An automobile having a mass of 1330 kg

is driven into a brick wall in a safety test.
The bumper behaves like a spring of constant
4.12 × 106 N/m and compresses 4.99 cm as
the car is brought to rest.
What was the speed of the car before impact,
assuming no energy was lost during impact
with the wall?
Answer in units of m/s

To find the speed of the car before impact, we can use the conservation of mechanical energy principle. The kinetic energy of the car before impact is equal to the potential energy stored in the compressed bumper spring.

The potential energy of a spring is given by the formula:

PE = (0.5) * k * x^2

Where:
PE is the potential energy
k is the spring constant
x is the displacement of the spring from its equilibrium position

In this case, the spring constant is 4.12 × 10^6 N/m and the displacement of the bumper is 4.99 cm, which is equal to 0.0499 m.

Plugging in the values, we have:

PE = (0.5) * (4.12 × 10^6 N/m) * (0.0499 m)^2

Simplifying:

PE = 0.5 * 4.12 × 10^6 N/m * (0.0499 m)^2
= 0.5 * 4.12 × 10^6 N/m * 0.00249 m^2
= 5.12 × 10^3 J

Since the potential energy PE is equal to the initial kinetic energy of the car, we can equate this to the formula for kinetic energy:

KE = (0.5) * m * v^2

Where:
KE is the kinetic energy
m is the mass of the car
v is the velocity of the car

Plugging in the values, we have:

5.12 × 10^3 J = (0.5) * 1330 kg * v^2

Simplifying:

5.12 × 10^3 J = 0.5 * 1330 kg * v^2
1.024 × 10^4 J = 665 kg * v^2

Now, let's solve for v:

v^2 = (1.024 × 10^4 J) / (665 kg)
v^2 = 15.39 m^2/s^2

Taking the square root of both sides:

v = √15.39 m^2/s^2
v ≈ 3.92 m/s

Therefore, the speed of the car before impact, assuming no energy was lost during impact with the wall, is approximately 3.92 m/s.

(1/2) M V^2 = (1/2) K X^2

1330 v^2 = 4.12*10^6 (.0499)^2