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Mysticmin
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#181
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#181
(Original post by imasillynarb)
Thats what I did, it wanted the angle of DEFLECTION. Therefore you need to find the angle between the resultant, and the its initial velocity. You get an angle of 40.something degrees if you do that.

Everyone seems to think it was easy, but I dont think its as easy as people think.

For example, with the restituion question, partb) v3 might not be greater than v1, as f could be between 0 < f < 1, and if f was something small like 0.1, then they wouldnt of collided.
But it's inital was just a straight line, so it wants the angle between the new one and the inital, therefore the angle between the new velocity and the horizontal.
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Barny
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#182
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#182
(Original post by Mysticmin)
But it's inital was just a straight line, so it wants the angle between the new one and the inital, therefore the angle between the new velocity and the horizontal.
Yes, but I dont think anyone used the new velocity did they? They used its components...
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Savioahang
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#183
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#183
Question 1 for Jonnymcc2003 and others

A lorry of mass 1500kg moves along a straight horizontal road. The resistance to the motion of the lorry has magnitude 750N and the lorry's engine is working at a rate of 36kw
(a) Find the acceleration of the lorry when its speed is 20m/s

The lorry comes to a hill inclined a an angle x to the horizontal,where sinx = 1/10.The magnitude of the resistance to motion from non-gravitational forces remains 750N

The lorry moves up the hill at a constant speed of 20m/s.
(b)Find the rate at which the lorry's engine is now working.

Correct Solution:
(a)P=FV ,F=P/V=36000/20=1800N
1800-750=ma=1500a ,a=0.7m/s^2

(b)F=Fr+mgsinx=750+1500*9.8*1/10=2220N
P=FV=2220*20=44400W=44.4KW
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Mysticmin
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#184
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#184
(Original post by imasillynarb)
Yes, but I dont think anyone used the new velocity did they? They used its components...
Same thing, the angle between the two components is the angle of the new speed to the horizontal, think of projectiles. As the initial velocity was just horizontal, it makes no difference.
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m:)ckel
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#185
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#185
(Original post by imasillynarb)
Thats what I did, it wanted the angle of DEFLECTION. Therefore you need to find the angle between the resultant, and the its initial velocity. You get an angle of 40.something degrees if you do that.

Everyone seems to think it was easy, but I dont think its as easy as people think.
i was thinking about the 'deflection' part of the question while doing it, but i realised that the initial velocity was 30i, i.e. just a straight line to the right. the new velocity was 10i + 20j (or that's what i got at least). if you imagine the direction of this new vector, it is going in a north east(ish) direction. Therefore the particle (or whatever it was) has been deflected by an angle of arctan(2) degrees in an anti-clockwise direction
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Barny
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#186
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#186
(Original post by mockel)
i was thinking about the 'deflection' part of the question while doing it, but i realised that the initial velocity was 30i, i.e. just a straight line to the right. the new velocity was 10i + 20j (or that's what i got at least). if you imagine the direction of this new vector, it is going in a north east(ish) direction. Therefore the particle (or whatever it was) has been deflected by an angle of arctan(2) degrees in an anti-clockwise direction
Oh well, thats another question Ive done wrong. So thats minus 15 marks so far.
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Mysticmin
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#187
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#187
(Original post by mockel)
i was thinking about the 'deflection' part of the question while doing it, but i realised that the initial velocity was 30i, i.e. just a straight line to the right. the new velocity was 10i + 20j (or that's what i got at least). if you imagine the direction of this new vector, it is going in a north east(ish) direction. Therefore the particle (or whatever it was) has been deflected by an angle of arctan(2) degrees in an anti-clockwise direction
Lol, you nearly got me hyperventilating there imasillynarb Glad someone agrees with me.
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mensandan
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#188
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#188
(Original post by jonnymcc2003)
For question 1b, 44.4 kw doesn't seem to ring quite true with me.

Can anyone post the question?
Okay dude

1) A lorry of mass 1500kg moves along a straight horizontal road. The resistance to the motion of the lorry has magnitude 750 N and the lorry's engine is working at 36kW.

(a) Find the acceleration of the lorry when its speed is 20ms¯1 (4)

The lorry comes to a hill inclined at an angle ß to the horizontal, where sin ß = 1/10. The magnitude of the resistance to motion from non-gravitational forces remains 750 N.

The lorry moves up the hill at a constant speed of 20ms¯1

(b) Find the rate at which the lorry's engine is now working. (3)

By the way I agree with your answer. Cool
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m:)ckel
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#189
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#189
(Original post by imasillynarb)
For example, with the restituion question, partb) v3 might not be greater than v1, as f could be between 0 < f < 1, and if f was something small like 0.1, then they wouldnt of collided.
what was the actual question for 3b)? can't really remember....was there a part c)? and if so was that the part where you had to find the range of 'f'?
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Ralfskini
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#190
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#190
(Original post by Mysticmin)
Lol, you nearly got me hyperventilating there imasillynarb Glad someone agrees with me.

hyperventilating over two marks?
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jonnymcc2003
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#191
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#191
I know what I have done. are you sure the angle wasn't sin 1/20 ?

How many marks you reckon I'll lose if I just got the angle wrong, which then made the answer wrong, despite getting all the method right?
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m:)ckel
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#192
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#192
(Original post by imasillynarb)
Oh well, thats another question Ive done wrong. So thats minus 15 marks so far.
where were the other marks?
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m:)ckel
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#193
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#193
(Original post by jonnymcc2003)
I know what I have done. are you sure the angle wasn't sin 1/20 ?

How many marks you reckon I'll lose if I just got the angle wrong, which then made the answer wrong, despite getting all the method right?
i think it was 1/10, but not quite sure...
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mensandan
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#194
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#194
(Original post by jonnymcc2003)
I know what I have done. are you sure the angle wasn't sin 1/20 ?

How many marks you reckon I'll lose if I just got the angle wrong, which then made the answer wrong, despite getting all the method right?

I'm certain it was sin A = 1/10 but I guess it will only affect the answer not method therefore 1 mark
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jonnymcc2003
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#195
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#195
For the collisions questions part b, did you say they would collide simply because they were heading towards each other, and e could not be negative?
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mensandan
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#196
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#196
(Original post by mockel)
what was the actual question for 3b)? can't really remember....was there a part c)? and if so was that the part where you had to find the range of 'f'?

In part b you had to show that there would be a second collision given that e=0.4

In part c you had to find the range of values of f to force a second collision given e = 0.8
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m:)ckel
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#197
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#197
(Original post by jonnymcc2003)
For the collisions questions part b, did you say they would collide simply because they were heading towards each other, and e could not be negative?
well, i substituted e=0.4 into my eqautions for v1 and v3 (velocity of 2nd one after hitting the wall), and i found that v3 must be negative, and v1 must be positive, so they are moving towards each other and will therefore collide
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m:)ckel
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#198
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#198
(Original post by mensandan)
In part b you had to show that there would be a second collision given that e=0.4

In part c you had to find the range of values of f to force a second collision given e = 0.8
oh yeah, i remember know, thanks
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Silly Sally
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#199
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#199
I thought overall the paper was ok

I didn't get all the marks for that 7 mark ladder question. Also i couldn't do parts b and c for the collisions question in question 6 - the part involving f - that was just confusing!!! :confused:
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mensandan
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#200
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#200
(Original post by jonnymcc2003)
For the collisions questions part b, did you say they would collide simply because they were heading towards each other, and e could not be negative?
Yeah that makes sense, I used the fact that they were moving in opposite directions given that f was in the velocity of Q therefore a collision was inevitable. I just related everything back to the original direction of projection of P.
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