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During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell?

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Final answer:

The mass of the barbell is calculated using the conservation of momentum to be 4.00 kg. The total kinetic energy gained through the maneuver is 210 J, which originates from the work done by the clown's muscles converting chemical energy into kinetic energy.

Step-by-step explanation:

The scenario provided is an application of the conservation of momentum in physics. Since the clown and the barbell are initially at rest, their combined momentum is zero. According to the law of conservation of momentum, the total momentum after the clown throws the barbell must also be zero. This can be represented by the equation:

initial momentum = final momentum

0 = (mass of clown x velocity of clown) + (mass of barbell x velocity of barbell)

Solving for the mass of the barbell (mbarbell), we get:

mbarbell = -(mass of clown x velocity of clown) / velocity of barbell

Plugging in the given values:

mbarbell = -(80.0 kg x 0.500 m/s) / 10.0 m/s

mbarbell = -40.0 kg m/s / 10.0 m/s = 4.00 kg

The kinetic energy gained by this maneuver can be calculated for both the clown and the barbell. The collective kinetic energy is given by:

Ktotal = (1/2 x mass of clown x (velocity of clown)2) + (1/2 x mass of barbell x (velocity of barbell)2)

Substituting the known values:

Ktotal = (1/2 x 80.0 kg x (0.500 m/s)2) + (1/2 x 4.00 kg x (10.0 m/s)2)

Ktotal = 20 J + 200 J = 210 J

The kinetic energy comes from the clown's muscles during the throwing action. The muscles do work by converting chemical potential energy found in ATP molecules into kinetic energy for the motion of the clown and the barbell.

User Dylan Vester
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