Question

A projectile of mass $m$ is fired with a velocity $u$ from a point $P$ as shown. Neglecting air resistance, find the magnitude of the change in momentum of the projectile between the points $P$ and $Q$.

• A. $mu\sec \beta \sin (\alpha -\beta )$
• B. $mu\sin \alpha (\tan \beta -1)$
• C. $mu(\tan \beta -1)$
• D. $mu\cos \alpha$

Answer 1

The correct option is A $mu\sec \beta \sin (\alpha -\beta )$

In the horizontal direction, velocity is constant. Hence,
$u\cos \alpha =V\cos \beta$
$\begin{array}{}\Rightarrow V=\frac{u\cos \alpha }{\cos \beta }& \text{(1)}\end{array}$
In the vertical direction :
$\Delta p_y=m(V_{Qy}-V_{Py})$
$=m(V\sin \beta -u\sin \alpha )$
Substituting the value of $V$ from $\left(1\right)$, we get
$\Delta p_y=m(\frac{u\cos \alpha }{\cos \beta }\times \sin \beta -u\sin \alpha )$
$\Delta p_y=mu\sec \beta (\cos \alpha \sin \beta -\sin \alpha \cos \beta )$
$\Delta p_y=-mu\sec \beta \sin (\alpha -\beta )$
In terms of magnitude,
$\Delta p_y=mu\sec \beta \sin (\alpha -\beta )$