Question

A 24.0 g ball is fired horizontally with initial speed toward a 100 g ball that is hanging motionless from a 1.10 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle = 52.0.
What was the initial velocity ?

Answer 1

Question

Given:

$$\begin{array}{l} m_{1} = 24.0 \text{ g} \\ m_{2} = 100 \text{ g} \\ l = 1.10 \text{ m} \\ \alpha = 52.0{}^{\circ} \\ V_{2} = 0 \\ \end{array}$$

Need to find: initial velocity $V_{1}$.

Solution:

At first we can find final velocity $v_{2}$ of the 100 g ball:

$$\frac{m_{2} \cdot v_{2}^{2}}{2} = m_{2} \cdot g \cdot h = m_{2} \cdot g \cdot l \cdot (1 - \cos \alpha) \Rightarrow \text{we can find final velocity and it equal to}$$ $$v_{2}^{2} = 2 \cdot g \cdot l \cdot (1 - \cos \alpha) \Rightarrow v_{2} = \sqrt{2 \cdot g \cdot l \cdot (1 - \cos \alpha)}.$$

We will get:

$$v_{2} = \sqrt{2 \cdot g \cdot l \cdot (1 - \cos \alpha)} = \sqrt{2 \cdot 9.81 \cdot 1.10 \cdot (1 - \cos 52{}^{\circ})} = 2.88 \frac{m}{s}.$$

Applying conservation of momentum principle, we get:

$$m_{1} \cdot V_{1} + m_{2} \cdot V_{2} = m_{1} \cdot v_{1} + m_{2} \cdot v_{2} \Rightarrow |V_{2} = 0| \Rightarrow v_{1} = \frac{m_{1} \cdot V_{1} - m_{2} \cdot v_{2}}{m_{1}} = V_{1} - v_{2} \cdot \frac{m_{2}}{m_{1}}.$$

Also applying principle of kinetic energy conservation, we get:

$$\begin{array}{l} \frac{1}{2} \cdot m_{1} \cdot V_{1}^{2} + \frac{1}{2} \cdot m_{2} \cdot V_{2}^{2} = \frac{1}{2} \cdot m_{1} \cdot v_{1}^{2} + \frac{1}{2} \cdot m_{2} \cdot v_{2}^{2} \Rightarrow \left| \begin{array}{l} V_{2} = 0 \\ v_{1} = V_{1} - v_{2} \cdot \frac{m_{2}}{m_{1}} \end{array} \right| \Rightarrow \\ \Rightarrow m_{1} \cdot V_{1}^{2} = m_{1} \cdot V_{1}^{2} + m_{1} \cdot v_{2}^{2} \cdot \frac{m_{2}^{2}}{m_{1}^{2}} - 2 \cdot m_{1} \cdot V_{1} \cdot v_{2} \cdot \frac{m_{2}}{m_{1}} + m_{2} \cdot v_{2}^{2} \Rightarrow \\ \Rightarrow v_{2}^{2} \cdot \frac{m_{2}^{2}}{m_{1}} + m_{2} \cdot v_{2}^{2} = 2 \cdot V_{1} \cdot v_{2} \cdot m_{2} \Rightarrow V_{1} = \frac{v_{2}^{2} \cdot m_{2}^{2} + m_{2} \cdot m_{1} \cdot v_{2}^{2}}{2 \cdot m_{1} \cdot v_{2} \cdot m_{2}} = \frac{v_{2} \cdot (m_{2} + m_{1})}{2 \cdot m_{1}}. \end{array}$$

So, we have the formula: $V_{1} = \frac{v_{2} \cdot (m_{2} + m_{1})}{2 \cdot m_{1}}$. We will get: $V_{1} = \frac{2.88 \cdot (0.1 + 0.024)}{2 \cdot 0.024} = 7.44 \frac{m}{s}$.

Answer: 7.44 $\frac{m}{s}$.