A 950 kg car strikes a huge spring at a speed of 25 m/s (Fig. 11–57), compressing the spring 4.0 m. (a) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before it bounces off in the opposite direction?

The kinetic energy is calculated in terms of the mass and the velocity of the object as:

\(E = \frac{1}{2}m{v^2}\)

Here,\(v\)is the velocity.

The mass of the car is\({\rm{m}} = 950\;{\rm{kg}}\).

The speed of the car is\({\rm{v}} = 25\;{\rm{m/s}}\).

The compression of spring is \({\rm{x}} = 4.0\;{\rm{m}}\).

Part (a)

According to conservation of energy principle, total energy of the system always remains conserved. It can however, change its form, from stored energy to energy used in motion.

The frequency\(f\)of SHM is

\(f = \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{k}{m}} \)

Here,\(k\)is the spring constant and\(m\)is the mass.

The elastic potential energy is

\(E = \frac{1}{2}k{A^2}\)

Here,is spring constant and\(A\)is the amplitude of the oscillation.

As shown in the figure,

Using conservation of energy principle, initial kinetic energy of the car in motion will be transformed into elastic potential energy, when spring is compressed.

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

Initially, spring is not compressed and after the collision, spring gets compressed but the car stops. Hence, initial extension is zero, so does the final velocity of the car.

Thus, spring stiffness constant is calculated as:

\(\begin{aligned}\frac{1}{2}mv_1^2 &= \frac{1}{2}kx_2^2\\k &= \frac{{mv_2^2}}{{x_2^2}}\\ &= \left( {950\;{\rm{kg}}} \right){\left( {\frac{{25\;{\rm{m/s}}}}{{4.0\;{\rm{m}}}}} \right)^2}\\ &= 3.7 \times {10^4}\;{\rm{N/m}}\end{aligned}\)

Hence, the spring stiffness constant of the huge spring is\(3.7 \times {10^4}\;{\rm{N/m}}\).

Part (b)

The car will collide with the spring and it will disturb its stable position. So, from its equilibrium position, spring will compress and then comes back to its equilibrium position again. So, car will remain in contact with the spring for half of total time. Thus,

\(\begin{aligned}\frac{1}{2}T &= \frac{1}{2}2{\rm{\pi }}\sqrt {\frac{m}{k}} \\ &= {\rm{\pi }}\sqrt {\frac{{950\;{\rm{kg}}}}{{3.7 \times {{10}^4}\;{\rm{N/m}}}}} \\ &= 0.50\;{\rm{s}}\end{aligned}\)

Hence, the car will remain in contact with the spring for 0.50 s.