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s.h_math s.h_math
wrote...
Posts: 293
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2 years ago
A 25 kg turkey is fired from a 1.1x10^3 kg turkey launcher. If the horizontal velocity of the turkey is 325 m/s east, what is the recoil of the launcher?

Hi there, I'm kind of confused on how to solve for the recoil of the launcher, I have to show a step-by-step solution, but not sure how to solve this question.
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wrote...
Staff Member
2 years ago
Another law of conservation of linear momentum problem:

Thankfully, it is being fired horizontally, otherwise the problem would get annoying.

Turkey: 25 kg
Launcher: 1100 kg
Turkey and launcher combined:  1125 kg
Initial speed turkey and launcher: 0 m/s
Speed of turkey: 325 m/s

1125 kg ( 0 m/s ) = 25 kg ( 325 m/s ) + 1100 kg ( v )

0 = 8125 kg*m/s + 1100 kg ( v )

Bring 8125 kg*m/s to the left:

-8125 kg*m/s = 1100 kg ( v )

Solve for v:

-8125 kg*m/s / 1100 kg = v

v = -7.4 m/s, which means 7.4 [west].
- Master of Science in Biology
- Bachelor of Science
s.h_math Author
wrote...
2 years ago
Do you think you could possibly share the equation to what these numbers were plugged in 1125 kg ( 0 m/s ) = 25 kg ( 325 m/s ) + 1100 kg ( v )? Thanks!
wrote...
Staff Member
2 years ago Edited: 2 years ago, duddy
Yes, of course. Remember I made reference to the law of conservation of linear momentum

I used this law here initial: https://biology-forums.com/index.php?topic=1998690.msg5283630#msg5283630

The main idea is that momentum = momentum, and depending on the situation, you may have to modify what you place on the left-side and what is placed on the right side. It's not a one-size fits all situation.

How's that?
- Master of Science in Biology
- Bachelor of Science
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