AQA Mechanics 2 Unofficial Mark Scheme

Did anyone get 2 rootlg?

No, tension was root(4gl) which simplified, was 2root(gl). Which gives non negative tension.


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Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
1. a) 9.6
b) i) 24.3
ii) 12.7
iii) No Air Resistance
2.
a) $a = (8-4x^3)i - 18e^{-3x}j$
b) i) $F = (16-8x^3)i - 36e^{-3x}j$
ii) 8.20
c) $r = (4x^2- \frac{x^5}{5} + 3)i - (3+ 2e^{-3x})j$
3. a) By symmetry
b)10.4 cm
c) 16.5 degrees
d) Explanation Q
4. a)78.4
b) 41.4 degrees
c) r=2.89
5. Lots of very long answers, will upload later if I have time. Part d) was $sqrt(5gl/2)$
6. a) Show that Q
b) $\frac{g-(g-\lambda\mu)e^{-t\lambda} }{\lambda}$
7. a) Diagram
b) $tanx = \frac{1-2\mu^2}{4\mu}$
8. a) 17.0
b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?

Guys, it's $\sqrt{5gl}$ since we have $u^2=v^2+4gl$ and we require $T \geq 0$ at the top of the circle, hence $\Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}$.

How can so many people not see that it's just 2 root lg? I'm confused

Mate this exact question has come up before word for word and the answer is 2rootlg

Guys, it's $\sqrt{5gl}$ since we have $u^2=v^2+4gl$ and we require $T \geq 0$ at the top of the circle, hence $\Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}$.

Guys, it's $\sqrt{5gl}$ since we have $u^2=v^2+4gl$ and we require $T \geq 0$ at the top of the circle, hence $\Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}$.

Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
1. a) 9.6
b) i) 24.3
ii) 12.7
iii) No Air Resistance
2.
a) $a = (8-4x^3)i - 18e^{-3x}j$
b) i) $F = (16-8x^3)i - 36e^{-3x}j$
ii) 8.20
c) $r = (4x^2- \frac{x^5}{5} + 3)i - (3+ 2e^{-3x})j$
3. a) By symmetry
b)10.4 cm
c) 16.5 degrees
d) Explanation Q
4. a)78.4
b) 41.4 degrees
c) r=2.89
5. Lots of very long answers, will upload later if I have time. Part d) was $sqrt(5gl/2)$
6. a) Show that Q
b) $\frac{g-(g-\lambda\mu)e^{-t\lambda} }{\lambda}$
7. a) Diagram
b) $tanx = \frac{1-2\mu^2}{4\mu}$
8. a) 17.0
b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?

Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from

Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from

Guys, it's $\sqrt{5gl}$ since we have $u^2=v^2+4gl$ and we require $T \geq 0$ at the top of the circle, hence $\Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}$.

Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from