3 \includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-3_408_1164_248_493} A particle \(P\) is released from rest at a point \(A\) which is 7 m above horizontal ground. At the same instant that \(P\) is released a particle \(Q\) is projected from a point \(O\) on the ground. The horizontal distance of \(O\) from \(A\) is 24 m . Particle \(Q\) moves in the vertical plane containing \(O\) and \(A\), with initial speed \(50 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and initial direction making an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac { 7 } { 24 }\) (see diagram). Show that the particles collide.

## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $24 = 50 \times 0.96t$ | M1 | For using $x = (V\cos\theta)t$ to find $t$ |
| $t = 0.5$ | A1 | |
| $[y = 50 \times 0.5 \times (7/25) - \frac{1}{2} \cdot 10 \times 0.5^2$ or $y = (7/24)24 - 10 \times 24^2/(2\times50^2\times0.96^2)]$ | M1 | For substituting $t = 0.5$ into $Y = Vt\sin\theta - \frac{1}{2}gt^2$ or $x = 24$ into the equation of the trajectory |
| $y_Q = 5.75$ | A1 | |
| $y_P = 7 - \frac{1}{2}g \times 0.5^2$ | M1 | For evaluating $y_P$ when $t = 0.5$ |
| $y_P = y_Q$ at time $t = 0.5 \rightarrow$ particles collide | A1 | **Total: 6** |

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\includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-3_408_1164_248_493}

A particle $P$ is released from rest at a point $A$ which is 7 m above horizontal ground. At the same instant that $P$ is released a particle $Q$ is projected from a point $O$ on the ground. The horizontal distance of $O$ from $A$ is 24 m . Particle $Q$ moves in the vertical plane containing $O$ and $A$, with initial speed $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and initial direction making an angle $\theta$ above the horizontal, where $\tan \theta = \frac { 7 } { 24 }$ (see diagram). Show that the particles collide.

\hfill \mbox{\textit{CAIE M2 2009 Q3 [6]}}