# Thread: an o-level physics prelim MCQ...

1. Originally Posted by bearycute
er... if the string is in tact (not broken) and in tension (not slack), can u tell me how 1 mass can move without affecting the other 2?
They move wrt each other, not without affecting each other...

The whole problem is like looking at the 2kg mass sitting on the table top without the other two 1kg mass. Thus it's like asking what will happen when there's a 8N force applied to the left of the 2kg mass.

The 1 kg mass P and Q are there to confuse the students. But then again, if the mass at both end are different, then we have to consider these mass when working out the problems.

2. Originally Posted by wacko
LOL, kira, are(were) you a teacher's nightmare in school?
Yes I was.... And so was Einstein when he's in school....

I'm not saying I'm right... But this is the way we learn. To question things that we don't understand...

3. bearycute : just wondering.. are you a physics tutor @ hcjc? Mr Quek?

4. Originally Posted by wacko
good, finally most of you see the light. think the most dangerous assumption many people make is that the 8N push downwards on block P will translate into a leftwards pull of 8N on the trolley, which is INCORRECT.
If this is incorrect, then what is the force on the trolley when 8N is applied to mass P and why is this so? I'm really keen to know this...

I also don't understand the diagram listed by wacko. Can anyone explain how he come up with those value?

p.s. The idea here is to learn, not to agrue for the sake of argument...

5. For left side
------------
Weight - Tension = ma
(10 + 8) - T1 = 1(a)
T1 = 18 - 1(a) - (1)

For trolley
---------
2a = T1 - T2 - (2)

For right side
-------------
ma = T2 - mg
1a = T2 - 10 - (3)

Thus..

from (3) T2 = 1a + 10 - (4)
Sub (4) into (2)

Thus, 2a = T1 - 1a - 10
T1 = 3a + 10 (5)

Sub (5) into (1)

3a + 10 = 18 - 1a
a = 2

6. Originally Posted by Kira
They move wrt each other, not without affecting each other...

The whole problem is like looking at the 2kg mass sitting on the table top without the other two 1kg mass. Thus it's like asking what will happen when there's a 8N force applied to the left of the 2kg mass.

The 1 kg mass P and Q are there to confuse the students. But then again, if the mass at both end are different, then we have to consider these mass when working out the problems.
Let me put it this way... u pulled 2kg of mass with force F

u applied the same force F to another 2kg mass attached to another 1 kg mass

Do u think the 2 separate system have the same accln?

7. Originally Posted by junyang
For left side
------------
Weight - Tension = ma
(10 + 8) - T1 = 1(a)
T1 = 18 - 1(a) - (1)

For trolley
---------
2a = T1 - T2 - (2)

For right side
-------------
ma = T2 - mg
1a = T2 - 10 - (3)

Thus..

from (3) T2 = 1a + 10 - (4)
Sub (4) into (2)

Thus, 2a = T1 - 1a - 10
T1 = 3a + 10 (5)

Sub (5) into (1)

3a + 10 = 18 - 1a
a = 2
Nicely done... was hoping someone will do this... me too lazy...haha

8. Originally Posted by junyang
For left side
------------
Weight - Tension = ma
(10 + 8) - T1 = 1(a)
T1 = 18 - 1(a) - (1)

For trolley
---------
2a = T1 - T2 - (2)

For right side
-------------
ma = T2 - mg
1a = T2 - 10 - (3)

Thus..

from (3) T2 = 1a + 10 - (4)
Sub (4) into (2)

Thus, 2a = T1 - 1a - 10
T1 = 3a + 10 (5)

Sub (5) into (1)

3a + 10 = 18 - 1a
a = 2

Correct!
Good solution provided. Hmm, now how I wish all the TCHS students can do physics like you.

9. Okie... I think I see the point now....

So overall accln at the instant of 1s is 2m/s^2.

My question then is, when the masses are in motion, then with 2m/s^2, the resultant force on each item will be 2N on mass p acting downwards, 2N on mass q acting upwards and 4N on mass of 2kg acting to the left. (excluding gravity force). So is it safe to say that the applied 8N has disperse into a ratio of 1:2:1 in the order of P:2kg:Q?

Also, what will happen if we consider the moment of inertia of each of the object?

Lastly, why can't we consider that the applied 8N be transmitted directly to the 2kg mass using equbilium of forces since mass P and Q cancel out each other? I'm assuming that this is what the teacher did to get v=4m/s^2....

10. Because they are connected together, you cannot cancel out P and Q?

When an 8N force is applied to the left (P) it has to move the trolley, as well as pull Q up....

Originally Posted by Kira
Okie... I think I see the point now....

So overall accln at the instant of 1s is 2m/s^2.

My question then is, when the masses are in motion, then with 2m/s^2, the resultant force on each item will be 2N on mass p acting downwards, 2N on mass q acting upwards and 4N on mass of 2kg acting to the left. (excluding gravity force). So is it safe to say that the applied 8N has disperse into a ratio of 1:2:1 in the order of P:2kg:Q?

Also, what will happen if we consider the moment of inertia of each of the object?

Lastly, why can't we consider that the applied 8N be transmitted directly to the 2kg mass using equbilium of forces since mass P and Q cancel out each other? I'm assuming that this is what the teacher did to get v=4m/s^2....

11. Originally Posted by bearycute
Let me put it this way... u pulled 2kg of mass with force F

u applied the same force F to another 2kg mass attached to another 1 kg mass

Do u think the 2 separate system have the same accln?

Let me ask u a qn. If u drag three boxes on the floor, each tied to the next by a piece of string. Will they have the same acceleration?

12. Originally Posted by innovas1
Let me ask u a qn. If u drag three boxes on the floor, each tied to the next by a piece of string. Will they have the same acceleration?
Yes.. if the string is taut.. and not slack.... they will have the same acc. but different tensions in the strings

13. Originally Posted by innovas1
Let me ask u a qn. If u drag three boxes on the floor, each tied to the next by a piece of string. Will they have the same acceleration?

hehe, are the strings all "light and inextensible" ?

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