Answer:
His speed vf after the collision is vf = 0.10m/s
Complete Question;
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 10.7 . Olaf's mass is 73.4 .
If the ball hits Olaf and bounces off his chest horizontally at 7.90 m/s in the opposite direction, what is his speed vf after the collision?
Explanation:
According to the law of conservation of momentum;
m1u1 + m2u2 = m1v1 + m2v2 .....1
Where;
m1 and m2 are masses of the boy and ball respectively
u1 and u2 are the initial velocity of the boy and ball respectively
v1 and v2 are the final velocity of the boy and ball respectively
Given;
m1 = 73.4kg
m2 = 0.4kg
u1 = 0
u2 = 10.7;m/s
v1 = vf
v2 = -7.90m/s
Substituting into equation 1, we have
73.4(0) + 0.4(10.7) = 73.4vf + 0.4(-7.90)
vf = [0.4(10.7) + 0.4(7.90)]/73.4
vf = 0.10m/s
