Particle projected up a rough slope
By Anonymous on Tuesday, November 21, 2000 - 10:23 am :

Hello,

My brain simply won't tick on this question. Part (a) was ok, but part (b) tried using constant acceleration but gave a silly answer. Can some one give a hand here please.

A particle P is projected up a line of greatest slope of a rough plane inclined at an angle arctan(5/12) to the horizontal. The coefficient of friction between the particle and the plane is 1/3. The particle is projected from point A with a speed of 30 ms-1 and comes to an instantaneous rest at point B.

(a) Show that, whilst P is moving up the plane, its acceleration is of magnitude (9/13)g and is directed down the plane.

Ok, this is what I did:
Resolve perpendicular to the plane, R=mgCos(theta )
So, F = (4/13)mg
Hence, using eqn of motion parallel to the motion, -(4/13)mg - mgSin(theta ) = ma
==> a = -(9/13)g

I can't do the parts below.....
(b) Find the distance AB, in metres to 3 s.f.

(c) Find the time, in seconds to 3 s.f. taken for P to move from A to B.

(d) Find the speed in ms-1 of the particle when it returns to A.

Thank you in advance!
By James Lingard (Jchl2) on Tuesday, November 21, 2000 - 10:57 am :
Hi,

Your answer to part (a) is fine.

For part (b), you can do it using the formulae for constant acceleration - have a look at it again and if you still can't do it then let me know what you did and what the 'silly answer' you got was, and I'll try and see what you've done wrong.

Part (c) can also be done with the constant acceleration formulae - in fact I did part (b) by first finding the answer to this part.

Part (d) is harder - here's one way of doing it.

If we call the acceleration down the slope a, then we have

m a = m gsinq- F

(remember the friction acts in the opposite direction now, as the motion is in the opposite direction), so

m a = m gsinq- m m gcosq

a = gsinq- m gcosq

a = g((5/13) - (1/3)(12/13))

a = g((5/13) - (4/13))

a = g(1/13) = g/13

Then you can use the appropriate constant acceleration formula to give you v2 and so v.

I hope that is of some help. If you're still stuck then please say.

James.

By Anonymous on Tuesday, November 21, 2000 - 02:43 pm :

Hi James,

Here's my attempt at (b).

u=30 (initial speed)
v=0 (at instant rest)
a= -(9/13)g (from part .a)
x=s=unknown (or is it component xmgCos(theta)?)

Using v2 = u2 + 2as
s = 650g = 6370m !!! (silly answer!!)

I then thought,
u=30
v=0
a= -(9/13)g (from part .a)
s= xmgCos(theta)
using v2 = u2 + 2as, but cannot cancel the m's????

then I tried energy,
KE lost = 0.5m302
PE gained = mgh = m x g x xmgSin(theta)
but again m's don't cancel.....

Any ideas?
Thanks for the help.

By James Lingard (Jchl2) on Tuesday, November 21, 2000 - 03:12 pm :
OK,

The first way you tried is probably the easiest way of doing this question, but you're just making a small mistake somewhere - the answer should be 650/g - here's my working:

v2 = u2 + 2a s

0 = 302 + 2(-9g/13)s

18g s/13 = 900

s = 900.13/(18g) = 100.13/(2g) = (1300/2)/g = 650/g

I hope you can see where you've gone wrong.

In your second method, I don't know what you mean when you write s = x m gcosq - what do you mean by x and by s?

Your third method will not work, unfortunately, because in this example, and indeed in any example where there is friction, energy is not conserved. Some of the particle's kinetic energy is used up in overcoming the friction, and will be converted into heat. It is for this reason that the answer to part (d) isn't 30ms-1, as it would be if there were no friction. (Can you see why this is true?)

Actually, the way I did this question first time round (actually second time, because I messed up the first time :-)) was slightly different (although essentially the same as your first method). I went:

v = u + at

0 = 30 - (9g/13)t

t = 130/(3g)

(t is the time it takes the particle to travel from A to B.)

then

s = ((u + v)/2)t

s = 15t

s = 15(130/(3g))

s = 5.130/g

s = 650/g

which (helpfully) gives you the answer to part (c) as well.

James.

By Anonymous on Tuesday, November 21, 2000 - 03:42 pm :

Hi James,

Thanks, I see where I made a mistake... I think. I haven't got my full working out in front of me but will check it.

I'll keep the fact that when friction is involved, that energies can not be used, as energy is not conserved. Thanks once again James :-))