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garik1379 [7]
3 years ago
13

Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and

facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.91 m/s, while Jo moves off in the opposite direction at a speed of 1.1 m/s. Assuming that friction is negligible, find Al's mass.
Physics
1 answer:
Keith_Richards [23]3 years ago
6 0

Answer: 91.94 kg

Explanation: Supoose Al's mass is m. The combined mass is 168kg, so Jo's mass is 168 - m.

The Law of Conservation of Momentum states that when two objects collide, the total momentum of both objects before the collision is the same as the total momentum of both objects after the collision.

At the beginning, Al and Jo are united without movement, so their initial momentum is zero.

After the release of their hands, Al goes in one direction and Jo moves to the opposite direction. Suppose the direction Al is moving is positive. Conservation of momentum will be

m_{1}v_{1}_{(f)}+m_{2}v_{2}_{(f)}=m_{1}v_{1}_{(i)}+m_{2}v_{2}_{(i)}

where

index i referes to initial momentum

index f to final momentum

index 1 refers to Al

index 2 to Jo

Calculating:

m(0.91)-[(168-m)(1.1)]=0

0.91m+1.1m-184.8=0

2.01m=184.8

m = 91.94

Al has a mass of 91.94 kg.

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Answer:

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Explanation:

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The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

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W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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