Question 3 - Motion

[Keyzer]

For this question, why is everyone saying that the acceleration is 2 m/s/s? The weight is providing a tension force of 1N which is pulling the cart along, and so the mass of the cart multiplied by its acceleration would give 2.5 m/s/s. However, I have been told that both masses need to be added together as the whole system is being accelerated. That does not make any sense though since the mass is being accelerated due to gravity and the trolley is accelerating due to the tension in the rope.

[/0]

Since we assume that the rope is inelastic, our basic assumption is that at any instant in time the masses must move with the same velocity, and hence they must have the same acceleration.

You can deal with this problem in two ways.

The simpler way is to consider the masses as a system. Then





However, if you deal with the masses separately, then for the mass hanging down you will have



And for the mass on the table,



where T is the tension force (which is the same throughout the rope)

Hence, by adding equations.

[Keyzer]

Since we assume that the rope is inelastic, our basic assumption is that at any instant in time the masses must move with the same velocity, and hence they must have the same acceleration.

You can deal with this problem in two ways.

The simpler way is to consider the masses as a system. Then





However, if you deal with the masses separately, then for the mass hanging down you will have



And for the mass on the table,



where T is the tension force (which is the same throughout the rope)

Hence, by adding equations.

Okay, so how many marks would I receive for this question? The maximum is two, but my answer is 2.5 m/s/s. However, I used F = ma and I labeled the diagram with tension force. Will I at least get 1 mark?

[/0]

I think you would, for finding the external force on the system. It's likely to be 1 mark for the force and the other for the acceleration.

[a.zirek]

You can also think of it like this:
If M1 has 1N weight force acting on it and 1N of tension acting on it in the opposite direction, it has a net force of zero and hence no acceleration. That's why you need to add both weights.

[Keyzer]

You can also think of it like this:
If M1 has 1N weight force acting on it and 1N of tension acting on it in the opposite direction, it has a net force of zero and hence no acceleration. That's why you need to add both weights.

Yeah, but wouldn't that mean that the M1 is just stationary, which it is at the beginning?

[a.zirek]

You can also think of it like this:
If M1 has 1N weight force acting on it and 1N of tension acting on it in the opposite direction, it has a net force of zero and hence no acceleration. That's why you need to add both weights.

But it also means that the system would never move. And if M1 is stationary then so is M2. But if you say that the rope has a tension of 0.8N on it, then M1 has 0.2N acting down on it. This means that both objects would accelerate at 2m/s down.

Yeah, but wouldn't that mean that the M1 is just stationary, which it is at the beginning?

[kenhung123]

That question is weird. Since you considered no tension, you must consider both weights. Fnet=Mg
(m1+m2)a=m1g (since the net force acts on both bodies)

[Akirus]

It's not a weird question. For the record, you learn this in year 11.

[kenhung123]

Sorry Einstein..

[ddaly]

You don't have to be Einstein to work it out. 2m/s^2 and it is year 11 course work.

[superflya]

^^ lol

[ddaly]

My teacher is the chief examiner and I have conferred with him.

[kenhung123]

Well I'm sorry if year 12 physics is too easy for you guys...

[Akirus]

Well I'm sorry if year 12 physics is too easy for you guys...

What are you getting so defensive about? We are merely pointing out that it was a reasonable question, not something beyond the scope of the course or excessively difficult.