Answer:
Ve(m) = sqrt (19.2/m)
Ve(0.35) = 7.407 m/s
Explanation:
Given:
- The ball has a mass = m
- The entry speed of the ball is Vi = Ve
- The final speed of the ball Vf = 0.5*Ve
- The constant frictional force on ball due to hay is F = 6 N
- The thickness of hay-stack is s = 1.2 m
- Assume the throw is in horizontal direction and neglect gravity forces
Find:
Derive an expression for the typical entry speed as a function of the inertia of the ball
What is the typical entry speed if the ball has an inertia of a 0.35 kg?
Solution:
- To determine the entry speed as a function of inertia we will use third equation of motion as follows:
Vf^2 = Vi^2 + 2*a*s
Where, a is acceleration of the ball through hay stack. We will use Newton's Law of motion to determine this:
F_net = m*a
The only force acting on the ball in its journey through hay-stack is the frictional force F:
- F = m*a
a = -F/m
- Input all the quantities in the third equation of motion:
(0.5Ve)^2 = Ve^2 - 2*F*s / m
0.75Ve^2 = 2*F*s / m
Ve = sqrt (8*F*s/3*m)
Plug in values:
Ve(m) = sqrt (8*6*1.2/3*m)
Ve(m) = sqrt (19.2/m)
- The entry speed for the inertia of the ball m = 0.35 kg is:
Ve(0.35) = sqrt(19.2/0.35)
Ve(0.35) = 7.407 m/s
