Question

A green ball (ball 1) of mass M collides with an orange ball (ball 2) of mass 1.26. 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 the direction of the final speed of the orange ball?

Answer 1

Given data:

The mass of ball 1 is m.

The mass of ball 2 is 1.26m.

The initial speed of ball 1 is u=5.4 m/s.

The final speed of the ball 1 U=2.6 m/s.

The angle at which the ball 1 moves from x-axis is θ=36.9.

Applying the conservation of momentum in x-direction,

`\begin{gathered} mu=mU\cos \theta+(1.26m)V\cos \alpha \\ u=U\cos \theta+(1.26)V\cos \alpha \\ 5.4=2.6\cos 36.9+(1.26)V\cos \alpha \\ V\cos \alpha=2.63\ldots\ldots\text{.}(1) \end{gathered}`

Here, V is the final speed of ball 2, and α is the angle of ball 2 with x-axis after the collision.

Applying the conservation of momentum in y-direction,

`\begin{gathered} 0=mU\sin \theta+(1.26m)V\sin \alpha \\ 0=U\sin \theta+(1.26)V\sin \alpha \\ 0=2.6\sin 36.9+(1.26)V\sin \alpha \\ V\sin \alpha=-1.56\ldots\ldots\text{.}(2) \end{gathered}`

Dividing equation (2) and (1),

`\begin{gathered} (V\sin \alpha)/(V\cos \alpha)=(-1.56)/(2.63) \\ \tan \alpha=0.593 \\ \alpha=30.6\degree \end{gathered}`

Subsitute the value of α in equation (1),

`\begin{gathered} V\cos \alpha=2.63 \\ V\cos 30.6\degree=2.63 \\ V=3.05\text{ m/s} \end{gathered}`

Thus, the final speed of the ball 2 (orange ball) is 3.05 m/s, and the direction of the orange ball is 30.6⁰.