What 2 types of energy are we examining for conservation in this lab? In what ty

Question

What 2 types of energy are we examining for conservation in this lab? In what type(s) of collisions is momentum conserved? In what type(s) of collisions is energy conserved? If the potential energy of a system increases by 210 J, how much does the kinetic energy of that system change (if no work is done on the system)? If a 2.00 kg mass traveling 6.00 m/s collides with a stationary 1.00 kg mass and they stick together: What is their velocity after the collision? What is the kinetic energy of the system before the collision? What is the kinetic energy of the system after the collision? What is the percent loss in kinetic energy?

Explanation / Answer

2)Momentum is conserved in all types of collisions namely elastic collision , inelastic ( semi elastic) collision and perfectly inelastic collisions

3)In elastic collisions , both total mechanical energy and kinetic energy are conserved

In semi elastic and perfectly inelastic collisions total energy of the system is conserved but kinetic energy is not conserved

4)Let the change in the potential energy of the object be U = 210 J

Then change in kinetic energy be K

Work done is equal to change in energy of the system

                 W = U + K

Since W = 0

Then        U + K = 0

                 U = - K

Finally      K = - 210 J

So magnitude of change in kinetic energy is equal to the change in potential energy

Negative sign indicates that kinetic energy is decreased.

So we can conclude that kinetic energy is converted to potential energy

5)

a)Let the mass of the object A be M = 2 kg

Velocity of the object be v = 6 m/s

Mass of the object B be m = 1 kg

Velocity of the object B is u = 0

The two objects stick after collision and let their common velocity be V

On applying law of conservation of momentum ,

M * v + 0 = ( M + m ) * V

    2 * 6    = ( 2 + 1 ) * V

         3V   = 12

           V   = 4 m/s

So the common velocity V = 4 m/s

b)kinetic energy of the system before collision is

           Kb = ½ * M * v2 + 0

                = ½ * 2 * 6 * 6

                = 36 J

c)kinetic energy after collision

          Ka = ½ * ( M + m ) * V2

                = ½ * ( 2 + 1 ) * 4 * 4

                = 24 J

So we can observe that kinetic energy of the system is decreased.

d)percentage loss in kinetic energy is given by

           Kp = change in kinetic energy / initial kinetic energy * 100

                = [ (36 – 24) / 36 ] * 100

                = ( 12 / 36 ) * 100

                = 33.33 %

So kinetic energy is decreased by 33.33 %

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