Pulley System Problem with String Impulse (Atwood's Machine)

[jmlibunao]

Homework Statement

Two particles of masses 8 kg and 5 kg are connected to the two ends of a light inextensible string which passes over a fixed smooth pulley. Initially each of the two particles are held at a position which is 5 m above a horizontal ground. The objects are then released from rest. Assuming that the particles never reach the pulley, and also that the particles do not rebound when they strike the ground,

#1 Find the tension in the string and the acceleraion of each particle
#2 Find the distance above the ground of the point H, the highest point reached by the 5 kg mass;
#3 Find the speed with which the system is jerked into motion and the impulse experienced by the string during the jerk.

Homework Equations

g = 10 m/s^2
F1 = m1a = m1g - T
F2 = m2a = T - m2g

The Attempt at a Solution

The acceleration of both particles would be the same.
I can easily solve for the acceleration, a, and tension T by:
(8 kg)(a) = (8 kg)(10 m/s^2) - T -- (1)
(5 kg)(a) = T - (5 kg)(10 m/s^2) -- (2)

I'd let both (1) and (2) be equal to T and then solve for a, which I got to be 2.3. I would then plug this in either (1) or (2) and then solve for T, which I got to b 61.54

My question lies in #2 and #3.
For #2, wouldn't the highest point the 5 kg mass would reach be 10 m (5 m + 5 m)?
For #3, can I use one of the kinematics equation to solve for a velocity v? I'm considering the string itself to be massless, so the total mass of the system would be 8 kg + 5 kg = 13 kg. Wouldn't I be able to get the impulse by multiplying 13 kg by g = 10 m/s^2 and my computed v (assuming that I do compute for a v)?

 

[Sunil Simha]

For #2, wouldn't the highest point the 5 kg mass would reach be 10 m (5 m + 5 m)?
For #3, can I use one of the kinematics equation to solve for a velocity v? I'm considering the string itself to be massless, so the total mass of the system would be 8 kg + 5 kg = 13 kg. Wouldn't I be able to get the impulse by multiplying 13 kg by g = 10 m/s^2 and my computed v (assuming that I do compute for a v)?

The 5 kg mass would still have some velocity after the 8 kg mass touches the ground. Thus it will go higher than 10 m.

I didn't understand what the question asks in part 3. Could you please clarify?

 

[jmlibunao]

I didn't understand what the question asks in part 3. Could you please clarify?

Thanks for your reply for #2!

For #3, when you release the system from rest, the 8-kg mass would go down while the 5-kg mass would go up, thus there would be a speed/velocity that would cause this movement. Following that, it's also asking for the impulse experienced by the string. I think that's what it is saying.

 

[haruspex]

Thanks for your reply for #2!

For #3, when you release the system from rest, the 8-kg mass would go down while the 5-kg mass would go up, thus there would be a speed/velocity that would cause this movement. Following that, it's also asking for the impulse experienced by the string. I think that's what it is saying.

Since the system is released from rest (presumably with the string taut) there is no impulse at that time. There will be impulsive tension when the 5kg mass descends and brings the string taut again. Have you missed part of the problem statement?

 

[jmlibunao]

Have you missed part of the problem statement?

I put the complete problem. That is everything given. :\

Impulse is change in momentum right? At the start momentum is 0 since there was no movement. After the release from rest, then you would have momentum so shouldn't there be impulse?

 

[Sunil Simha]

I put the complete problem. That is everything given. :\

Impulse is change in momentum right? At the start momentum is 0 since there was no movement. After the release from rest, then you would have momentum so shouldn't there be impulse?

Yes, there will be a change in momentum because the bodies are being accelerated. However, to find the impulse, you do need the final velocity and that is a function of time. So unless the time at which the (average) impulse is to be found is given, I don't know how one can find the impulse.

 

[haruspex]

Impulse is change in momentum right? At the start momentum is 0 since there was no movement. After the release from rest, then you would have momentum so shouldn't there be impulse?

The OP asks for "the speed with which the system is jerked into motion". This clearly refers to a sudden change in speed, as in an impact. The only such in this scenario is when the 5kg mass falls back down, tautening the string. Since part 2 refers to the 5kg mass reaching its highest point, it seems reasonable to expect part 3 to discuss something that happens later. So although it could have been worded more clearly, I'm fairly confident that this tautening is the event part 3 refers to.