# Conservation of Momentum

## 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

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?

BvU
Homework Helper
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 !

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.

BvU
Homework Helper
Well, apparently you thought wrong: a negative mass is unphysical !

Well, apparently you thought wrong: a negative mass is unphysical !
Good to know

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?

haruspex
Homework Helper
Gold Member
2020 Award
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.

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.

haruspex
Homework Helper
Gold Member
2020 Award
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.

BvU