Question

A toy cannon uses a spring to project a 5.39-g soft rubber ball. The spring is originally compressed by 4.99 cm and has a force constant of 8.03 N/m. When the cannon is fired, the ball moves 14.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 3 N on the ball. At what point does the ball have maximum speed?

Answer 1

A toy cannon uses a spring to project a 5.39-g soft rubber ball. The spring is originally compressed by 4.99 cm and has a force constant of 8.03 N/m. When the cannon is fired, the ball moves 14.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 3 N on the ball. At what point does the ball have maximum speed?

The law of conservation of energy:

$$\mathrm{T}_1 + \mathrm{U}_1 = \mathrm{T}_2 + \mathrm{U}_2 + \mathrm{A}$$ $$\mathrm{T} = \frac{\mathrm{mv}^2}{2} - \text{kinetic energy}, \quad \mathrm{m} - \text{mass of the body}, \quad \mathrm{v} - \text{speed}$$ $$\mathrm{U} = \frac{\mathrm{kx}^2}{2} - \text{potential energy}, \quad \mathrm{k} - \text{force constant of spring}, \quad \mathrm{x} - \text{deformation}$$

$\mathrm{A} = \mathrm{Fl} - \text{work of constant friction force}, \mathrm{F} - \text{magnitude of force}, \mathrm{l} - \text{position of ball}.$

So, for our case we have:

$$\frac{\mathrm{kx}_0^2}{2} + 0 + 0 = \frac{\mathrm{mv}^2}{2} + \frac{\mathrm{k(x_0 - x)}^2}{2} + \mathrm{Fx} \quad \text{(for } x < x_0, \text{ if } x \geq x_0, \text{ ball will slow down)}$$

$x_0 - \text{initial spring deformation}, \quad x - \text{current position of ball}$

$$\frac{\mathrm{v}^2}{2} = -\frac{\mathrm{k}}{2\mathrm{m}} \mathrm{x}^2 + \left(\frac{\mathrm{k}}{\mathrm{m}} \mathrm{x}_0 - \frac{\mathrm{F}}{\mathrm{m}}\right) \mathrm{x}$$

Maximum value at point $\frac{\mathrm{dv}}{\mathrm{dx}} = 0$:

$$-\frac{\mathrm{k}}{\mathrm{m}} \mathrm{x}_{\max} + \frac{\mathrm{k}}{\mathrm{m}} \mathrm{x}_0 - \frac{\mathrm{F}}{\mathrm{m}} = 0$$ $$\mathrm{x}_{\max} = \mathrm{x}_0 - \frac{\mathrm{F}}{\mathrm{k}} = 4.99 \, \mathrm{cm} - \frac{0.0313 \, \mathrm{N}}{0.0803 \, \mathrm{mm} / \mathrm{cm}} = 4.99 \, \mathrm{cm} - 0.39 \, \mathrm{cm} = 4.6 \, \mathrm{cm}$$

Answer: 4.6 cm from initial point