A particle $$$A$$$ is projected vertically upwards from level ground with an initial speed of $$$30 \mathrm{~m} \mathrm{~s}^{-1}$$$. At the same instant a particle $$$B$$$ is released from rest $$$15 \mathrm{~m}$$$ vertically above $$$A$$$. The mass of one of the particles is twice the mass of the other particle. During the subsequent motion $$$A$$$ and $$$B$$$ collide and coalesce to form particle $$$C$$$. 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Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_s21_qp_42 Year：2021 Question No：6

Answer:

$ s_{A}=\pm\left(30 t-5 t^{2}\right) \text { or } s_{B}=\pm 5 t^{2}
$
$s_{A}+s_{B}=15$ leading to $15=30 t$ leading to $t=0.5$
$t=0.5$ leading to $v_{A}=\pm 25$ and $v_{B}=\pm 5$
$t=0.5$ leading to $h_{A}=\pm\left(30 \times 0.5-\frac{1}{2} g \times 0.5^{2}\right)=\pm 13.75$
$ \begin{array}{l}
25 \times(2 m)-5(m)=(3 m) v \rightarrow v_{1}=15 \\
25(m)-5 \times(2 m)=(3 m) v \rightarrow v_{2}=5
\end{array}
$

Particle $C_{I}-13.75=15 t-5 t^{2}$
Particle $C_{2}-13.75=5 t-5 t^{2}$
$t_{C_{1}}, t_{C_{2}}=3.74,2.23$ leading to $T=1+\sqrt{5}-\sqrt{3}=1.50$
Alternative method for the final two marks
$0=15-g t_{1}, 0=5-g t_{2} \rightarrow t_{1}=1.5, t_{2}=0.5$
Total heights $h_{1}=13.75+11.25=25$
Or $\quad h_{2}=13.75+1.25=15$
$25=5 T_{1}^{2}$ and $15=5 T_{2}^{2} \rightarrow T_{1}=\sqrt{5}, T_{2}=\sqrt{3}$
$T=1.5+\sqrt{5}-(0.5+\sqrt{3})=1+\sqrt{5}-\sqrt{3}=1.50$

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.）

4.3.2 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.）

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.

Answer:

Knowledge points:

Solution:

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