suppose a cart of mass m = 969 g is initial moving at speed v0 = 5.17 m/s. It strikes a spring attached to a fixed wall, comes to a stop, and bounces back at speed vf = 4.48 m/s. Assume the track is frictionless.
a) Find the magnitude of the impulse exerted by the spring on the cart.
b) Find the energy lost in this collision.
c) If the cart was in contact with the spring for 232 ms, find the average force exerted by the spring on the cart.
To find the answers to these questions, we will use the principle of conservation of momentum and conservation of energy.
a) The impulse exerted by the spring on the cart can be found by using the principle of conservation of momentum. The initial momentum of the cart is given by the product of its mass (m) and initial velocity (v0), which is p_initial = m * v0. The final momentum of the cart is given by the product of its mass and final velocity (vf), which is p_final = m * vf. The impulse exerted by the spring can be calculated as the change in momentum, which is the difference between the final and initial momentum: impulse = p_final - p_initial.
b) The energy lost in this collision can be found by using the principle of conservation of energy. The initial kinetic energy of the cart is given by (1/2) * m * v0^2, and the final kinetic energy is (1/2) * m * vf^2. The energy lost in the collision is the difference between the initial and final kinetic energy: energy_lost = (1/2) * m * v0^2 - (1/2) * m * vf^2.
c) The average force exerted by the spring on the cart can be found by dividing the impulse exerted by the spring (which we found in part a) by the contact time between the cart and the spring. The contact time is given as 232 ms, which is 0.232 s. Therefore, average force = impulse / contact time.
Using the provided values, we can now calculate the answers to each part of the question.