Professional golfers can drive the ball over 200 m. While air resistance is sign

Question

Professional golfers can drive the ball over 200 m. While air resistance is significant, its effects are nearly canceled because lift is gained from backspin on the ball, and so a reasonable description of the ball’s trajectory is obtained if air resistance is neglected. Suppose a golfer hits a ball 200 m on level ground, giving it an initial velocity at an angle of 45.0o. The duration of the impact of the club and ball has been measured to be 1.00 ms.

            a)         Calculate the velocity of the ball just after it is struck.

b)         What is the 0.0450 kg ball’s change in momentum?

c)         Find the average force exerted by the club on the ball.

Explanation / Answer

(a)
Let the initial velocity be v @ angle of 450
Horizontal component = v*cos(45)
Horizontal distance travelled = 200 m

Distance = Speed * time
200 = v*cos(45) * t
t = 200/ v*cos(45)

Vertical component = v *sin(45)
vertical acceleration = 9.8 m/s^2

S = v *sin(45) * t - 0.5*9.8*t^2
0 = v *sin(45) * t - 4.9*t^2
v *sin(45) = 4.9 *t

Substituting value of t,
v *sin(45) = 4.9 *  200/ v*cos(45)
v = 31.3 m/s
Velocity of the ball just after it is struck, v = 31.3 m/s

(b)
Change in Momentum = m*v
Vf = 31.3 m/s
Vi = 0 m/s
Change in Momentum = 0.045 * 31.3
Change in Momentum = 1.41 kg m/s

(c)
Impulse = F*t =  m*v

We know, t = 1 * 10^-3 s
F = 1.41/(1 * 10^-3)
F = 1.41 * 10^3 N
Average force exerted by the club on the ball, F = 1.41 * 10^3 N

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