Asked by jay

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?
I used p1+p2=P1'+p2' but I can't seem to get a correct answer.
Answered by bobpursley
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.
Answered by Scott
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
Answered by Henry
M1 = 68kg, V1 = ?.
M2 = 0.057kg, V2 = 38 m/s.

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