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IrinaK [193]
2 years ago
9

Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you

would need to impose an additional condition on the problem. What is that additional condition
Physics
1 answer:
hjlf2 years ago
6 0

The problem referred to in this question is missing and it is;

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left. It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

Answer:

The condition is that p_f - p_i which is the change in momentum will not be equal to zero but equal to the impulse (Ft).

Explanation:

In the problem described, by inspection, we can say that since there is no friction, we have a closed system and thus momentum is conserved.

Since momentum is conserved, we can say that;

Initial momentum(p_i) = final momentum(p_f)

Now, in this question we are told that some friction wants to be introduced on the ice and it's possible to still use conservation of momentum.

From impulse - momentum theory, we know that;

Impulse = change in momentum

Impulse is zero when no force is acting on the ice and we have; 0 = p_f - p_i

This will yield initial momentum = final momentum.

Now, since a force is applied, we know that impulse is; J = F × t

Thus;

Ft = p_f - p_i

Where F is the force due to friction.

Thus, the condition is that p_f - p_i will not be equal to zero

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Answer:

U_{b}=+7.3*10^{-8}J

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Given data

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Now  substitute the given values

So

U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J

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A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
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The acceleration of the box is approximately 1 m/s^2

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

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m = 12.0 kg is the mass of the box

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The net force can be written as

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where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

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Answer:

\theta=15.70^\circ

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