Suppose a 0.340 kg ball is thrown at 20.0 m/s to a motionless person standing on

Question

Suppose a 0.340 kg ball is thrown at 20.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in the figure below. (a) Calculate the final linear velocity (in m/s) of the person, given his mass is 90.0 kg. _____ m/s (b) What is his angular velocity (in rad/s) if each arm is 4.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.880 m) and the rest of his body as a uniform cylinder of radius 0.190 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. _______ rad/s (c) Compare the initial and final total kinetic energies. Initial kinetic energy is the same as final kinetic energy. Initial kinetic energy is greater than final kinetic energy. Initial kinetic energy is less than final kinetic energy.

Explanation / Answer

Accoring to the given problem,

The diagram is important.

(a) Here you must conserve linear momentum. If the person catches the ball while it's velocity is w/r/t horizontal, then
0.340kg * 20m/s * cos = 90 kg * v

v = 0.0756 m/s

(b) Again, it's the horizontal component of the velocity that's important.
Here you'll conserve angular momentum:
mvr = I

I_body = ½mr² = ½ * 82 * (0.190m)² = 1.4801 kg·m²
I_arms = mL²/3 = 8 kg * (0.88m)² / 3 = 1.291 kg·m²
I_ball = mr² = 0.340kg * (0.88m)² = 0.263296 kg·m²
total I = 3.0344 kg·m²

0.340kg * 20m/s * cos * 0.88m = 3.0344kg·m² *

That's the best I can do without the diagram. Of course, if = 0º, then cos = 1.

= 1.972 rad/s

(c) initial KE = ½ * 0.340kg * (20m/s)² = 68
final KE = ½ * 3.0344kg·m² * ² = 5.9004

Answer: B Intial is grater than final.

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