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trapecia [35]
3 years ago
14

A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding

without any friction on a smooth surface. a. Find the momentum of the box with respect to the floor. b. Find the momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface.
Physics
1 answer:
Yuki888 [10]3 years ago
8 0

For the part a) we need only the momentum of the box and we have the data to find it.

Momentum is given by,

p=mv

where clearly, p is the momentum, m the mass of the box and v is the velocity.

Substituting,

p=(5kg)(5m/s)\\p=25kg.m/s

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

p=(5kg)(5m/s)\\p=25kg.m/s

<em>It is the same that part a)</em>

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Answer:

<em>Hello, The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).</em>

Explanation:

The equation for the velocity of an object thrown upward is the following:

v = v0 + g · t

Where:

v = velocity of the ball.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:

v = v0 + g · t

v = 3.0 m/s - 9.8 m/s² · 0.40 s

v = -0.92 m/s

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

<em>Hope That Helps!</em>

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