A 45 g golf ball is hit by a 150 g driver with an impulsive force of 6000 N for

Question

A 45 g golf ball is hit by a 150 g driver with an impulsive force of 6000 N for 0.5ms. The ball initially travels at an angle of 25 degrees above the horizontal a) The speed of the golf ball just after impact by the driver is ______ m/s. b) Assuming no air resistance or other rotational effects, the ball will go to a maximum height of ____ m. c) The work done by gravity on the ball to reach its maximum height is ___ J. d) The speed of the ball at its maximum height is____ e) assuming no air resistance the kinetic energy of ball when lands is ___ J f) max potential energy of ball is ___ J

Explanation / Answer

(a)

For ball,

Final momentum - Initial momentum = impulse force = F*delta_t = 6000N*0.5 ms

Final moentum = m*v=3

Speed of of the ball,

v' =3/0.045=66.67m/s

(b)

Vertical speed

Vy = v*sin 25 = 28.17

vertical height

h = vy^2/2*g = 40.46 m/s

(c)

Work done by gravity = m*g*h = 0.045*9.81*40.46 =17.86 J

(d)

Speed of the ball will be original horizontal speed = v*cos 25 =60.42 m/s

(e)

It is same as answer in (c) [ using conservation of energy law)

(f)

Max potential energy is same as (c).

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