Particles $$\(P\)$$ and $$\(Q\)$$ have masses $$\(m \mathrm{~kg}\)$$ and $$\(2 m \mathrm{~kg}\)$$ respectively. The particles are initially held at rest $$\(6.4 \mathrm{~m}\)$$ apart on the same line of greatest slope of a rough plane inclined at an angle $$\(\alpha\)$$ to the horizontal, where $$\(\sin \alpha=0.8\)$$ (see diagram). Particle $$\(P\)$$ is released from rest and slides down the line of greatest slope. Simultaneously, particle $$\(Q\)$$ is projected up the same line of greatest slope at a speed of $$\(10 \mathrm{~m} \mathrm{~s}^{-1}\)$$. The coefficient of friction between each particle and the plane is $$\(0.6\)$$. Find the time for which the particles are in motion before they collide. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w21_qp_42 Year：2021 Question No：7(b)
Answer:
For \(P: \quad m g \sin \alpha-0.6 R=m a\), leading to \(8 m-3.6 m=m a\) \(\left[R=m g \cos \alpha=6 m, a=4.4 \mathrm{~ms}^{-2}\right]\)
\(Q\) comes to rest when \(10-11.6 T_{1}=0, \quad\left[T_{1}=\frac{25}{29}=0.862\right]\)
For \(P \quad s_{P(\text { down })}=\frac{1}{2} \times 4.4 \times T_{1}^{2} \quad[=1.635]\)
For \(Q \quad s_{Q(\text { up })}=10 T_{1}+\frac{1}{2} \times(-11.6) \times T_{1}^{2} \quad[=4.31]\)
\(
d=6.4-s_{P(\text { down })}-s_{Q(u p)} \quad[=0.455]
\)
and to find \(T_{2}[=0.12]\) by using \(d=s_{P 2}-s_{Q 2}=\left(4.4 T_{1}\right) \times T_{2}\) \(\left[s_{P 2}\right.\) and \(s_{Q 2}\) are distances travelled by \(P\) and \(Q\) in time \(\left.T_{2}\right]\)
Time before collision \(=\left[t=T_{1}+T_{2}=0.862+0.12=\right] 0.982\)
Alternative method for Question
For \(P: \quad m g \sin \alpha-0.6 R=m a\), leading to \(8 m-3.6 m=m a\) \(\left[R=m g \cos \alpha=6 m, a=4.4 \mathrm{~ms}^{-2}\right]\)
\(Q\) comes to rest when \(10-11.6 T_{1}=0, \quad\left[T_{1}=\frac{25}{29}=0.862\right]\)
For \(P \quad s_{P(\text { down })}=\frac{1}{2} \times 4.4 \times t^{2}\)
For \(Q \quad s_{Q(\text { up })}=10 T_{1}+\frac{1}{2} \times(-11.6) T_{1}^{2}-\frac{1}{2} \times 4.4\left(t-T_{1}\right)^{2}\)
\(\frac{1}{2} \times 4.4 t^{2}+10 T_{1}+\frac{1}{2} \times(-11.6) T_{1}^{2}-\frac{1}{2} \times 4.4\left(t-T_{1}\right)^{2}=6.4\)
Time before collision is \(t=0.982 \mathrm{~s}\)
Special case for those who do not take into account the fact that \(Q\) comes to rest and then changes its direction(only 4 marks)
For \(P: \quad m g \sin \alpha-0.6 R=m a\), leading to \(8 m-3.6 m=m a\) \(\left[R=m g \cos \alpha=6 m, a=4.4 \mathrm{~ms}^{-2}\right]\)
For \(P \quad s_{p(\text { down })}=(\pm) \frac{1}{2} \times 4.4 t^{2}\)
For \(Q \quad s_{q(\mathrm{up})}=(\pm) 10 t+\frac{1}{2} \times(-11.6) t^{2}\)
\(s_{p}+s_{q}=6.4\) leading to \(\frac{1}{2} \times 4.4 t^{2}+10 t+\frac{1}{2} \times(-11.6) t^{2}=6.4\)
Time that particles are in motion before collision \(=t=1 \mathrm{~s}\)
\(Q\) comes to rest when \(10-11.6 T_{1}=0, \quad\left[T_{1}=\frac{25}{29}=0.862\right]\)
For \(P \quad s_{P(\text { down })}=\frac{1}{2} \times 4.4 \times T_{1}^{2} \quad[=1.635]\)
For \(Q \quad s_{Q(\text { up })}=10 T_{1}+\frac{1}{2} \times(-11.6) \times T_{1}^{2} \quad[=4.31]\)
\(
d=6.4-s_{P(\text { down })}-s_{Q(u p)} \quad[=0.455]
\)
and to find \(T_{2}[=0.12]\) by using \(d=s_{P 2}-s_{Q 2}=\left(4.4 T_{1}\right) \times T_{2}\) \(\left[s_{P 2}\right.\) and \(s_{Q 2}\) are distances travelled by \(P\) and \(Q\) in time \(\left.T_{2}\right]\)
Time before collision \(=\left[t=T_{1}+T_{2}=0.862+0.12=\right] 0.982\)
Alternative method for Question
For \(P: \quad m g \sin \alpha-0.6 R=m a\), leading to \(8 m-3.6 m=m a\) \(\left[R=m g \cos \alpha=6 m, a=4.4 \mathrm{~ms}^{-2}\right]\)
\(Q\) comes to rest when \(10-11.6 T_{1}=0, \quad\left[T_{1}=\frac{25}{29}=0.862\right]\)
For \(P \quad s_{P(\text { down })}=\frac{1}{2} \times 4.4 \times t^{2}\)
For \(Q \quad s_{Q(\text { up })}=10 T_{1}+\frac{1}{2} \times(-11.6) T_{1}^{2}-\frac{1}{2} \times 4.4\left(t-T_{1}\right)^{2}\)
\(\frac{1}{2} \times 4.4 t^{2}+10 T_{1}+\frac{1}{2} \times(-11.6) T_{1}^{2}-\frac{1}{2} \times 4.4\left(t-T_{1}\right)^{2}=6.4\)
Time before collision is \(t=0.982 \mathrm{~s}\)
Special case for those who do not take into account the fact that \(Q\) comes to rest and then changes its direction(only 4 marks)
For \(P: \quad m g \sin \alpha-0.6 R=m a\), leading to \(8 m-3.6 m=m a\) \(\left[R=m g \cos \alpha=6 m, a=4.4 \mathrm{~ms}^{-2}\right]\)
For \(P \quad s_{p(\text { down })}=(\pm) \frac{1}{2} \times 4.4 t^{2}\)
For \(Q \quad s_{q(\mathrm{up})}=(\pm) 10 t+\frac{1}{2} \times(-11.6) t^{2}\)
\(s_{p}+s_{q}=6.4\) leading to \(\frac{1}{2} \times 4.4 t^{2}+10 t+\frac{1}{2} \times(-11.6) t^{2}=6.4\)
Time that particles are in motion before collision \(=t=1 \mathrm{~s}\)
Knowledge points:
4.1.1 identify the forces acting in a given situation； e.g. by drawing a force diagram.
4.1.2 understand the vector nature of force, and find and use components and resultants； Calculations are always required, not approximate solutions by scale drawing.
4.1.4 understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component
4.1.6 understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = nR or F G nR, as appropriate
4.2.4 use appropriate formulae for motion with constant acceleration in a straight line. （Questions may involve setting up more than one equation, using information about the motion of different particles.）
4.4.1 apply Newton’s laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
4.4.3 solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
Solution:
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