# [M2] - Edexcel Jan '04 - Question 3Watch

This discussion is closed.
#1
(Original post by Edexcel - January 2003 - Question 3)
A particle P of mass 2kg is projected from a point A up a lone of greates slope AB of a fixed plane. The plane is inclined at an angle of 30 degrees to the horizontal and AB = 3m with B above A. The speed of P at A is 10ms^-1

Assuming the plane is smoote

(a) Fine the speed of P at B
What I did for this question was:

I decided to go with the "Motion with Constant Acceleration" way of solving it and said:

u = 10
v = ?
a = -(2g Sin 30) = -g
s = 3

then I used the equation: v^2 = u^2 + 2as

and got:

v^2 = 10^2 + 2 (-9.8)(3)
v^2 = 100 - 58.8
=> v = 6.4187 = 6.42 ms^-1 (3 sf)

Now the answer says it should be 8.4
they say the value of Acceleration (a) should be -(g Sin 30) - and not -(2g Sin 30)

I can't work out how the Acc. could be = -(g Sin30) - can someone check this? thanks.

MdSalih
0
14 years ago
#2
(Original post by MdSalih)
What I did for this question was:

I decided to go with the "Motion with Constant Acceleration" way of solving it and said:

u = 10
v = ?
a = -(2g Sin 30) = -g
s = 3

then I used the equation: v^2 = u^2 + 2as

and got:

v^2 = 10^2 + 2 (-9.8)(3)
v^2 = 100 - 58.8
=> v = 6.4187 = 6.42 ms^-1 (3 sf)

Now the answer says it should be 8.4
they say the value of Acceleration (a) should be -(g Sin 30) - and not -(2g Sin 30)

I can't work out how the Acc. could be = -(g Sin30) - can someone check this? thanks.

MdSalih
Ans is 8.4

Energy considerations

Pe gain= ke loss
2g*1.5=100-1v^2
29.4=100-v^2
v=8.4

will check other method, but someone will arrive first!

Aitch
0
14 years ago
#3
(Original post by MdSalih)
What I did for this question was:

I decided to go with the "Motion with Constant Acceleration" way of solving it and said:

u = 10
v = ?
a = -(2g Sin 30) = -g
s = 3

then I used the equation: v^2 = u^2 + 2as

and got:

v^2 = 10^2 + 2 (-9.8)(3)
v^2 = 100 - 58.8
=> v = 6.4187 = 6.42 ms^-1 (3 sf)

Now the answer says it should be 8.4
they say the value of Acceleration (a) should be -(g Sin 30) - and not -(2g Sin 30)

I can't work out how the Acc. could be = -(g Sin30) - can someone check this? thanks.

MdSalih
S is 1.5, not 3

It is the height gained.

Aitch
0
#4
Ohhh... I know what I did wrong in mine

I forgot to use N2L (F=ma) to work out a... and just used the Force as a...

Silly me

Thanks Aitch

MdSalih
0
#5
(Original post by Aitch)
S is 1.5, not 3

It is the height gained.

Aitch
The displacement of particle P is 3 along the slope... I'm applying the eqn on motion parallel to the plane.

MdSalih
0
14 years ago
#6
(Original post by MdSalih)
The displacement of particle P is 3 along the slope... I'm applying the eqn on motion parallel to the plane.

MdSalih
The vertical distance travelled is 1.5. This is what matters. Since the plane is smooth, the situation is the same as that of a projectile travelling vertically upward for a distance of 1.5. No force other than gravity involved in this model.

Aitch
0
14 years ago
#7
(Original post by Aitch)
The vertical distance travelled is 1.5. This is what matters. Since the plane is smooth, the situation is the same as that of a projectile travelling vertically upward for a distance of 1.5. No force other than gravity involved in this model.

Aitch
...or..

g is a vertical force, but resolve it along the slope and perp. to the slope and you get (-)g sin 30 along the slope...

...THEN you can use s=3.

Aitch
0
14 years ago
#8
(Original post by Aitch)
Ans is 8.4

Energy considerations

Pe gain= ke loss
2g*1.5=100-1v^2
29.4=100-v^2
v=8.4

will check other method, but someone will arrive first!

Aitch
I've just noticed that you multiplied the acceleration by the mass. The mass of the particle is not relevant to v^2=u^2+2as.

Only if you use energy considerations do you need the mass

PE gained = KE lost

here mgh = (m/2)(u^2 -v^2)

note that m/2 = 1 so here

mgh = (u^2 -v^2)

Aitch
0
#9
(Original post by Aitch)
I've just noticed that you multiplied the acceleration by the mass. The mass of the particle is not relevant to v^2=u^2+2as.

Only if you use energy considerations do you need the mass

PE gained = KE lost

here mgh = (m/2)(u^2 -v^2)

note that m/2 = 1 so here

mgh = (u^2 -v^2)

Aitch
I haven't include the mass in my eqn...

The various ways you have mentioned are all correct...

However, the problem with my specific method of solving it is that I resolved the resistive force due to gravity... and then forgot to use N2L to get the acceleration from that resitive force (you need to include mass for N2L - if you were refering to that). I just subbed in the force as acc.. which is a big no no :P

MdSalih
0
14 years ago
#10
(Original post by MdSalih)
I haven't include the mass in my eqn...

The various ways you have mentioned are all correct...

However, the problem with my specific method of solving it is that I resolved the resistive force due to gravity... and then forgot to use N2L to get the acceleration from that resitive force (you need to include mass for N2L - if you were refering to that). I just subbed in the force as acc.. which is a big no no :P

MdSalih
OK. I think it's a good idea when you're doing questions like this (for homework) to do both methods, partly to get a check, but mainly so that you're confident with both routes.

Aitch
0
#11
(Original post by Aitch)
OK. I think it's a good idea when you're doing questions like this (for homework) to do both methods, partly to get a check, but mainly so that you're confident with both routes.

Aitch
Yup.. true point...

MdSalih
0
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