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

Answer:

522.84 N

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

$R=\frac{u^{2}Sin2\theta }{g}$

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

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

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

We know that range of a projectile is given by

$R=\frac{v_{o}^{2}sin(2\theta )}{g}$

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

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

Thus the force is obtained as

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

Thus force equals $F=338.66N$