Question

two identical balls traveling parallel to the x-axis have speed of 30cm/s and are oppositely directed.they collide perfectly elastically. after the collision, one ball is moving at an angle of 30 degrees above the x-axis.find its speed and the velocity of the other balls.

Answer 1

QUESTION:

Two identical balls traveling parallel to the x-axis have speed of $30\mathrm{cm / s}$ and are oppositely directed. they collide perfectly elastically. after the collision, one ball is moving at an angle of 30 degrees above the x-axis.find its speed and the velocity of the other balls

SOLUTION:

According to the energy and momentum conservation laws:

$$\left\{ \begin{array}{l} \mathrm {m} \cdot \mathrm {v} _ {1} - \mathrm {m v} _ {1} = 0 \\ \mathrm {m} \cdot \mathrm {v} _ {2} \cos \alpha - \mathrm {m} \cdot \mathrm {v} _ {3} \cos \beta = 0 \quad (\text {projection on x axis}) \\ \mathrm {m} \cdot \mathrm {v} _ {2} \sin \alpha - \mathrm {m} \cdot \mathrm {v} _ {3} \sin \beta = 0 \quad (\text {projection on y axis}) \\ \frac {\mathrm {m} \cdot \mathrm {v} _ {1} ^ {2}}{2} + \frac {\mathrm {m} \cdot \mathrm {v} _ {1} ^ {2}}{2} = \frac {\mathrm {m} \cdot \mathrm {v} _ {2} ^ {2}}{2} + \frac {\mathrm {m} \cdot \mathrm {v} _ {3} ^ {2}}{2} \end{array} \right.$$ \left\{ \begin{array}{l} v _ {2} \cos \alpha = v _ {3} \cos \beta \\ v _ {2} \sin \alpha = v _ {3} \sin \beta \Rightarrow \left\{ \begin{array}{l} v _ {2} \sin \alpha = v _ {3} \sin \beta \\ 2 v _ {1} ^ {2} = v _ {2} ^ {2} + v _ {3} ^ {2} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \frac {v _ {2} \cos \alpha}{v _ {2} \sin \alpha} = \frac {v _ {3} \cos \beta}{v _ {3} \sin \beta} \\ v _ {2} \sin \alpha = v _ {3} \sin \beta \Rightarrow \left\{ \begin{array}{l} c t g \alpha = c t g \beta \\ v _ {2} \sin \alpha = v _ {3} \sin \beta \Rightarrow \\ 2 v _ {1} ^ {2} = v _ {2} ^ {2} + v _ {3} ^ {2} \end{array} \right. \Rightarrow \right. \end{array} \right. $$\Rightarrow \left\{ \begin{array}{l} \alpha = \beta \\ v _ {2} = v _ {3} \\ 2 v _ {1} ^ {2} = v _ {2} ^ {2} + v _ {3} ^ {2} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \alpha = \beta \\ v _ {2} = v _ {3} \\ 2 v _ {1} ^ {2} = v _ {2} ^ {2} + v _ {2} ^ {2} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \alpha = \beta \\ v _ {2} = v _ {3} \\ 2 v _ {1} ^ {2} = 2 v _ {2} ^ {2} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \alpha = \beta \\ v _ {2} = v _ {3} \\ v _ {2} = v _ {1} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \beta = 3 0 {}^ {\circ} \\ v _ {2} = 3 0 \mathrm {c m / s} \\ v _ {3} = 3 0 \mathrm {c m / s} \end{array} \right.$$

ANSWER:

$$\left\{ \begin{array}{l} \beta = 3 0 {}^ {\circ} \\ v _ {2} = 3 0 \mathrm {c m / s} \\ v _ {3} = 3 0 \mathrm {c m / s} \end{array} \right.$$

The second ball moves at angle $\beta = 30{}^{\circ}$ below the -x-axis (opposite to the first ball)