# Edexcel M2

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#181

(Original post by

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.

**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.

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#182

(Original post by

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.

**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.

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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

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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#184

(Original post by

Yes, but I dont think anyone used the new velocity did they? They used its components...

**imasillynarb**)Yes, but I dont think anyone used the new velocity did they? They used its components...

**speed**to the horizontal, think of projectiles. As the initial velocity was just horizontal, it makes no difference.

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#185

(Original post by

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.

**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.

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#186

(Original post by

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

**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

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#187

**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

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#188

(Original post by

For question 1b, 44.4 kw doesn't seem to ring quite true with me.

Can anyone post the question?

**jonnymcc2003**)For question 1b, 44.4 kw doesn't seem to ring quite true with me.

Can anyone post the question?

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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#189

(Original post by

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.

**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.

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#190

(Original post by

Lol, you nearly got me hyperventilating there imasillynarb Glad someone agrees with me.

**Mysticmin**)Lol, you nearly got me hyperventilating there imasillynarb Glad someone agrees with me.

hyperventilating over two marks?

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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?

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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#192

(Original post by

Oh well, thats another question Ive done wrong. So thats minus 15 marks so far.

**imasillynarb**)Oh well, thats another question Ive done wrong. So thats minus 15 marks so far.

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#193

(Original post by

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?

**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?

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#194

**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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#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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#196

(Original post by

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'?

**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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#197

(Original post by

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?

**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?

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#198

(Original post by

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

**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

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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!!!

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!!!

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#200

**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?

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