Question

A ball of mass 1.5 kg rolling to the right with a speed of 3.6 ms^-1
collides head-on with
a spring with a spring constant of 2.0 Nm^-2
Determine the maximum compression of the
spring and the speed of the ball when the compression of the spring is 0.10 m.

Answer 1

Answer on Question #40207 – Physics – Physics, Mechanics | Kinematics | Dynamics

1. A ball of mass $1.5 \, \text{kg}$ rolling to the right with a speed of $3.6 \, \text{ms}^{-1}$ collides head-on with a spring with a spring constant of $2.0 \, \text{Nm}^{-2}$. Determine the maximum compression of the spring and the speed of the ball when the compression of the spring is $0.10 \, \text{m}$.

$$\begin{array}{l} m = 1.5 \, \text{kg} \\ v_0 = 3.6 \, \frac{\text{m}}{\text{s}} \\ k = 2 \, \frac{\text{N}}{\text{m}} \\ x_1 = 0.1 \, \text{m} \\ \hline x_{\max}, v_1 - ? \\ \end{array}$$ $$\begin{array}{l} \text{Solution.} \\ \text{The initial kinetic energy of the ball can be found according to Steiner's} \\ \text{theorem: } W = \frac{I \omega^2}{2} + m r^2, \quad \text{where } I = \frac{2 m r^2}{5} \text{ is momentum of inertia regarding to} \\ \text{the axis across the center of the ball, } \omega = \frac{v_0}{r} \text{ is the initial angular speed of the ball.} \\ \text{So, } W = \frac{6 m v_0^2}{5}. \\ \end{array}$$

As the total mechanical energy of the system is conserved, then

$$W = \frac{k x^2}{2} + \frac{6 m v^2}{5},$$

where $x$ and $x$ are the compression of the spring and the velocity of the ball at any moment.

The maximum compression of the spring can be found assuming $\nu = 0$:

$$\frac{6 m v_0^2}{5} = \frac{k x_{\max}^2}{2}, \quad x_{\max} = 2 v_0 \sqrt{\frac{3 m}{5 k}}.$$

The speed of the ball, when the compression of the spring is $x_1$:

$$\frac{6 m v_0^2}{5} = \frac{k x_1^2}{2} + \frac{6 m v_1^2}{5}, \quad v_1 = \sqrt{v_0^2 - \frac{5 k x_1^2}{12 m}}.$$

Let check the dimensions:

$$\left[ x_{\max} \right] = \frac{m}{s}, \quad \left( \frac{\text{kg}}{\text{N}} \right) = \frac{m}{s}, \quad \left( \frac{\text{kg} \cdot m}{\text{kg} \cdot \frac{m}{s^2}} \right) = m, \quad \left[ v_1 \right] = \sqrt{ \left( \frac{m}{s} \right)^2 - \frac{ \frac{N}{m} \cdot m^2 }{ \text{kg} } } = \sqrt{ \frac{m^2}{s^2} - \frac{ \left( \text{kg} \cdot \frac{m}{s^2} \right) \cdot m }{ \text{kg} } } = \frac{m}{s}.$$

Let evaluate the quantities:

$$x_{\max} = 2 \cdot 3.6 \cdot \sqrt{ \frac{3 \cdot 1.5}{5 \cdot 2} } = 4.83 \, \text{(m)}, \quad v_1 = \sqrt{3.6^2 - \frac{5 \cdot 2 \cdot 0.1^2}{12 \cdot 1.5}} = 3.60 \, \text{(m/s)}.$$

Answer: $4.83 \, \text{m}$, $3.60 \, \text{m/s}$.

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