Hunter's Recoil Speed from Rifle Shot

[henry3369]

Homework Statement

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 950m/s . The mass of the hunter (including his gun) is 72.5kg , and the hunter holds tight to the gun after firing it.

Find the recoil speed of the hunter if he fires the rifle at 52.0∘ above the horizontal.

Homework Equations

The Attempt at a Solution

Vx = 950cos(52)
Vy = 950sin(52)

Conservation of Momentum in the x-Direction:
m1v1' = -m2v2'
(72.5)v1' = (-0.0042)(950cos(52))
v1' = -0.033882611 m/s

Conservation of Momentum in the y-Direction:
m1v1' = -m2v2'
(72.5)(v1') = (-0.0042)(950sin(52))
v1' = -0.043367764 m/s

vf = sqrt(0.033882611²+0.043367764²) = 0.055 m/s

The correct answer is 0.0339 m/s

What am I doing incorrectly?

 

[gneill]

Is the hunter free to recoil in the Y-direction?

 

[henry3369]

Is the hunter free to recoil in the Y-direction?

I know that the hunter wouldn't move in the y-direction, but wouldn't there still be an initial velocity in the y-direction? Similar to how if an object is dropped, it has a y-component of velocity right as it hits the ground.

 

[gneill]

I know that the hunter wouldn't move in the y-direction, but wouldn't there still be an initial velocity in the y-direction? Similar to how if an object is dropped, it has a y-component of velocity right as it hits the ground.

The hunter is already on the ice surface. Any velocity in the downward direction would have to include the ice and the rest of the planet.