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aev [14]
3 years ago
10

A red cart has a mass of 4 kg and a velocity of 5 m/s. There is a 2-kg blue cart that is parked and not moving, thus its velocit

y is 0 m/s.
The red cart hits the blue cart.


The blue cart starts to move forward with a velocity of 6 m/s. The red cart bounces off of the blue cart and moves backwards, at a velocity of 2 m/s.


Name the TYPE of collision that occurred.

Calculate the BEFORE momentum of the red cart. Use correct units.

Calculate the BEFORE momentum of the blue cart. Use correct units.

Calculate the AFTER momentum of the red cart. Use correct units.

Calculate the AFTER momentum of the blue cart. Use correct units.

Calculate the SYSTEM’S TOTAL MOMENTUM before the collision. Use correct units.

Calculate the SYSTEM’S TOTAL MOMENTUM after the collision. Use correct units.

EXPLAIN how this example supports the Law of Conservation of Momentum. Use the evidence to support your answer of how this example supports the Law of Conservation of Momentum. Use correct units. Be specific.
Physics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:dam hold up

Explanation:

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(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

a_r=\omega^2R

Plug in the given values and solve for a_r. This gives,

a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2

Now, tangential acceleration is given by the formula:

a_t=R\alpha

Plug in the given values and solve for a_t. This gives,

a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

a_{Total}=\sqrt{(a_r)^2+(a_t)^2}

Plug in the given values and solve for total acceleration, a_{Total}. This gives,

a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

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3 years ago
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