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AS Physics P1 MCQs Preparation Thread.

I

IR Germany

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sagar65265 said:
Okay, so the concept here is that for a string or rubber band or any other elastic material that exerts a tension force, the assumption can be made that the tension is the same at any and all points on the string/rubber band. Another thing is that the force of tension on the object due to the rubber band always acts along the rubber band, i.e. acts parallel to the rubber band at the point where the rubber band is connected to the object. This tension always tries to bring the string back to it's original length.

Therefore, the forces on the object try to move the object in such a way that the rubber band is restored to a position where it has less tension, until it eventually reaches a position where it has no tension within itself. In this case, the force of tension cannot be acting from the center of each 50 mm section towards the object, because then that would accelerate the object in the leftwards direction; this will increase the tension in the rubber band, but this cannot happen, otherwise the elastic potential energy stored in the rubber band will be increasing without any external force doing work.

After all this, we can conclude that the tension acts parallel to the rubber band in the forwards direction, i.e. the tension forces on the object will in acting from X towards the center of the line XY, as shown in the attached diagram.

Now, since we know the force magnitudes and directions, we can solve the equations for the motion of the object. Since the force diagram is symmetric about the line passing along the trolley (from X in the direction of P) we can conclude that the net force in the direction from Y to Z is zero. Now, we need to find the force components in the direction along P.

Since tension is 4N throughout the rubber band, we can write:

4cos(theta) + 4cos(theta) = Net forward force

Where theta is the angle between the line of symmetry in the direction of P and the rubber band segments.
Therefore, since we have a classic 3-4-5 triangle here, where
i) the hypotenuse, XY, is 50mm
ii)half the length of YZ, is 40 mm
iii)the other side has length 30 mm by pythagoras's theorem.

Therefore, cos(theta) = (30 mm)/(50 mm) = 3/5
So Net forward force = 8cos(theta) = 8 * 3/5 = 24/5 = 4.8 Newtons = C

Hope this helped!
Good luck for all your exams!
Click to expand...


Can you help with this one too? number 31 http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf
 
mnotanerd

mnotanerd

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please help me with this mcq

32. A motorist travelling at 10 m s–1 can bring his car to rest in a
braking distance of 10 m. In what distance could he bring the car
to rest from a speed of 30 m s–1 using the same braking force?
A 17 m B 30 m C 52 m D 90 m answer is C
 
N

naoomi

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mnotanerd said:
please help me with this mcq

32. A motorist travelling at 10 m s–1 can bring his car to rest in a
braking distance of 10 m. In what distance could he bring the car
to rest from a speed of 30 m s–1 using the same braking force?
A 17 m B 30 m C 52 m D 90 m answer is C
Click to expand...

Are you sure the answer is C? I am getting D as an answer though i think I've done it correct.
 
Moi

Moi

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can someone please tell me when is voltage proportional to resistance and inversly proportional? same with current and resisitance / and Voltage and current ? i get so confused
 
enchanted era

enchanted era

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Two spheres A and B approach each other along the same
straight line with speeds uA and uB.
The spheres collide and move off with speeds vA and vB, both in
the same direction as the initial direction of sphere A, as shown
below.
Which equation applies to an elastic collision?
A uA + uB = vB – vA
B uA – uB = vB – vA
C uA – uB = vB + vA
D uA + uB = vB + vA..............i know the answer but can't understand how it is so. will be grateful if helped :')
 
Thought blocker

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Font wen hue said:
Anyone please help me with Winter 2010 p12 question 8 (kinematics)
Click to expand...
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37
 
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enchanted era said:
Two spheres A and B approach each other along the same
straight line with speeds uA and uB.
The spheres collide and move off with speeds vA and vB, both in
the same direction as the initial direction of sphere A, as shown
below.
Which equation applies to an elastic collision?
A uA + uB = vB – vA
B uA – uB = vB – vA
C uA – uB = vB + vA
D uA + uB = vB + vA..............i know the answer but can't understand how it is so. will be grateful if helped :')
Click to expand...
C ?
look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..
 
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Moi said:
can someone please tell me when is voltage proportional to resistance and inversly proportional? same with current and resisitance / and Voltage and current ? i get so confused
Click to expand...
Look there is one formula that says V = IR
It means V ∝ I and V 1/∝ to R
Keep this formula in MIND, V = IR so I = V/R, so R = V/I any formula you have same thing --> V ∝ I and V 1/∝ to R
 
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