Question

A golf club exerts an average horizontal force of 1005 N in the –x direction on a 453 grams golf ball that is initially at rest on the tee. The club is in contact with the ball for 1.89 milliseconds. Calculate (i) the impulse on the club and (ii) the final velocity of the golf ball just as it leaves the tee (after it breaks contact with the club head)?

Answer 1

Answer:

The impulse is 1.89 Ns and final velocity of the golf 4.17 $\frac{m}{s}$

Explanation:

Given:

Average force $F = 1005$ N

Contact time $t = 1.89 \times 10^{-3}$ sec

Mass of golf ball $m = 453 = 0.453$ kg

(a)

Impulse is given by,

   $J = Ft$

   $J = 1005 \times 1.89 \times 10^{-3}$

   $J = 1.89$ Ns

(b)

For finding the final velocity,

  $J = \Delta P$

  $J = mv$

$1.89 = 0.453 v$

  $v = 4.17 \frac{m}{s}$

Therefore, the impulse is 1.89 Ns and final velocity of the golf 4.17 $\frac{m}{s}$