you strike a huge spring at a speed of 30m/s while strapped in a steel cage. When you hit the spring with combined mass of 320kg it is compressed 9.0m. A) What is the spring constant of the spring? B) How long are you in contact with the spring before it bounces off in the opposite direction?

KE = EPE

1/2 m v^2 = 1/2 kx^2 so
k =mv^2/x^2

Offhand I don't know how to find the time. The force is constantly changing so eq of motion and impulse momentum are out. Unless throw in some calculus.

To find the spring constant of a spring and the contact time with the spring, we can use the principles of conservation of energy and basic physics equations.

A) To find the spring constant (k) of the spring, we can use the formula for potential energy stored in a spring:

Potential Energy (PE) = (1/2) * k * x^2

Where PE is the potential energy, k is the spring constant, and x is the distance the spring is compressed.

In this case, the potential energy is equal to the kinetic energy before the collision. The kinetic energy (KE) can be calculated using the formula:

Kinetic Energy (KE) = (1/2) * m * v^2

Where m is the mass of the cage and the combined mass of 320kg, and v is the speed of 30m/s.

Setting the potential energy equal to the kinetic energy, we have:

(1/2) * k * x^2 = (1/2) * m * v^2

Now we can solve for the spring constant (k):

k = (m * v^2) / x^2

Plugging in the given values:

k = (320kg * (30m/s)^2) / (9.0m)^2

k = 320kg * 900m^2/s^2 / 81m^2

k = 320kg * 11.11...

k ≈ 3555.56 N/m (rounded to two decimal places)

Therefore, the spring constant of the spring is approximately 3555.56 N/m.

B) To find the contact time with the spring before bouncing off, we can use the equation for time:

t = 2 * sqrt(m/k)

Where t is the contact time, m is the combined mass of 320kg, and k is the spring constant we calculated in part A.

Plugging in the given values:

t = 2 * sqrt(320kg / 3555.56 N/m)

t = 2 * sqrt(0.09 kg/m)

t = 2 * 0.3 s

t = 0.6 s

Therefore, you are in contact with the spring for approximately 0.6 seconds before it bounces off in the opposite direction.