Question

A green ball (ball 1) of mass M, collides with an orange (ball 2) of mass 1.26 m. The initial speed of the green ball is 5.4 m/s the final speed of the green ball is 2.6 m/s and theta = 36.9° A find the magnitude of the final speed of the orange ball? B. what is a direction of the final speed of the orange ball?

Answer 1

We know that in a collision the momentum is conserved, that is:

$\vec{p}_0=\vec{p}_f$

Since this is vector equation we can divide it in two scalar equations, one for x and for y. Then we have:

$\begin{gathered} p_{0x}=p_{fx} \\ \text{and} \\ p_{0y}=p_{fy} \end{gathered}$

Then we have for the x direction:

$\begin{gathered} 5.4m=m(2.6)\cos 36.9+1.26mv_o\cos \phi \\ 5.4=2.6\cos 36.9+1.26v_o\cos \phi \end{gathered}$

and for the y direction:

$\begin{gathered} 0=-m2.6\sin 36.9+1.6mv_o\sin \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}$

Hence, we have the system of equations:

$\begin{gathered} 5.4=2.6\cos 36.9+1.26v_o\cos \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}$

From the second equation we have:

$v_o=\frac{2.6\sin 36.9}{1.6\sin \phi}$

Plugging this in the first equation:

$\begin{gathered} 5.4=2.6\cos 36.9+1.26(\frac{2.6\sin36.9}{1.6\sin\phi})\cos \phi \\ 1.26(\frac{2.6\sin36.9}{1.6})\tan \phi=5.4-2.6\cos 36.9 \\ \tan \phi=\frac{1.6(5.4-2.6\cos 36.9)}{(1.26)(2.6\sin 36.9)} \\ \phi=\tan ^{-1}(\frac{1.6(5.4-2.6\cos36.9)}{(1.26)(2.6\sin36.9)}) \\ \phi=69.69 \end{gathered}$

Now that we know the value of the angle we plug it in the expression for the velocity, then we have:

$\begin{gathered} v_o=\frac{2.6\sin 36.9}{1.6\sin 69.69} \\ v_0=1.04 \end{gathered}$

Therefore, the magnitude of the final speed of the orange ball is 1.04 m/s and the direction is 69.69°