A tennis player places a 68kg ball machine on a frictionless surface. the machine fires a 0.057 kg tennis ball horizontally with a velocity of 38m/s toward the north. What is the final velocity of the machine?

Question

I used p1+p2=P1'+p2' but I can't seem to get a correct answer.

Answer 1

The sum of momentum change is zero
Initial+Final=0
.057*38+68*V=0
solve for V. The negative sign means it is in the opposite direction as the original v.

Answer 2

the tennis ball and the machine have the same momentum, but in opposite directions
... Newton's 3rd law

68 * v = 0.057 * 38

TWO significant figures

Answer 3

M1 = 68kg, V1 = ?.
M2 = 0.057kg, V2 = 38 m/s.

M1*V1 = -M2*V2.
68*V1 = -(0.057*38),
V1 =