Question

In order to make a split while bowling, it is determined that one of the pins needs to be hit and forced to move at a 70deg angle from the direction of the incoming ball. The ball's mass is 5times the pins mass. It is moving at 12 m/s when the ball strikes the pin. It can be assumed the collision is perfectly elastic, find the final speed of the ball and the pin and the ancgle at which the ball ends up moving.

Answer 1

In order to make a split while bowling, it is determined that one of the pins needs to be hit and forced to move at a 70deg angle from the direction of the incoming ball. The ball's mass is 5 times the pin's mass. It is moving at $12\mathrm{m / s}$ when the ball strikes the pin. It can be assumed the collision is perfectly elastic, find the final speed of the ball and the pin and the angle at which the ball ends up moving.

Solution

$$V = 12 \frac{m}{s}, \alpha = 70{}^{\circ}, \mathrm{M} = 5\mathrm{m}$$

the final speed of the ball - $V_{2}$

the final speed of the pin - $V_{1}$

the angle at which the ball ends up moving - $\beta$

According to the law of conservation of momentum for projections on the axes

$$m V_{1} * \sin \alpha = 5 m V_{2} * \sin \beta \rightarrow V_{1} \sin \alpha = 5 V_{2} \sin \beta.$$ $$5 m V = 5 m V_{2} \cos \beta + m V_{1} * \cos \alpha \rightarrow 5 V = 5 V_{2} \cos \beta + V_{1} \cos \alpha$$

According to the law of conservation of energy when the collision is perfectly elastic

$$\frac{5 m V^{2}}{2} = \frac{5 m V_{2}^{2}}{2} + \frac{m V_{1}^{2}}{2} \rightarrow 5 V^{2} = 5 V_{2}^{2} + V_{1}^{2}.$$ $$\left\{ \begin{array}{c} V_{1} \sin \alpha = 5 V_{2} \sin \beta \\ 5 V - V_{1} \cos \alpha = 5 V_{2} \sin \beta \end{array} \right. \rightarrow (V_{1} \sin \alpha)^{2} + (5 V - V_{1} \cos \alpha)^{2} = 25 V_{2}^{2}.$$ $$V_{1}^{2} \sin^{2} \alpha + 25 V^{2} + V_{1}^{2} \cos^{2} \alpha - 2 * 5 V V_{1} \cos \alpha = 25 V_{2}^{2}. \rightarrow$$ $$\left\{ \begin{array}{c} V_{1}^{2} + 25 V^{2} - 10 V V_{1} \cos \alpha = 25 V_{2}^{2} \\ 5 V^{2} = 5 V_{2}^{2} + V_{1}^{2} \end{array} \right. \rightarrow V_{1}^{2} + 25 V^{2} - 10 V V_{1} \cos \alpha = 25 V^{2} - 5 V_{1}^{2}$$ $$6 V_{1}^{2} - 10 V V_{1} \cos \alpha = 0 \rightarrow 6 V_{1} - 10 V \cos \alpha = 0 \rightarrow V_{1} = \frac{10 V \cos \alpha}{6} = \frac{5}{3} V \cos \alpha = 6.84 \frac{m}{s}$$ $$V_{2}^{2} = \frac{1}{5} (5 * 12^{2} - 6.84^{2}) = 134.64 \frac{m^{2}}{s^{2}}$$ $$V_{2} = 11.6 \frac{m}{s}$$ $$\sin \beta = \frac{V_{1} \sin \alpha}{5 V_{2}} = \frac{6.84 * 0.94}{5 * 11.6} = 0.11$$ $$\beta = 6.36{}^{\circ}$$

Answer: $V_{2} = 11.6 \frac{m}{s}, V_{1} = 6.84 \frac{m}{s}, \beta = 6.36{}^{\circ}$