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luda_lava [24]
3 years ago
11

As a consultant to the soft-drink industry, Dr. J is given the task of conducting the ultimate Pepsi taste test. This is Dr. J's

tenth taste test, which puts him seven up on his nearest consultant, who had only done three. Of course Dr. J is very qualified, having been hooked on soft drinks (especially orange soda) since he was Nehi to a pop bottle. Dr. J mounts a rather large container of Pepsi on a ledge some 3 meters above the ground. A bullet of mass 5 grams is then fired into the container, thus killing the taste. Not only that, but the Pepsi falls through the bullet hole onto the ground below (causing the taste to go flat). The wall of the container is 2 cm thick. The velocity of the bullet changes from an initial value of 500 m/sec just before striking the container wall to 5 m/sec upon leaving the container wall and entering the Pepsi. It finally fizzles out at a point 25 cm from the container wall.A. How much work does the container wall do on the bullet?
B. How much work does the Pepsi do on the bullet?
C. At what velocity does the Pepsi hit the floor?
Physics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

A) How much work does the Pepsi do on the bullet = 0.0625J

B) At what velocity does the Pepsi hit the floor = 7.67m/s

Explanation:

  • Given mass of bullet = 5g
  • initial velocity = 500 m/sec
  • final velocity = 5 m/sec

From work done = Force X Distance

  • WD = Mad
  • a = v - u /t

Workdone ; m(v - u )/t

A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:

  • but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²

= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²

= 624.94J

  • Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J

B) At what velocity does the Pepsi hit the floor?

  • From conservation of energy principle; PE = KE
  • mgh = 1/2mv²
  • 0.98 X 3 = 1/2v²
  • velocity = 7.67m/s

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