Question

A 50 kg skater is travelling due east at a speed of $9m/s$. Another 70 kg skater is moving due south at a speed of $7m/s$. They collide and hold on to each other after the collision. Determine the magnitude and direction of their combined velocity after collision.

Answer 1

1. Given, $m_1=50kg,v_1=9m/s,m_2=70kg,v_2=7m/s$
   After collision, $m_3=50+70=120kg$
2. $P_1=m_1{v}_1=450kg.m/s$
   $P_2=m_2{v}_2=490kg.m/s$
   $P_1$ and $P_2$ are perpendicular so their resultant
   $P=\sqrt{{450}^2+{490}^2}=665.3$
   From the momentum conservation theorem,
   $P=m_3{v}_3=665.3$
   Hence,$v_3=\frac{665.3}{120}=5.5$
   Hence the magnitude of the velocity is $5.5m/s.$
   $\tan \theta =\frac{450}{490}$
   Hence, $\theta ={\tan }^{-1}0.9=42.5$

Hence, the combined magnitude of the velocity is$5.5m/s$ . And the direction is $42.5^0$ east of south.