# an o-level physics prelim MCQ...

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#### innovas1

##### New Member
physics qn i think u need to save the jpeg to yr harddisk and then open it. The pic will be clearer..

can anyone tell me what is the velocity? is it 2 or 4m/s?

haiz..

#### bearycute

##### New Member
Acceleration for the whole 1 s is 2 ms^-2 and the velocity at the end of 1 s is 2 ms^-1 :embrass:

#### innovas1

##### New Member
the acceleration is 0 at the 2nd sec as the force is applied for only 1 sec..

what is the velocity?? I say its 4, my bro says it is 2 and argued with me. Should we lump the 3 masses together to calculate velocity or just calculate vel using the 2kg mass?

some say vel is 2, some say 4...
wat is yr view? ANy physics teacher or expert here?

#### bearycute

##### New Member
innovas1 said:
the acceleration is 0 at the 2nd sec as the force is applied for only 1 sec..

what is the velocity?? I say its 4, my bro says it is 2 and argued with me. Should we lump the 3 masses together to calculate velocity or just calculate vel using the 2kg mass?

some say vel is 2, some say 4...
wat is yr view? ANy physics teacher or expert here?
u r right that the accln is zero after 1 s. thus the whole systemwill travel at constant velocity. so whether u r considering the whole system or just one single mass, the velocity is the same provided the string connecting them remains taut and do not snap. ans is 2 for velocity...

#### innovas1

##### New Member
if we just consider the mass on the table, then the tension to the left is 18N and 10N to the right. Resultant force of 8N acts on the 2kg mass. so, a=4,v=4 at that instant.

#### bearycute

##### New Member
innovas1 said:
if we just consider the mass on the table, then the tension to the left is 18N and 10N to the right. Resultant force of 8N acts on the 2kg mass. so, a=4,v=4 at that instant.
Er... not right. the tension to the right if 10, then the 1 kg mass will not acc up. after 1 s then i agree w u. but after 1 s the tension to the left can't be 18 as the 8N force no longer exerts... hope that helps... accln is 2 during the first second...

#### wacko

##### New Member

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.

#### bearycute

##### New Member
wacko said:

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.
the 6N tension in red should be 16N and the 2N tension in red should read 12. I think u forget to consider the weight of the 1 kg mass.

#### wacko

##### New Member
bearycute said:
the 6N tension in red should be 16N and the 2N tension in red should read 12. I think u forget to consider the weight of the 1 kg mass.
erm, i did say "minus gravity and reactional forces"...

#### Kira

##### New Member
wacko said:

here is the answer if you look at the breakdown of forces. red is forces acting on the individual objects (minus the gravity and reactional forces), blue is the resultant force on each object. thus it's clear to see that the answer is 2m/s for velocity. as to how do you get the breakdown of the forces shown, think about it slowly, heh.
Okie my version...

Assume the whole set up is in equibrum...

Thus, before 8N force is applied, tension in both ropes are 9.81N.

When 8N force is applied, looking at the FBD of the 2kg mass. We'll have, 8+9.81N acting on the left and 9.81N still acting on the right... Therefore, the resultant force on the 2kg mass is still 8N to the left. (The 9.81N cancel out each other)

Thus overall resultant force acting on the 2kg mass is 8N to the right... We all know that F=ma. Therefore, at 1s the acceration is thus 8/2=4m/s^2. Velocity also 4m/s

After 1s. When 8N is taken away, applying newton's law of motion, assuming zero frictional force. We'll have Acceration=0, velocity=4m/s.

My 2 cent's worth...

#### wacko

##### New Member
Kira said:
Okie my version...

Assume the whole set up is in equibrum...

Thus, before 8N force is applied, tension in both ropes are 9.81N.

When 8N force is applied, looking at the FBD of the 2kg mass. We'll have, 8+9.81N acting on the left and 9.81N still acting on the right... Therefore, the resultant force on the 2kg mass is still 8N to the left. (The 9.81N cancel out each other)

Thus overall resultant force acting on the 2kg mass is 8N to the right... We all know that F=ma. Therefore, at 1s the acceration is thus 8/2=4m/s^2. Velocity also 4m/s

After 1s. When 8N is taken away, applying newton's law of motion, assuming zero frictional force. We'll have Acceration=0, velocity=4m/s.

My 2 cent's worth...
erroneous assuption that the tension in the string is 17.81N. let's put it this way, if the tension IS 17.81N, then the FBD of the block P will have resultant force of 0N, and if there is zero resultant force, why is the block P accelerating in the first second?

#### Kira

##### New Member
wacko said:
erroneous assuption that the tension in the string is 17.81N. let's put it this way, if the tension IS 17.81N, then the FBD of the block P will have resultant force of 0N, and if there is zero resultant force, why is the block P accelerating in the first second?
17.81N acts to the left. 9.81N acts to the right.... Resultant force is 8N acting to the left. Thus block move left...

#### wacko

##### New Member
Kira said:
17.81N acts to the left. 9.81N acts to the right.... Resultant force is 8N acting to the left. Thus block move left...
oh man, i dunno which part of the sahara desert i lost you at, i was referring to block P in that post...

#### bearycute

##### New Member
wacko said:
erm, i did say "minus gravity and reactional forces"...
paiseh... my bad... cock eye haha... my apologies....

#### bearycute

##### New Member
Kira said:
17.81N acts to the left. 9.81N acts to the right.... Resultant force is 8N acting to the left. Thus block move left...
er... u r right that the resultant force is 8... but that is the resultant force on the system which consist of 3 masses adding to 4kg... thus accln is 2...hope that is clear... guess i might have to draw free body diagram on each of the masses to show if still not clear...

#### Kira

##### New Member
bearycute said:
er... u r right that the resultant force is 8... but that is the resultant force on the system which consist of 3 masses adding to 4kg... thus accln is 2...hope that is clear... guess i might have to draw free body diagram on each of the masses to show if still not clear...
But doesn't the 1kg mass cancel out each other? i.e. P aids motion, Q resist motion?????

:dunno:

#### wacko

##### New Member
guys guys, quit bickering and reflect upon my diagram will ya? the answer is in there... :sweatsm:

(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)

#### bearycute

##### New Member
Kira said:
But doesn't the 1kg mass cancel out each other? i.e. P aids motion, Q resist motion?????

:dunno:
to a certain extent yes... which is y initially it was not accelerating without the application of the 8N force. (newton's first law: an object tends to stay at its rest state or uniform motion in a straight line unless a net external force acts on it).

W the application of a 8N force on the system ( which consist of 3 masses), a net force results. Applying newton's second law: F = ma, the resultant force F = 8N. m = (1+2+1)= 4 kg. Thus the accln is 2. all the masses share the same accln = 2 for the 1st second in which the force is applied.

v = u + at. Since initial speed is zero, final speed v = 0 + 2t => v = 2 since t = 1. After one second when the 8N force is removed, the system moves at constant speed as again now the net force acting on the system is zero(newton's 1st law) ignoring all resistive forces....

#### innovas1

##### New Member
:dunno: which one is right?

#### sevenWAYS

##### New Member
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 ?

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