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=(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((2.6\sin36.9)/(1.6\sin\phi))\cos \phi \\ 1.26((2.6\sin36.9)/(1.6))\tan \phi=5.4-2.6\cos 36.9 \\ \tan \phi=(1.6(5.4-2.6\cos 36.9))/((1.26)(2.6\sin 36.9)) \\ \phi=\tan ^(-1)((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=(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°