# A golf ball of mass 46 g is struck a blow by a club

#### Waylander

A golf ball of mass 46 g is struck a blow by a club....

Hi,
I've been going through some past exam papers for the physics paper I'm doing (I got my exam tomorrow), and I came across this question:

A golf ball of mass 46 g is struck a blow by a club such that the impulse imparted to the ball makes an angle of 25˚ with the horizontal. The ball lands 200 m away on a level fairway, bouncing inelastically off the fair way at an angle of 20˚ for a second 10 m hop and successively moving on until it comes to rest. Assume air resistance is negligible and ignore any effects due to spin in what follows below.

(a) If the golf club and ball are in contact for 7.0 ms, determine the average force of impact

I can't seem to work out how to get the answer, I'm thinking it's got to do with impulse and projectile motion... yet... I am totally lost. Any guidance would be greatly appreciated.

Cheers

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#### OlderDan

Homework Helper
Waylander said:
Hi,
I've been going through some past exam papers for the physics paper I'm doing (I got my exam tomorrow), and I came across this question:

A golf ball of mass 46 g is struck a blow by a club such that the impulse imparted to the ball makes an angle of 25˚ with the horizontal. The ball lands 200 m away on a level fairway, bouncing inelastically off the fair way at an angle of 20˚ for a second 10 m hop and successively moving on until it comes to rest. Assume air resistance is negligible and ignore any effects due to spin in what follows below.

(a) If the golf club and ball are in contact for 7.0 ms, determine the average force of impact

I can't seem to work out how to get the answer, I'm thinking it's got to do with impulse and projectile motion... yet... I am totally lost. Any guidance would be greatly appreciated.

Cheers
Impulse is the integral of force over the time interval that force is applied. The average force is the total impulse divided by the time interval. The impulse is equal to the change in momentum of the ball due to that force, which you can calculate from the launch angle and the initial distance to landing.

Last edited:

#### Waylander

Thanks for your quick reply OlderDan! I think I know what you are saying... Is the following correct?

So is this calculation correct:
from
$R=\frac{v^2sin2\theta}g$

$v=\sqrt{\frac{Rg}{sin2\theta}}$

so v = (200*9.8/sin50)^(1/2)
v = 50.582 m/s

$J=P_f - P_i = m(v_f - v_i)$
so

J = 0.046(50.582 - 0)
J = 2.3268

$F_{av} = \frac{J}{{\Delta}t}$

F = 2.3268/(7*10^-3)
F = 332.396 N

Edit: Messed up Unit conversion! XD

Last edited:

#### OlderDan

Homework Helper
Waylander said:
Thanks for your quick reply OlderDan! I think I know what you are saying... Is the following correct?

So is this calculation correct:
from
$R=\frac{v^2sin2\theta}g$

$v=\sqrt{\frac{Rg}{sin2\theta}}$

so v = (200*9.8/sin50)^(1/2)
v = 50.582 m/s

$J=P_f - P_i = m(v_f - v_i)$
so

J = 0.046(50.582 - 0)
J = 2.3268

$F_{av} = \frac{J}{{\Delta}t}$

F = 2.3268/(7*10^-3)
F = 332.396 N

Edit: Messed up Unit conversion! XD
Looks OK to me

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