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Natasha_Volkova [10]
2 years ago
5

A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling e

ast at a speed of 10.0 m/s. After the collision, the 0.75 kg model car is now moving east at 11 m/s. What is the speed and direction of the model truck after the collision?
Physics
1 answer:
Lyrx [107]2 years ago
3 0

Answer:

  2.5 m/s east

Explanation:

Let east be the positive direction for velocity.

The change in momentum of the 0.75 kg model car is ...

  m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)

  = (0.75 kg)(20 m/s) = 15 kg·m/s

The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.

  m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s

The required relation is ...

  15 kg·m/s = -2(v2 -10) kg·m/s

  -7.5 = v2 -10 . . . . divide by -2

  2.5 = v2 . . . . . . . add 10

The velocity of the model truck after the collision is 2.5 m/s east.

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I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)
Helga [31]
Hello! I can help you with this! 

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
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Calculate the force generated by a car that hits the wall at an
Makovka662 [10]

This is a defective question. It was WRITTEN by someone who is unclear on the concepts.  DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard.  Several times.

Your mass is the same every time.  Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one.  Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time.  But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.  

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