Chapter 11, Problem 68GP

To determine

To Calculate: The maximum compression of the bumper.

Answer to Problem 68GP

  $0.11489\text{cm}$

Explanation of Solution

Given Information:

Stiffness constant, $\mathrm{k}=550\text{N/m}$

Mass, $\mathrm{m}=1500\text{kg}$

Speed before collision, ${\mathrm{v}}_1=2.2\text{m/s}$

Formula used:

Using the law of conservation of energy,

  $\mathrm{E}{}_1={\mathrm{E}}_2$

  $\frac{1}{2}\mathrm{m}{\mathrm{v}}_1{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_1{}^2=\frac{1}{2}\mathrm{m}{\mathrm{v}}_2{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_2{}^2$

Here,

  $\mathrm{k}$ is spring stiffness constant,

  ${\mathrm{x}}_1,{\mathrm{x}}_2$ is compression of bumper,

  $\mathrm{m}$ is the mass of car,

  ${\mathrm{v}}_1,{\mathrm{v}}_2$ is the speed of car.

Energy before the collision is equal to energy after the collision.

As there is no potential energy before the collision

  ${\mathrm{x}}_1=0$

Because there is no kinetic energy after the collision

  ${\mathrm{v}}_2=0$

Calculation:

from the law of conservation of energy

  $\mathrm{E}{}_1={\mathrm{E}}_2$

  $\frac{1}{2}\mathrm{m}{\mathrm{v}}_1{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_1{}^2=\frac{1}{2}\mathrm{m}{\mathrm{v}}_2{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_2{}^2$

By putting the values

  $\frac{1}{2}\mathrm{m}{\mathrm{v}}_1{}^2$ = $\frac{1}{2}\mathrm{k}{\mathrm{x}}_2{}^2$

So,

  ${\mathrm{x}}_2=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}{\mathrm{v}}_1$

  ${\mathrm{x}}_2=\sqrt{\frac{\text{1500}\text{kg}}{{\text{550×10}}^{\text{3}}\text{N/m}}}\text{(2}\text{.2}\text{m/s)}$

  ${\mathrm{x}}_2=0.11489\text{cm}$

Conclusion:

The maximum compression of the bumper due to the collision is $0.11489\text{cm}$ .