Question

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The clay compresses the spring a certain maximum amount before momentarily stopping. What is the maximum compression of the spring

Answer 1

Answer:

Approximately $0.560\; {\rm m}$, assuming that:

• the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
• air resistance on the clay sphere is negligible,
• the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$, and
• the clay sphere did not deform.

Explanation:

Notations:

• Let $k$ denote the spring constant of the spring.
• Let $m$ denote the mass of the clay sphere.
• Let $v$ denote the initial speed of the spring.
• Let $g$ denote the gravitational field strength.
• Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$, the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$.

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$.

When the spring is at the maximum compression:

• The clay sphere would be right on top of the spring.
• The top of the spring would be below the original position (when the spring was uncompressed) by $x$.
• The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$.

Thus, the initial position of the clay sphere ($h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$.

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$.

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$.

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$.

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$:

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$.

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$, $m = 4.87\; {\rm kg}$, $v = 5.21\; {\rm m\cdot s^{-1}}$, $g = 9.81\; {\rm m \cdot s^{-2}}$, and $h = 3.21\; {\rm m}$. Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$:

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$.

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$.