Question

16. (II) A golf ball of mass 0.045 kg is hit off the tee at a speed

of 38 m/s. The golf club was in contact with the ball for
3.5 x 10-ºs. Find (a) the impulse imparted to the golf ball,
and (b) the average force exerted on the ball by the golf club.​

Answer 1

(a) 1.71 kgm/s

(b) 488.57 N

Step-by-step explanation:

Given:

Mass of the golf ball (m) = 0.045 kg

Change in velocity of the golf ball (Δv) = 38 m/s

Time of contact of the golf ball with the golf club (t) =
`3.5* 10^(-3)\ s`

(a)

Now, impulse imparted to the golf ball is equal to the change in momentum of the golf ball.

Change in momentum is given as:

`\Delta p=m* (\Delta v)\\\Delta p = 0.045* 38 = 1.71\ kgm/s`

Therefore, the impulse imparted to the golf ball = Δp = 1.71 kgm/s

(b)

Impulse imparted is also equal to the product of average force acting during the contact and the time of contact.

Therefore, Impulse = Average force × Time of contact.

Rewriting in terms of average force, we get:

`Average\ force=(Impulse)/(Time)`

Now, plug in the given values and solve. This gives,

`Average\ force=(1.71)/(3.5* 10^(-3))\ N\\\\Average\ force=488.57\ N`

Therefore, the average force exerted on the ball by the golf club is 488.57 N