a = g, we have

kx0 + (M+m)g = (M+m)g

or x0 = 0

or h - x -17.8 cm = 0

or h = x + 17.8 cm = 12.2 cm + 17.8 cm = 30 cm

So, the ball leaves the tray at a point 30 cm above point A.

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(b) for the system is,

= [k/(M+m)] = [(185 N/m)/(2.3 kg)] = 8.97 rad/s

Amplitude for the motion is A' = 17.8 cm

So equation for simple harmonic motion is

x = A'cost, considering downward from point B as positive

So when the separation takes place, x = -12.2cm at time t, thus we have

-12.2 cm = (17.8 cm)cos(8.97)t

or (8.97)t = 2.326

or t = 0.26 s

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(c) As we saw above, the ball leaves the tray at 30cm from A. Distance from B or the mean postion of the oscillatin is 12.2 cm as we saw that the elongation in the spring becomes zero at this point.

Now, let the point A be the zero level of potential energy, Then initial energy of the system when the spring is compressed by 17.8 cm will be,

PE = (1/2)k(17.8 cm + 12.2 cm)2

When the position of system is at the point of separation, the elongation in the spring is zero. Let the speed at this point be v. The heigh of this point from A will be H = 30cm

or So total enery when the system is at point of separation is,

E = (1/2)(M + m)v2 + (M + m)gH,

Now applying law of conservation of energy we get,

PE = E

or (1/2)k(17.8 cm + 12.2 cm)2 = (1/2)(M + m)v2 + (M + m)gH

or (0.5)(185 N/m)(0.09 m2) = 1.15v2 + 6.762

or v = 1.166 m/s

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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