How to Handle the Collision of Projectiles Explained by Fully Worked Solutions

Collision of Projectile Motion

Collision of Projectile Motion is handled by setting the same vertical height and horizontal distance at a given time.

Worked Examples of Collision of Projectile Motion

Two Points \(A\) and \(B\) are \(d\) metres apart. A particle is projected from \(A\) towards \(B\) with initial velocity \(u\) m/s at angle \( \alpha \) to the horizontal. Concurrently, another particle is projected from \(B\) towards \(A\) with initial velocity \(w\) m/s at \(\beta\) to the horizontal. The particles collide with projectile motion when they both reach their maximum height.

The equation of projectile motion projected from the origin with initial velocity \(V\) m/s at angle \(\theta\) to the horizontal is \( x = Vt \cos \theta \) and \( \displaystyle y=Vt \sin \theta – \frac{gt^2}{2} \).

(a)    Find the time taken from \(A\) to reach its maximum height.

The maximum height occurs when \( \dot{y} = 0\).
\( \begin{aligned}
\dot{y} &= u \sin \alpha-gt \\
0 &= u \sin \alpha-gt \\
gt &= u \sin \alpha \\
\therefore t &= \frac{u \sin \alpha}{g}
\end{aligned} \)

(b)    Show that \( u \sin \alpha = w \sin \beta \).

The particle from \(B\) reaches its maximum height after \( \displaystyle \frac{w \sin \beta}{g} \). Both particles reach their maximum height concurrently upon collision.
\( \begin{aligned} \displaystyle
\frac{u \sin \alpha}{g} = \frac{w \sin \beta}{g} \\
\therefore u \sin \alpha = w \sin \beta
\end{aligned} \)

(c)    Show that \( \displaystyle d = \frac{uw}{g} \sin(\alpha + \beta) \).

\( \begin{aligned} \displaystyle \require{color}
x_A &= V t \cos \theta \\
&= u \times \frac{u \sin \alpha}{g} \times \cos \alpha \require{AMSsymbols} &\color{red} \text{horizontal distance from } A \text{ at } t = \frac{u \sin \alpha}{g} \\
&= \frac{u^2}{g} \sin \alpha \cos \alpha \\
x_B &= V t \cos \theta \\
&= w \times \frac{w \sin \beta}{g} \times \cos \beta&\color{red} \text{horizontal distance from } B \text{ at } t = \frac{w \sin \beta}{g} \\
&= \frac{w^2}{g} \sin \beta \cos \beta \\
d &= x_A + x_B \\
&= \frac{u^2}{g} \sin \alpha \cos \alpha + \frac{w^2}{g} \sin \beta \cos \beta \\
&= \frac{1}{g}(u \sin \alpha \times u \cos \alpha + w \sin \beta \times w \cos \beta) \\
&= \frac{1}{g}(w \sin \beta \times u \cos \alpha + u \sin \alpha \times w \cos \beta) &\color{red} u \sin \alpha = w \sin \beta \\
&= \frac{uw}{g}(\sin \beta \times \cos \alpha + \sin \alpha \times \cos \beta) \\
\therefore d &= \frac{uw}{g} \sin(\alpha + \beta)
\end{aligned} \)

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