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Mashcka [7]
3 years ago
7

A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in

the block with half of its original speed. what is the velocity of the block right after the collision?
Physics
1 answer:
sesenic [268]3 years ago
7 0
Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0 

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v 
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.
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I attached the picture of the missing table.
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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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Part 3
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v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}
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Part 5
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Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

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