High-speed stroboscopic photographs show that the head of a 214 g golf club is trav- eling at 36.7 m/s just before it strikes a 45.9 g golf ball at rest on a tee. after the collision, the club head travels (in the same direction) at 26.9 m/s. find the speed of the golf ball immediately after impact. answer in units of m/s.

v = 45.7 m/s

Explanation:

It is given that the head of a golf club has a mass of 214 g and is travelling at 36.7 m.s before the impact with a golf ball which has a mass of 45.9 g and is at rest on a tee.

After the club hits the ball, its speed decreases to 26.9 m/s and we are to determine the speed of the golf ball after the impact.

So, the initial momentum () = 36.7 * 214 = 7853.8 g; and

the final momentum () = 45.9 * v + 214 * 26.9

Since () = (),

7853.8 = 45.9v + 5756.6

45.9v = 2097.2

v = 45.7 m/s

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.