Question

A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a flat fairway. The acceleration of gravity is 9.8 m/s 2 . If the golf club and ball are in contact for 6.3 ms, what is the average force of impact? Neglect air resistance. Answer in units of N.

Answer 1

522.84 N

Step-by-step explanation:

m = 59.2 g = 59.2 x 10^-3 kg

θ = 50.5

R = 130 m

g = 9.8 m/s^2

t = 6.3 m s = 6.3 x 10^-3 s

Let u be the velocity of projection

Use the formula for the range

`R=(u^(2)Sin2\theta )/(g)`

130 = u^2 x Sin 2(50.5) / 9.8

u = 36.06 m/s

The vector form of the initial velocity

u = 36.05 (Cos 50.5 i + Sin 50.5 j)

u = 22.93 i + 27.82 j

The velocity of the body as it strikes with the surface is same but the direction is downward.

v = 22.93 i - 27.82 j

Change in momentum, Δp = m (v - u)

Δp = 59.2 x 10^-3 x (22.93 i - 27.82 j - 22.93 i - 27.82 j)

Δp = 59.2 x 10^-3 x ( - 55.64 j)

Δp = - 3.294 j

Δp = 3.24

Force = Rate of change in momentum

F = Δp / Δt = 3.294 / (6.3 x 10^-3) = 522.84 N

Answer 2

The force is calculated as 338.66 N

Step-by-step explanation:

We know that force is given by

`\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=(m(v_(f)-v_(o)))/(\Delta t)`

We know that range of a projectile is given by

`R=(v_(o)^(2)sin(2\theta ))/(g)`

it is given that R=130 m applying values in the above equation we get

`R=(v_(o)^(2)sin(2\theta ))/(g)\\\\v_(0)=\sqrt{(Rg)/(sin(2\theta ))}\\\\\therefore v_(o)=\sqrt{(130* 9.81)/(sin(2* 50.5_(o)) )}\\\\v_(o)=36.04m/s`

Thus the force is obtained as

`\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2* 10^{}-3(36.04-0)}{6.3* 10^(-3)}`

Thus force equals
`F=338.66N`