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Conservation of Momentum

  • #1
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Homework Statement


After shooting a 28g arrow with an initial velocity of 92m/s[forward], an archer standing on a frictionless surface travels in the opposite direction at a speed of 0.039m/s. Calculate the combined mass of the archer and the bow.

Given: **Subscript of 1 indicates values for the arrow and subscript of 2 indicates values for the archer

m1=0.028kg
vi1=92m/s[forward]
vf1=0m/s

vi1=0.039m/s[backward]
vf2=0m/s

Homework Equations


m1vi1+m2vi2=m1vf1+m2vf2

The Attempt at a Solution


The only way this problem works is if you treat the archer and the bow as one body/one object and using the above formula and values, solve for m2:

(0.029)(92)+m2(0.039)=0
2.576=-m2(0.039)
-66.05128205kg=m2

I got the correct answer aside from the negative sign in front of my answer. Why is my answer negative? Does that mean I did something wrong?

Also, why do you treat the archer and the bow as one collective object? I thought the law of conservation of momentum, which gives us the equation I used, applied to situations when two objects collide in an isolated system. If so, then what collision occurs in this problem; how are both the bow and the archer colliding with the arrow?
 

Answers and Replies

  • #2
BvU
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With your initial and final you describe a fully inelastic collision of an arrow and an archer plus bow. Rather fatal for the archer ...

Is it strange you get a negative mass ? The recoil from shooting the arrow makes the archer go backwards (that's why it's called recoil) wrt the arrow ! Yet you have the same sign for both speeds !
 
  • #3
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With your initial and final you describe a fully inelastic collision of an arrow and an archer plus bow. Rather fatal for the archer ...

Is it strange you get a negative mass ? The recoil from shooting the arrow makes the archer go backwards (that's why it's called recoil) wrt the arrow ! Yet you have the same sign for both speeds !
I thought that it was okay not to plug in negative signs indicating direction for vector values when doing calculations to find a scalar value, like mass.
 
  • #4
BvU
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Well, apparently you thought wrong: a negative mass is unphysical !
 
  • #5
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Well, apparently you thought wrong: a negative mass is unphysical !
Good to know
 
  • #6
226
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With your initial and final you describe a fully inelastic collision of an arrow and an archer plus bow. Rather fatal for the archer ...

Is it strange you get a negative mass ? The recoil from shooting the arrow makes the archer go backwards (that's why it's called recoil) wrt the arrow ! Yet you have the same sign for both speeds !
But why are the bow and archer one mass?
 
  • #7
haruspex
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But why are the bow and archer one mass?
Because the archer is assumed to hold the bow rigidly. In a different model, the archer could let go of bow and arrow at the same instant. The arrow would travel a smallish distance one way, and the bow a smaller distance the other.
Note that with a firearm that second model is closer to what happens. The impulse is so great that it is not possible (and maybe not a good idea to try) to hold the weapon rigidly.
 
  • #8
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Because the archer is assumed to hold the bow rigidly. In a different model, the archer could let go of bow and arrow at the same instant. The arrow would travel a smallish distance one way, and the bow a smaller distance the other.
Note that with a firearm that second model is closer to what happens. The impulse is so great that it is not possible (and maybe not a good idea to try) to hold the weapon rigidly.
What is the collision that happens, the bow and the arrow are not colliding, they were already in contact.
 
  • #9
haruspex
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What is the collision that happens, the bow and the arrow are not colliding, they were already in contact.
Conservation of momentum is not limited to collisions.
 
  • #10
BvU
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What is the collision that happens
Physically the time reversal of a collision happens: first they are together (sum of momenta = 0) and then they separate -- without any external forces acting. So the center of mass does not accelerate and sum of momenta remains zero.

Newton 3: action = - reaction makes that afterwards the individual momenta are equal and opposite.
 

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