A golf club of mass 0.215 kg has a clubhead speed of 55 m/s (~120mph) strikes a golf ball of mass 0.046 kg at rest on a tee. After the collision, the clubhead has a speed of 42 m/s (same direction of swing). Use the conservation of momentum law to find the speed of the ball just after impact.

For the Clubhead:

m1 = 0.215 kg

initial velocity(u1) = 55 m/s

Final velocity (v1) = 42 m/s

For the Ball:

m2 = 0.046 kg

initial velocity (u2) = 0 m/s

final velocity (v2) = x m/s

Using the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

replacing the variables

(0.215)(55) + (0.046)(0) = (0.215)(42) + (0.046)(x)

11.825 = 9.03 + 0.046x

0.046x = 2.795

x = 2.795 / 0.046

x = 60.76 m/s

Therefore, the ball will be projected at a velocity of 60.76 m/s

the answer for apex is the lowest

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form