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Vikki [24]
2 years ago
9

A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag

e force due to the collision is correct? The small car and the truck experience the same average force. The small car experiences the greater average force. The truck experiences the greater average force. It is impossible to tell since the masses are not given. It is impossible to tell since the velocities are not given. 1 points Saved Question 1 of 5 Moving to another question will save this response.
Physics
1 answer:
andrey2020 [161]2 years ago
6 0

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

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Horizontally, the block is under the influence of

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It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)

where <em>m</em> is the mass of the block. Solve for <em>m</em> :

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<em>m</em> = (7 N) / (1 m/s²)

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(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:

<em>n</em> + (-<em>w</em>) = 0

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<em>n</em> = <em>m</em> <em>g</em>

where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.

<em>n</em> = (7 kg) (9.8 m/s²)

<em>n</em> = 68.6 N

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3 N = <em>µ</em> (68.6 N)

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