Question

3
012 10.0 points
A car with mass 1060 kg crashes into a wall.
The car's velocity immediately before the col-
lision was 14.6 m/s and the bumper is com-
pressed like a spring with spring constant
1.14 × 107 N/m.
What is the maximum deformation of the
bumper for this collision?
Answer in units of m.
013 10

Answer 1

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

$KE = \frac{1}{2}mv^2$

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

$U_s = \frac{1}{2}kx^2$

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

$E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

We can simplify and solve for 'x'.

$mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}$

Plug in the givens and solve.

$x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}$