Question

An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with constant 5.90 ✕ 106 N/m and is compressed 2.63 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost in the collision with the wall?

Answer 1

Answer:

$v=2.02\frac{m}{s}$

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

$K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}$

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

$v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}$

Answer 2

The speed of the car before impact, assuming no energy is lost in the collision with the wall is equal to 2.02 m/s.

Given the following data:

• Mass of automobile = 1000 kg

• Spring constant = $5.90 \times 10^6 \; N/m$

• Extension = 2.63 cm

Conversion:

Extension = 2.63 cm to m = 0.0263 meter

To find the speed of the car before impact, assuming no energy is lost in the collision with the wall:

Applying the law of conservation of energy, we have:

The kinetic energy possessed by the bumper is equal to the elastic potential energy possessed by the bumper.

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

Making v the subject of formula, we have:

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

Where:

• v is the speed.

• k is the spring constant.

• m is the mass.

Substituting the given parameters into the formula, we have;

$v = 0.0263\sqrt{\frac{5.90 \times 10^6 }{1000} } \\\\v = 0.0263\sqrt{5900}\\\\v = 0.0263 \times 76.81$

Speed, v = 2.02 m/s

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