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3 years ago

8

A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l

eft. The bowling ball barely slows down. What is the approximate velocity of the beach ball after the collision?

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

$v'_2=3V$

Explanation:

From the question we are told that:

Bowling ball Speed $v_1=5 m/s$

Beach ball Speed $v_2=2 m/s$

Let The Mass be equal i.e

$M_1=M_2$

Therefore

Generally the equation for Velocity of beach ball after collision $v'_2$ is mathematically given by

Since Velocity is Vector Quantity

Therefore

$v'_2=v_1-v_2$

$v'_2=5-2$

$v'_2=3V$

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Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

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<u>Equations</u>

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Now, at the end of the movement, $Y=0$ , then

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I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
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Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
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I will do all the calculations in the google sheet that I will share with you.
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Part 2
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I will do all the calculations in the google sheet that I will share with you.
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$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
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Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

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3 years ago

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Andrej [43]

i think its b cause if you were talking about ice it freezes the melts the evaporates to gas

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3 years ago

Read 2 more answers