Physics...Momentum. Help

A 50.0 kg archer, standing on frictionless ice, shoots a 200g arrow at a speed of 150 m/s to the right

A)What is the velocity of the archer after the shot?

please explain how you reach an answer. I am lost. THANKS

Well we know the momentum equation is p = m*v and that momentum is conserved (momentum before = momentum after)

So since the arches is standing on frinctionless ice, it just tells us that all the velocity he/she ends up with will be translated into motion instead of being wasted by friction.

So I'll first define the knowns
m1 = 50.0 kg (archer)
m2 = 200 g = 0.200 kg (arrow)
V1 = ???
V2 = 150 m/s

So since momentum is conserved...:
Pbefore = Pafter
m1*v01 + m2*v02 = m1*v1 + m2*v2

Neither the arrow or the archer have an initial velocity because before the arrow is shot they both have no velocity. So this means the left-hand side of the equation is 0.
0 = m1*v1 + m2*v2
Plug in our known variables:
0 = (50.0 kg)*v1 + (0.200 kg)*(150 m/s)
0 = (50.0 kg)*v1 + (30 kg*m/s)
-30 kg*m/s = (50.0 kg)*v1
(-30 kg*m/s) / 50.0 kg = v1
v1 = -0.6 m/s

That is the speed in which the archer will move on the frictionless ice. Since it is negative, that means he/she is moving opposite of the arrow so to the left.

200g = 0.2kg.
(0.2 x 150)/50kg = 0.6m/sec., in the opposite direction to the arrow.

Momentum = (mass x velocity). So you find arrow momentum, and divide it by archer's mass.That gives archer's velocity the other way.