Mechanics 2 (M2) Exam 22nd June 2012 - OCR (Not MEI)

Original post by Snakefingers13

Impulse was 4.16 Ns, KE lost was 5.82 J.
Acceleration of car was 0.125, max speed was 10.6.

Force caused by strut was 452.6 N
Force at A was 334N, 47.6 degrees below horizontal or bearing of around 223.

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I got my vertical force at A as 244N or something?

Reply 84

Original post by Snakefingers13

Impulse was 4.16 Ns, KE lost was 5.82 J.
Acceleration of car was 0.125, max speed was 10.6.

Force caused by strut was 452.6 N
Force at A was 334N, 47.6 degrees below horizontal or bearing of around 223.

This was posted from The Student Room's iPhone/iPad App

I get you now, the magnitude was 334N which if I remember is what I got too!

Reply 85

Original post by Snakefingers13

Impulse was 4.16 Ns, KE lost was 5.82 J.
Acceleration of car was 0.125, max speed was 10.6.

Force caused by strut was 452.6 N
Force at A was 334N, 47.6 degrees below horizontal or bearing of around 223.

This was posted from The Student Room's iPhone/iPad App

I remember getting all those answers I think so that's good

Reply 86

bbchen

My Answer1)

i) 4.16 Ns
ii) 5.824J

2 i) 0.125 ms^-2
ii) 10.56 ms^-1

3) i) 392N
ii) 333.5N --- 47.27 degree

4) i) 2.91m
ii) 10.2ms^-1

5) i) Ta=19.75N Tb= 44.25N
ii) 2.97 ms^-1

6) i) Va=0.2 ms^-1
ii) 4/15<e<2/5 (<+=)

7) i) show it
ii) 8.66 cm
iii) a<= 59.97 degree

Reply 87

lilipop

Should have done the last one correctly but I don't know where went wrong and I got some angle over 60, right answer should be 60 though.

Reply 88

bbchen

Original post by Anon 17

For minimum speed three of us at my school got v = 2.97m/s.

Not confident on 7iii) or 4ii), but other than that I though it was alright.

same

Reply 89

lilipop

Original post by bbchen

My Answer1)

i) 4.16 Ns
ii) 5.824J

2 i) 0.125 ms^-2
ii) 10.56 ms^-1

3) i) 392N
ii) 333.5N --- 47.27 degree

4) i) 2.91m
ii) 10.2ms^-1

5) i) Ta=19.75N Tb= 44.25N
ii) 2.97 ms^-1

6) i) Va=0.2 ms^-1
ii) 4/15<e<2/5 (<+=)

7) i) show it
ii) 8.66 cm
iii) a<= 59.97 degree

I remember that for Q5, A should be above B? so you probably wrote the answer other way round?

Reply 90

Original post by bbchen

My Answer1)

i) 4.16 Ns
ii) 5.824J

2 i) 0.125 ms^-2
ii) 10.56 ms^-1

3) i) 392N
ii) 333.5N --- 47.27 degree

4) i) 2.91m
ii) 10.2ms^-1

5) i) Ta=19.75N Tb= 44.25N
ii) 2.97 ms^-1

6) i) Va=0.2 ms^-1
ii) 4/15<e<2/5 (<+=)

7) i) show it
ii) 8.66 cm
iii) a<= 59.97 degree

6.ii) was different for me, I got e<2/3 I think.
Also the last one I got exactly 60 but perhaps you used a round answer or something.
Apart from those I got the same from what I remember.

Reply 91

bbchen

Original post by J.C

6.ii) was different for me, I got e<2/3 I think.
Also the last one I got exactly 60 but perhaps you used a round answer or something.
Apart from those I got the same from what I remember.

Hope you get an A in this paper

Reply 92

lilipop

Original post by J.C

6.ii) was different for me, I got e<2/3 I think.
Also the last one I got exactly 60 but perhaps you used a round answer or something.
Apart from those I got the same from what I remember.

yes, you are right

Reply 93

If it provides hope, then the last few grade boundaries for an A have been 52, 50 and 56 so it maybe it will be around 55 perhaps? I thought it was a still a tough paper..!

Reply 94

Sp3k7r0li7

Isn't e>2/3 as we know it doesn't bounce back otherwise there will be a colision between a and b no mater what as they are going towards each other (with no friction). So speed of b has to be greater than 2 to the right so it can't aford to lose to much speed. So e has to be 2/3<e<1

(edited 11 years ago)

Reply 95

Original post by Sp3k7r0li7

Isn't e>2/3 as we know it doesnt bounce back otherwise there will be a colision no mater what as they are going towards each other. So speed of b has to be greater than 2. So e has to be 2/3<e<1

I put e>2/3 too, but I think you ended up with negatives and by inequality laws the signs reverse when dividing by negatives so it would be e<2/3 I think anyway..

Reply 96

Original post by As_Dust_Dances_

I put e>2/3 too, but I think you ended up with negatives and by inequality laws the signs reverse when dividing by negatives so it would be e<2/3 I think anyway..

Yeah I switched it around before dividing to avoid that but I did it both ways and still got e<2/3. I didn't really think about it physically though, just relied on the algebra.

Reply 97

Sp3k7r0li7

Ok we know that both a and b are going to the right. We also know that b*e is the speed of b to the right. The smaller the e the smaller b*e. we don't want a to catch up to b as there will be another collision. So e has to be > than a certain value. So e>2/3. If there is a mistake in my reasoning please point it out.

Edit: Overall an easy paper I made 2 stupid mistakes. For the last part of last question I used "a" I worked out in the last part instead of x(bar). Also I didn't realise there were two strings so I treated the tension to be the same. Other that that I'm confident I did everything correct.

(edited 11 years ago)

Reply 98

Original post by Sp3k7r0li7

Ok we know that both a and b are going to the right. We also know that b*e is the speed of b to the right. The smaller the e the smaller b*e. we don't want a to catch up to b as there will be another collision. So e has to be > than a certain value. So e>2/3. If there is a mistake in my reasoning please point it out.

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I'm not sure but I'd have to look at the question again to be honest. I guess just wait for solutions and see who's right, then discuss it.

Reply 99

Mr M (jr)