Question

5. Describe the energy transfers that occur as a cart inelastically collides with an object of equal mass at rest. This should be a narrative description of the energy transfers that occur as the cart inelastically collides with a cart of equal mass. In your descriptions, address the following:
   • What was the kinetic energy of each cart before and after the collision?
   • What happens to the total energy of the system, now including both carts, as a result of the inelastic collision?
   • Describe how the principle of conservation of energy is applied in this situation.

Following the inelastic collision of the carts, the two carts fuse into an object with double the mass of the original cart. There is then a frictional section of the track to slow the cart to a stop over 20 meters. Describe the amount of work due to friction and frictional force exerted to stop both carts over 20 meters.

6. Calculate the work due to friction and frictional force. In your calculations, be sure to explicitly state the equations you use and what values you will be substituting to calculate the final value.
7. Describe the energy transfers that occur as the cart is brought to a stop. This should be a narrative description of the energy transfers—written to describe these concepts to a nontechnical audience—that occur as the cart is brought to a stop. In your descriptions, address the following:
   • What is the kinetic energy of the cart system before and after it has been brought to a stop?
   • What happens to the total energy of the system as a result of this change in motion?
   • Describe how the principle of conservation of energy is applied in this situation.

Answer 1

Let mass of cart =m = mass of colliding object,

velocity of cart = v

kinetic energy before collision, K_i = 1/2mv²

let the common velocity of cat and object is v'.

By momentum conservation ,

v' = v/2

there fore kinetic energy K_f = 21/2 mv'² = 1/4 mv²

^{as friction force is non-conservative force}

work done by frictional force = change in mechanical energy of common "cart+object"

or, work done by frictional force = change in kinetic energy of common "cart+object"

= 0-1/4 mv² = -1/4 mv²

^{as the frictional force do work such that kinetic energy of the moving cart+object system is dissipating;consequentaly cart+ object system stops.}

^{total energy of the system remains conserved; eventhough mechanical energy is not conseved.}

^{work done by frictional force =- kinetic enrgy of common "cart+object" before stop}

^{total energy before rest = total energy after rest}