In a carnival game, the player throws a ball at a haystack. For a typical throw, the ball leaves the hay with a speed exactly one-half of the entry speed. If the frictional force exerted by the hay is a constant 6.0 N and the haystack is 1.2 m thick, derive an expression for the typical entry speed as a function of the inertia of the ball. Assume horizontal motion only, and ignore any effects due to gravity. What is the typical entry speed if the ball has an inertia of a 0.35 kg?

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

answer; /// i believe that the correct answer is (c)

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂