Question

A 50 g golf ball is hit by a 200 g driver with a force of 8000 N at an angle of 20 degrees. The club is in contact for about 0.8 millisec. a) The speed of the golf ball just after impact is Blank 1 m/s. b) Assuming the ball simply travels as a point object (no air resistance, rotational effects),it will go to a height Blank 2 m. c) The work done by gravity on the ball to reach its maximum height is Blank 3 J. The ball (assuming no air resistance or other effects) will land at the same level on the fairway . d) The kinetic energy of the ball when it lands is Blank 4 J. If there is air resistance... e) what can you say about the distance the ball travels down the fairway Blank 5 ? f) what can you say about the speed of the ball when it lands Blank 6 ?

Answer 1

a)

change in momentum of the ball:

p = m*(v) = F*t = 8000*0.8*10^-3 = 6.4 kg.m/s

So, v = 6.4/0.2 = 32 m/s <------answer

b)

Using conservation of energy,

PE = KE

So, mgh = 0,5*m*(v*sin20)^2

So, gh = 0.5*v^2*sin^2(20)

So, 9.8*h = 0.5*32^2*(sin(20 deg))^2

So, h = 6.1 m <------answer

c)

Work done by gravity = -mgh = -0.2*9.8*6.1 = -11.96 J

d)

Kinetic energy remains same = 0.5*mv^2

KE = 0.5*0.2*32^2 = 102.4 J

e)

Using the equation of motion,

32*sin(20 deg) = -32*sin(20 deg) +9.8*t

So, t = 2.23 s

So, distance it traveles horizontally = 32*cos(20 deg)*2.23 = 67.1 m

f)

speed of the ball is same as initial = 32 m/s