Particles $$\(P\)$$ and $$\(Q\)$$ have masses $$\(m \mathrm{~kg}\)$$ and $$\(2 m \mathrm{~kg}\)$$ respectively. The particles are initially held at rest $$\(6.4 \mathrm{~m}\)$$ apart on the same line of greatest slope of a rough plane inclined at an angle $$\(\alpha\)$$ to the horizontal, where $$\(\sin \alpha=0.8\)$$ (see diagram). Particle $$\(P\)$$ is released from rest and slides down the line of greatest slope. Simultaneously, particle $$\(Q\)$$ is projected up the same line of greatest slope at a speed of $$\(10 \mathrm{~m} \mathrm{~s}^{-1}\)$$. The coefficient of friction between each particle and the plane is $$\(0.6\)$$. The particles coalesce on impact. Find the speed of the combined particle immediately after the impact. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w21_qp_42 Year：2021 Question No：7(c)
Answer:
\(u_{p(\text { down })}=0+4.4 \times 0.982[=4.3208]\)
\(u_{q(\mathrm{down})}=4.4 \times 0.12[=0.528]\)
\(\pm m \times 4.3208 \pm 2 m \times 0.528=\pm(m+2 m) v\)
[Correct equation is \(m \times 4.3208+2 m \times 0.528=\pm(m+2 m) v\) ]
Speed of combined particle immediately after impact \(=v=1.79 \mathrm{~ms}^{-1}\)
Special case for those who do not take into account the fact that \(Q\) comes to rest and then changes its direction
\(u_{p(\text { down })}=0+4.4 \times 1[=4.4]\)
\(u_{q(\mathrm{up})}=10-11.6 \times 1[=-1.6]\) so \(u_{q(\mathrm{down})}=1.6\)
\(\pm m \times 4.4 \pm 2 m \times 1.6=\pm(m+2 m) v\)
Speed of combined particle immediately after impact \(=v=2.53 \mathrm{~ms}^{-1}\)
\(u_{q(\mathrm{down})}=4.4 \times 0.12[=0.528]\)
\(\pm m \times 4.3208 \pm 2 m \times 0.528=\pm(m+2 m) v\)
[Correct equation is \(m \times 4.3208+2 m \times 0.528=\pm(m+2 m) v\) ]
Speed of combined particle immediately after impact \(=v=1.79 \mathrm{~ms}^{-1}\)
Special case for those who do not take into account the fact that \(Q\) comes to rest and then changes its direction
\(u_{p(\text { down })}=0+4.4 \times 1[=4.4]\)
\(u_{q(\mathrm{up})}=10-11.6 \times 1[=-1.6]\) so \(u_{q(\mathrm{down})}=1.6\)
\(\pm m \times 4.4 \pm 2 m \times 1.6=\pm(m+2 m) v\)
Speed of combined particle immediately after impact \(=v=2.53 \mathrm{~ms}^{-1}\)
Knowledge points:
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.）
4.3.1 use the definition of linear momentum and show understanding of its vector nature （For motion in one dimension only.）
4.3.2 use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies. （Including direct impact of two bodies where the bodies coalesce on impact. Knowledge of impulse and the coefficient of restitution is not required.）
Solution:
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