M1 question

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For part a), I done :

Conservation Of Momentum (→), (m1u1) + (m2u2) = (m1v1) + (m2v2)

(3x1.5) + (2.5x-4) = (1.5x-2.5) + (2.5x-v)
4.5 - 10 = -3.75 -2.5v
-5.5 = -3.75 -2.5v
-1.75 = -2.5v
0.7 m/s = v

Then for part b, how would I determine whether the motion changes or not?

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In the above question, I done part a) and b), and got:

Normal Reaction = 23.0 N
Force, X = 17.6 N

But I am confused on how you do part c)?

Reply 1

TenOfThem

since you have v>0 it must be moving in the +ve direction

Reply 2

TenOfThem

if x is removed friction wil act up the plane rather than down ... is the friction enough to hold it in place?

Reply 3

Chrisofsmeg

For question 4 (c), I think if you work out F_MAX (=

\mu

R), then use the relationship:

F \leq \mu

R

You should be able to work it out from there. If the force is smaller than F_MAX then will it remain in equilibrium?

Reply 4

the_flying_walrus

for c)

If the force is removed, then the force acting down the plane is the component of the weight acting down the plane 23 x sin (20).

We know that F <= uR, and as

uR = 0.4 x 23 = 9.2
23 sin (20) = 7.86...

Then the frictional force and the component of weight acting down the slope will be equal (as the weight acting down the slope does not exceed the maximum force which could occur due to friction, and thus it remains in equilibrium.

Hope this helps (and I hope its right :P) :smile:

Reply 5

raheem94

When X is removed, the particle will try to move down. As friction always opposes motion so it will act upward(parallel to the plane) to keep the ball on the plane.
Compare the downward(parallel to the plane) and upward(parallel to the plane) forces.
You don't need to use f=ma, just compare the forces.

Reply 6

xXxiKillxXx

Original post by TenOfThem

since you have v>0 it must be moving in the +ve direction

I defined the positive direction as being →(from left to right)..

But if you see my calculations, I done ''(2.5x-v)'' (I used a negative value for v), because I defined v as moving in the opposite direction to the positive direction..
So surely if it must be ''moving in the +ve direction'' then the particle's direction DOES change

Reply 7

TenOfThem

Original post by xXxiKillxXx

(edited 12 years ago)

Reply 8

xXxiKillxXx

Original post by TenOfThem

correct

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As you can see above, in the mark scheme, they defined v as moving in the same direction as the +ve direction, so they got:

v = –0.7 m/s

But they said that the direction remains unchanged; since their answer is negative, doesn't that mean that v will be in the opposite direction to the positive direction? (as in the direction does change)

so their answer was that the direction of motion of Q does not change..
I put it DOES change, so would they consider my answer and mark it

Reply 9

TenOfThem

Sorry I missed your second -v

so you had v = 0.7 but the particle moving at -v so -0.7 so no change in direction

Sorry

Reply 10

xXxiKillxXx

Original post by Chrisofsmeg

For question 4 (c), I think if you work out F_MAX (=

\mu

R), then use the relationship:

F \leq \mu

R

You should be able to work it out from there. If the force is smaller than F_MAX then will it remain in equilibrium?

I worked out Fmax and got 23x0.4= 9.2, and 'F' (=X) = 16.7 N

so Fmax< Force X ..

Reply 11

Chrisofsmeg

Original post by xXxiKillxXx

I worked out Fmax and got 23x0.4= 9.2, and 'F' (=X) = 16.7 N

so Fmax< Force X ..

Actually I think I made a boo boo - I misread the question. Sorry about that

If you remove the force X, then the only force that acts on it to move it down the slope will be its weight.

The weight of P acting down the slope will be 2.5cos(70) or 2.5sin(20), whichever you prefer.

Using that as your new F, you can show that

F \leq \mu

R

That make sense?

Reply 12

xXxiKillxXx

Original post by Chrisofsmeg

F \leq \mu

R

That make sense?

Ohhh, so since Force X has been removed, we now assume that uR acts opposite to the component of the weight right? (so uR acts upwards)

Btw I think you mean 2.5gcos70 or 2.5gSin20 ?

Reply 13

Chrisofsmeg

Original post by xXxiKillxXx

Ohhh, so since Force X has been removed, we now assume that uR acts opposite to the component of the weight right? (so uR acts upwards)

Btw I think you mean 2.5gcos70 or 2.5gSin20 ?