Question

A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible mass. Knowing that the arm is released from the horizontal position and that the coefficient of restitution between the foot and the 1-lb ball is 0.8, determine the exit velocity of the ball (a) if the ball is stationary, (b) if the ball is struck when it is rolling toward the foot with a velocity of 10 ft/s.

Answer 1

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) If we use the conservation of energy for the soccer arm between tap and lowest position:

$mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s$

The velocity of the ball in the tangencial direction is:

v₂*sinθ = v₁*sin30

If v₁ = 0

θ = 0º

The coefficient of restitution is:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}$

where O=θ

The total momentum is:

$mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)$, where O = θ

$mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)$

If we multiply by g:

$mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)$

Clearing v₂:

v₂ = 26.6 ft/s

b) The velocity of the ball is:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

v₂ * sinθ = 10 * sin30

sinθ = 5/v₂

The coefficient is:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}$

where O = θ

$mgv-mgv_{1} =mgv_{2} +mgv_{2} cos(O+30)\\5*19.66-1*0=5v_{2} +v_{2} cos(O+30)\\98.3=5v_{2}+v_{2} cosOcos30-v_{2} sinOsin30$

solving for v₂:

v₂ = 31.9 ft/s