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

1. ok. v in m/s, a in ms^-2.

the 4 choices are:

A: v=2, a=0
B: v=2, a=2
C:v=4, a=0
D: v=4, a=2.

By the way, the teacher gave the answer as C. This is a SAP school prelim papers 3 yrs back.

2. Originally Posted by sevenWAYS
dunno whether my answer rite or not .... just giving a try at this ....... from my point of view the entire system is in dynamic equilibrium....... so when no forces are acting on the system ..... the trolley doesn't move .....

so when the force of 8N is being applied.... it can be thought of juz as a 8N force pulling on the trolley........ (the downward forces that are contributed by P and Q can be ignored because both are acting downwards = in the opp. position so in the end they cancel out each other)

using formula v=u+at

u=0
t=2

F=MA
8=19.62(A)
A= 8m/s^2

therefore, v= 16m/s

since this is a MCQ.....
can post the options out ?
why yr mass became 19.62?

3. Originally Posted by wacko
guys guys, quit bickering and reflect upon my diagram will ya? the answer is in there...

(bearycute: though you may have got the right answer, i think you still have the fundamentals wrong as you seem to think the resultant force on the trolley is 8N when it's not)
Din see ur small note there... hee... err which statement did i say that implied that the 8N was acted solely on the trolley?

Anyway,
to junyang: I think if i remembered correctly, u r from hcjc... y not check out w ur physics tutor to clarify? HCJC promo is next wk... better get ur concepts right

to innovas:
THE CONFIRMED ANSWER IS V = 2

4. Originally Posted by innovas1
ok. v in m/s, a in ms^-2.

the 4 choices are:

A: v=2, a=0
B: v=2, a=2
C:v=4, a=0
D: v=4, a=2.

By the way, the teacher gave the answer as C. This is a SAP school prelim papers 3 yrs back.
The answer given by the teacher is wrong i think... unless i have missed out certain things 9not quite possible since i have gone thru the qn at 10 times )or the qn u gave was not complete... Correct answer is A

5. i think i got a simplier solution:

take the system as one body: total mass = 4kg
a 8N force is applied, the acc of the system is 2ms-2 in the first second.

the Net acc in the 2nd second is zero, because no net force acts on the system. therefore acc=0

the force is variating, so it is an impluse question (A level stuff)
it is defined as: Force x Time = mass x velocity

as Netwon's 2nd Law is defined as the rate of change of momentum (m x v)/t
if m is const, v/t = a, F is not changing... therefore, simplified => F=ma... (not to worry about this, you will get to learn if you go to A levels)

SO,

8N X 2 = 4Kg x Velocity
Therefore, velocity = 4ms-1

6. Originally Posted by bearycute
Din see ur small note there... hee... err which statement did i say that implied that the 8N was acted solely on the trolley?

Anyway,
to junyang: I think if i remembered correctly, u r from hcjc... y not check out w ur physics tutor to clarify? HCJC promo is next wk... better get ur concepts right

to innovas:
THE CONFIRMED ANSWER IS V = 2
what do you think of my solution above??
i'm also having promo this week, don't scare me... or else... i die.....

7. Originally Posted by yamcake
i think i got a simplier solution:

take the system as one body: total mass = 4kg
a 8N force is applied, the acc of the system is 2ms-2 in the first second.

the Net acc in the 2nd second is zero, because no net force acts on the system. therefore acc=0

the force is variating, so it is an impluse question (A level stuff)
it is defined as: Force x Time = mass x velocity

as Netwon's 2nd Law is defined as the rate of change of momentum (m x v)/t
if m is const, v/t = a, F is not changing... therefore, simplified => F=ma... (not to worry about this, you will get to learn if you go to A levels)

SO,

8N X 2 = 4Kg x Velocity
Therefore, velocity = 4ms-1

Er... y 8N X 2? hope u din think that that after 2s, that is y X2. Then if after 3 or 4 s leh? Since u are in A level, can now talk momentum to u... the momentum change only lasted 1 s (since a force is applied during 1 s). after 1 s, there is no change in momentum since no net force acts on the system now.

Try drawing the F-t graph... area under is the impulse aka change in momentum.
hope that helps...

8. Originally Posted by bearycute
Er... y 8N X 2? hope u din think that that after 2s, that is y X2. Then if after 3 or 4 s leh? Since u are in A level, can now talk momentum to u... the momentum change only lasted 1 s (since a force is applied during 1 s). after 1 s, there is no change in momentum since no net force acts on the system now.

Try drawing the F-t graph... area under is the impulse aka change in momentum.
hope that helps...

YA RITE,
the thought just struck me while i was in the toliet just now... hehe
hence, i fully agree with you, the velocity is 2ms-1..

sorry for the "wrong" subsitution of the values:

should be:

8N X 1s = 4kg X velo
Velocity = 2ms-1

hope that helps

velo =

9. mhmm.. ok.. haha.. i finally get it.. its 2ms^-2.. bleah

10. 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.

11. if this is the case, a=2, then too bad, it means the teacher marked the whole of that level's papers wrongly...

AT first i didnt want to believe its 2, coz I was thinking more of the isolation of the trolley from the two 1kg masses... That the tensile force is 8N, resultant force on mass of 2kg is 8N, so a=4, v=4 during the 1st sec.. Can someone explain why i shouldn't do this?

By conservation of momentum or energy, people lump the masses together as 4kg, so if thats the case, v=2...

Hahah.. I argued with my elder bro, even though he went for S-papers in physics and Maths at A-level..

12. Originally Posted by innovas1
if this is the case, a=2, then too bad, it means the teacher marked the whole of that level's papers wrongly...

AT first i didnt want to believe its 2, coz I was thinking more of the isolation of the trolley from the two 1kg masses... That the tensile force is 8N, resultant force on mass of 2kg is 8N, so a=4, v=4 during the 1st sec.. Can someone explain why i shouldn't do this?

By conservation of momentum or energy, people lump the masses together as 4kg, so if thats the case, v=2...

Hahah.. I argued with my elder bro, even though he went for S-papers in physics and Maths at A-level..
You cannot lump the 3 mass together because they are independent of each other. i.e. when the 2kg moves, it doesn't mean that the other 1kg masses move in the same direction too.... The 1kg masses actually neutrilze each other... therefore, the only force that contribute to the motion is the 8N acting on the 2kg mass.

So what is the correct answer according to your teacher then?

13. Originally Posted by Kira
You cannot lump the 3 mass together because they are independent of each other. i.e. when the 2kg moves, it doesn't mean that the other 1kg masses move in the same direction too.... The 1kg masses actually neutrilze each other... therefore, the only force that contribute to the motion is the 8N acting on the 2kg mass.

So what is the correct answer according to your teacher then?
erm, i think it's safe to say what the correct answer is. the more you are going to argue your calculations, the more silly you are going to think of yourself when you finally realise the right way. maybe you would like to read through all the correct answers proposed here, and slowly reflect on the physics behind it.

14. Originally Posted by Kira
You cannot lump the 3 mass together because they are independent of each other. i.e. when the 2kg moves, it doesn't mean that the other 1kg masses move in the same direction too.... The 1kg masses actually neutrilze each other... therefore, the only force that contribute to the motion is the 8N acting on the 2kg mass.

So what is the correct answer according to your teacher then?
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?

15. Originally Posted by innovas1
if this is the case, a=2, then too bad, it means the teacher marked the whole of that level's papers wrongly...

AT first i didnt want to believe its 2, coz I was thinking more of the isolation of the trolley from the two 1kg masses... That the tensile force is 8N, resultant force on mass of 2kg is 8N, so a=4, v=4 during the 1st sec.. Can someone explain why i shouldn't do this?

By conservation of momentum or energy, people lump the masses together as 4kg, so if thats the case, v=2...

Hahah.. I argued with my elder bro, even though he went for S-papers in physics and Maths at A-level..
ok... lets isolate the 3 individual masses and form the 3 equation. Taking g =10

1 kg mass (w 8N)

total force down = 8N + 1(10) = 18

18 - T1 = 1(a) [applying F = ma]

1kg mass (w/o 8N)

T2 - 10 = 1(a)

2kg trolley

T1 - T2 = 2(a)

Solving the three equation yields a = 2, T1 = 16, T2 = 12

Thus we can see that the resultant force on trolley (T2 - T1) = 4 N
Resultant force on 1 kg mass = 2N

16. Originally Posted by bearycute
ok... lets isolate the 3 individual masses and form the 3 equation. Taking g =10

1 kg mass (w 8N)

total force down = 8N + 1(10) = 18

18 - T1 = 1(a) [applying F = ma]

1kg mass (w/o 8N)

T2 - 10 = 1(a)

2kg trolley

T1 - T2 = 2(a)

Solving the three equation yields a = 2, T1 = 16, T2 = 12

Thus we can see that the resultant force on trolley (T2 - T1) = 4 N
Resultant force on 1 kg mass = 2N
a is not the same in all 3 masses.. So cannot do it this way... a for mass 2kg is wrt a for p mass. similiarly, a for mass of q is wrt mass p too... There is another relation wrt the 3 different a....

17. Originally Posted by Kira
a is not the same in all 3 masses.. So cannot do it this way... a for mass 2kg is wrt a for p mass. similiarly, a for mass of q is wrt mass p too... There is another relation wrt the 3 different a....
Pls read carefully... all 3 masses have the same accln BUT THEY HAVE DIFFERENT RESULTANT FORCES ACTING ON IT!!!

18. LOL, kira, are(were) you a teacher's nightmare in school?

19. To think that I've gone thru O levels before....... I've got no idea what is being discussed here. Not a single clue

20. Originally Posted by bearycute
Pls read carefully... all 3 masses have the same accln BUT THEY HAVE DIFFERENT RESULTANT FORCES ACTING ON IT!!!
How can each item have different resultant force acting on it? Assuming the accln is 2, then resultant force on 2kg mass is 2X2=4N.... What happen to the other 4N?

Resultant force on each item is 8N as the items are in equiblium before the applied force. Thus, at s=1sec, accln on each item is 8m/s^2 on p and q and 4m/s^2 on the 2kg mass. Using F=ma...

Therefore, v=u+at with u=0 and a=4m/s^2 and t=1s. Which gives v=4m/^s.

After 1s when 8N is taken away, accln=0 as no more applied force. Thus v(at 2s)=u+at with u=4m/s^s and a=0. Thus the answer given by the teacher is correct.....

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