A package of mass m is released from rest at a warehouse loading dock and slides down the h = 3.0 m - high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

a. What is the speed of the package of mass m right before the collision?
b. Suppose the packages stick together. What is their common speed after the collision? Is the total mechanical energy for the system including both packages conserved in this case? If not, what is the difference of the mechanical energies before and after the collision?
c. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Is the total linear momentum for the system including both packages conserved before and after this perfectly elastic collision? Why or why not?

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Respuesta :

Xema8 Xema8 · Answer 1 · 2019-11-25T06:32:11+01:00

Answer:

Explanation:

a ) Applying law of conservation mechanical energy

1/2 mv² = mgh

v =√ 2gh

= √ 2 x 9.8 x 3

= 7.67 m /s

b ) Applying law of conservation of momentum

m x 7.67 = ( m + 2m ) v

= 3m v

v = 2.55 m /s

In perfectly inelastic collision , mechanical energy is not conserved.

Initial kinetic energy

= 1/ 2 x m x v²

= .5 x m x 58.8

= 29.4 m

Kinetic energy after collision

= .5 x 3m x (2.55) ²

= 9.75375 m

Loss of kinetic energy

= 19.646 m

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