A $20g$ball of clay traveling east at $3.0m/s$ collides with a $30g$ ball of clay traveling north at $2.0m/s$. What are the speed and the direction of the resulting $50g$ ball of clay? Give your answer as an angle north of east.

Mass of clay ball 1,$m_{Clayball1}=20g$

Mass of clay ball 2,$m_{clayball2}=30g$

The velocity of clay ball 1,$v_1=3m/s$

The velocity of clay ball 2,$v_2=2m/s$

To evaluate speed of clay ball 3.

Clay 1 travels east (in positive x-direction), while clay 2 travels north (in a negative x direction) (positive y direction) clay no. 2 The velocity of the ensuing ball of clay (which we'll call clay 3), is represented as

$v_3=v_3\cos \theta ·\overrightarrow{i}+v_3\sin \theta ·\overrightarrow{j}$

As we all know, each element's momentum is conserved individually.

For $x-$component,

$$\begin{gathered} {\left(p_1\right)}_x+{\left(p_2\right)}_x={\left(p_3\right)}_x \\ 0.02kg\times 3m/s+0=0.05kg·v_3\cos \theta \end{gathered}$$

$\begin{array}{r}0.02kg·3m/s+0=0.05kg·v_3\cos \theta \\ \Rightarrow v_3\cos \theta =1.2m/s\end{array}$

For y-component,

$\begin{array}{r}{\left(p_1\right)}_y+{\left(p_2\right)}_y={\left(p_3\right)}_y\\ 0+0.03kg.2m/s=0.05kg·v_3\sin \theta \\ \Rightarrow v_3\sin \theta =1.2m/s\end{array}$

From both the components,

$$\begin{gathered} \tan \theta =1 \\ \theta ={45}^o \end{gathered}$$

Now evaluating the velocity,

$v_3\frac{\sqrt{2}}{2}=1.2\Rightarrow v_3=1.7m/s$