Question

An energy-absorbing car bumper has a spring constant of \(410\;{\rm{kN/m}}\). Find the maximum compression of the bumper if the car, with mass \(1300\;{\rm{kg}}\), collides with a wall at a speed of \(2.0\;{\rm{m/s}}\) (approximately \(5\;{\rm{mi/h}}\)).

Short answer

The maximum compression of the bumper is 0.1126 m.

Step-by-step solution

Concept of conservation of energy

According to the conservation of mechanical energy, total energy before and after the collision remains constant.

Given data

The spring constant of the bumper is\(k = 410\;{\rm{kN/m}}\).

The mass of the car is\(m = 1300\;{\rm{kg}}\).

The speed of the car is\({v_1} = 2.0\;{\rm{m/s}}\).

Calculation of maximum compression

Using law of conservation of mechanical energy,

\(\begin{aligned}{c}{E_1} = {E_2}\\\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}kx_2^2\end{aligned}\)

Initially, car will not face damage, so there will not be any compression, that is, \({x_1} = 0\),\({x_2}\)is the maximum compression of the bumper and after collision car will stop,\({v_2} = 0\).

Substitute the known values in the above relation to find the maximum compression.

\(\begin{aligned}{c}\frac{1}{2}mv_1^2 &= \frac{1}{2}kx_2^2\\{x_2} &= \sqrt {\frac{m}{k}} {v_1}\\{x_2} &= \sqrt {\frac{{1300\;{\rm{kg}}}}{{410 \times {{10}^3}\;{\rm{N/m}}}}} \left( {2.0\;{\rm{m/s}}} \right)\\{x_2} &= 0.1126\;{\rm{m}}\end{aligned}\)

Hence, the maximum compression would be 0.1126 m.