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Mechanics M1: Post your doubt here

qffdhruba

qffdhruba

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minie23 said:
sagar65265

would you mind explaining me no. 7 ? :s

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Click to expand...

U can find acceleration by:

a = -0.12/0.15 = -0.8 (through F=ma)

So there is a negative acceleration, due to the resistive frictional force of 0.12N

at 2s the velocity is:

v = 3 + 2(-0.8) = 1.4 (through v=u + at)

KE after collision:

(0.5(0.15)(1.4)^2) - 0.072 = 0.075

Velocity after collision:
root(2*0.075/0.15) = -1 (the velocity is negative because the motion is opposite to the initial motion) (through K.E=0.5mv^2)

Frictional force remains the same, therefore deceleration is the same:
Therefore:

t= 2 + (1/0.8) = 3.25

U can do the displacement time graph through finding the area under the velocity-time graph......both are uploaded......
 

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minie23

minie23

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qffdhruba said:
U can find acceleration by:

a = -0.12/0.15 = -0.8 (through F=ma)

So there is a negative acceleration, due to the resistive frictional force of 0.12N

at 2s the velocity is:

v = 3 + 2(-0.8) = 1.4 (through v=u + at)

KE after collision:

(0.5(0.15)(1.4)^2) - 0.072 = 0.075

Velocity after collision:
root(2*0.075/0.15) = -1 (the velocity is negative because the motion is opposite to the initial motion) (through K.E=0.5mv^2)

Frictional force remains the same, therefore deceleration is the same:
Therefore:

t= 2 + (1/0.8) = 3.25

U can do the displacement time graph through finding the area under the velocity-time graph......both are uploaded......
Click to expand...
Thank you soooo much (Y)
 
Talhakhan

Talhakhan

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Which Mechanics M1 paper has been the most difficult in the last decade.. I mean which had the lowest grade threshold. I want to attempt that paper plz if anyone have any idea plz share. ;-)
 
L

leosco1995

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JalalKaiser said:
Q5, part ii? :|

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Click to expand...
S1 = 30 N
S2 = 50 N

Take the system as a whole:

total mass * acceleration = net force
(3 + 2) * a = 50 - (1.6 + 4) (tension cancels out when taking the system as a whole, so don't bother considering it)

5a = 44.6
a = 8.88 ms^-2

Now just consider the motion of any particle, say A:

30 - T - 1.6 = 8.88(3)
T = 1.76 N

Or if you want to consider the motion of particle B,

T + 20 - 4 = 8.88(2)
T + 16 = 17.76
T = 1.76 N
 
Ahmedraza73

Ahmedraza73

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qffdhruba said:
U can find acceleration by:

a = -0.12/0.15 = -0.8 (through F=ma)

So there is a negative acceleration, due to the resistive frictional force of 0.12N

at 2s the velocity is:

v = 3 + 2(-0.8) = 1.4 (through v=u + at)

KE after collision:

(0.5(0.15)(1.4)^2) - 0.072 = 0.075

Velocity after collision:
root(2*0.075/0.15) = -1 (the velocity is negative because the motion is opposite to the initial motion) (through K.E=0.5mv^2)

Frictional force remains the same, therefore deceleration is the same:
Therefore:

t= 2 + (1/0.8) = 3.25

U can do the displacement time graph through finding the area under the velocity-time graph......both are uploaded......
Click to expand...
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_4.pdf
help me with question 6 (iii)
Its urgent
 
S

sagar65265

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minie23 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_41.pdf
A lil help for no. 7 please :(
Click to expand...

Man, is this a tough question! The examiner's report has made it clear that this question bugged students the most, but anyways, here's a stab at it:

Since we know the system is in equilibrium, the vector sum of the forces at any point in the system (any point on the rope, on the rod, on the ring, etc) should be equal to zero, else that point will accelerate, making the "equilibrium statement" a wrong one.
However, the easiest point to use is the point at which the 8N force is acting, since that is the only point where all three forces (both tension force and the 8N force) are acting.

If we resolve the forces at that point into horizontal components:
(If the Right Direction is positive and the Left Direction is negative)

8N = Tension(BC) * cosA + Tension(AB) * cosB
A is the angle between Section BC and the horizontal, and B is the angle between Section AB and the horizontal. Together the make up the right angle there.

A look at the diagram reveals that the angle between Section BC and the upwards direction is equal to the angle ACB - it's an angle along the transversal BC.
Similarly, the angle between Section AB and the downwards direction is equal to the angle CAB.

These transformations now give us a new equation:

8N = Tension(BC) * sin(ACB)+ Tension(AB) * sin(CAB)
(This step is just changing cos(90 - theta) to sin(theta))

Since hypotenuse = sprt(4 + 2.25) = 2.5,
sin(ACB) = 1.5/2.5 = 0.6
sin(CAB) = 2/2.5 = 0.8

So,
8N = 0.6 * Tension(BC) + 0.8 * Tension(AB)

One equation done. Phew!
Now we can take vertical components at B. Since the 8N force is a horizontal force, it has no component along the vertical axis. So the vertical components of the two Tension forces are equal and opposite to each other, since that point is at equilibrium:

Tension(BC) * cos(ACB) = Tension(AB) * cos(CAB)
cos(ACB) = 2/2.5 = 0.8
cos(CAB) = 1.5/2.5 = 0.6

So,
0.8 * Tension(BC) = 0.6 * Tension(AB)
Therefore, Tension(BC) = 0.75 * Tension(AB)

Putting this into the first equation,
8N = 0.6 * 0.75 * Tension(AB) + 0.8 * Tension(AB)
8N = Tension(AB) * 1.25
Tension(AB) = 8/1.25 = 6.4 Newtons.

Therefore, Tension(BC) = 0.75 * 6.4 Newtons = 4.8 Newtons

So the tension in BC = 4.8N and the tension in AB = 6.4N.

The next part is much simpler, i'll see if I can post it soon.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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minie23 said:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_41.pdf
A lil help for no. 7 please :(
Click to expand...

Okay, so for part (ii), you are again told that the system is in equilibrium, but that the ring is about to slide up the rod. Friction tries to prevent any relative motion between two surfaces, so if the ring is going to move upwards relative to the rod, the friction exerted BY the rod ON the ring is going to act in the other direction - straight downwards.
There are four forces acting on the rope, the tension from section AB of the string, the weight of the ring, the normal force from the rod and the frictional force between the rod and the ring.

The Frictional Force is acting downwards;
the Weight is acting downwards;
the normal force is acting perpendicular to the rod(and thus has no direct part to play in the vertical motion of the ring) and
the tension has a component upwards; we have to take the upwards component/ the positive y - component of the tension (up the rod) since the motion of the ring can only be up or down the rod, so the component of the tension causing the motion (or in situation, trying to cause the motion) of the ring has to be either up or down the rod; however, two forces are acting down the rod, so for equilibrium, the tension must act upwards.

Taking vertical forces into consideration,

Weight + Frictional Force = Tension(AB) * cos(CAB)
cos(CAB) = 1.5/2.5 = 0.6
Weight = 0.2 * 10 = 2N
Tension(AB) = 6.4N (from part (i))
Frictional Force = 0.6 * 6.4 - 2
Frictional force = 1.84N

Since the frictional force is dependent on two variables - the coefficient of friction (between the ring and the rod) and the normal force (acting on the ring), we need to find the normal force on the ring to find the coefficient of friction, which is asked in the question.

If we take components in the horizontal direction, Newton's Second Law effectively becomes

Normal Force = Tension(AB) * sin(CAB)
sin(CAB) = 2/2.5 = 0.8N
Tension(AB) = 6.4N
Normal Force = 6.4 * 0.8
Normal Force = 5.12N

Since Frictional Force/Normal Force = Coefficient of Friction,
Coef. of Friction = 1.84 / 5.12 = 0.359

Finally, done!

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_42.pdf

its 5 (ii) ..... the problem is why should we take (t-0.4) for Q if we take 't' for P ? I mean at any point the time taken by P to reach there is 't' then the time taken for Q to reach must t+0.4 cuz its coming later !! ?? am confused :\ :confused:
Click to expand...

You need to take a reference point in all these cases; by saying that you've taken "t" as time for Particle P, you're defining that t = 0 when P has just been thrown.

Imagine the situation; P is thrown at time t = 0, and 0.4 seconds later, Q is thrown upwards.
After time t passes, the variable of time is different for both of them - here's a thought:

After 0.4 seconds, where will P be? Somewhere in the air.
After 0.4 seconds, where will Q be? Still at the displacement s = 0! It won't be in the air!

Another way to see it:
If t = 10 seconds, P has been in the air for 10 seconds, whereas Q has been in the air only for 9.6 seconds - using the variable t for P defines the reference point for t = 0 as the time when P is thrown upwards. If Q is released later than P, it is always in the air for less time than P assuming that they are both still moving.

Hope this helped!
Good Luck for all your exams!
 
S

sagar65265

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JalalKaiser said:
Oh and, could you please explain the marking scheme method for the second part? I'd like to know both ways tbh.
Click to expand...

Sure, here it is:

Since the system is in equilibrium and just about to move, the friction applied by the rod on the ring is limiting friction.
Since the coefficient of friction is 0.4, the frictional force = 0.4 * Normal Force = 0.4 * (Tcos65 + 40) = 0.169T + 16 - that's where they get the right side of the equation from.

The force in the other direction is only the component of the force T along the rod, which is Tcos25 = 0.906T - and that's where they get the left side of the equation from.

Hope this helped!

Good Luck for all your exams!
 
I

Iadmireblue

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