The E 4ny theexpe (7%) Problem 11: Suppose that a 1 with his feet off the ground

Question

The E 4ny theexpe (7%) Problem 11: Suppose that a 1 with his feet off the ground, while both of them are moving horizontally ,73 kg fo th all player n ming at SS msa hesa044kgball moving datas mis as 25% Part ( ) Calculate the final spoed f the player in meters per econd, if the ball and player are initially ingin thesame direction Grade Sammary sin0 cota asino acsO atanO acotanO sin0 cosho tanho cotanh O Degrees O Radians Submit Hine Feedback: 1deduction per feedboack 0% deduction per hit. Hints remaining: - 25% Part (b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball. 25% Part (c) Calculate the final speed of the player, in meters per second, if he ball and player are initially moving in opposite directions 25% Part (d) Calculate the change in kinetic energy of the system, in joules, in this case. MacBook

Explanation / Answer

(a) Consider ball and player as a system
Initial momentum of the system
= mpVP + mBVB = 117.5*(5.5)+ 0.44*(26.5) = 657.91 kg-m/s -----------(1)
Final momentum of the system
= (mP + mB)V = (117.5+0.44)V -------------(1)
Where V is the final velocity of player and ball.
Applying conservation of momentum
(117.5+0.44)V =  657.91
V = 5.578 m/s
(b) Initial kinetic energy of the system
= (1/2)mPVP2 + (1/2)mBVB2
= (1/2)*(117.5)*(5.5)2 + (1/2)*0.44*(26.5)2 = 1777.1875 + 154.495 = 1931.6825 J
Final kinetic energy
= (1/2)(mP +mB)V2 = (1/2)(117.5+0.44)*(5.578)2 = 1834.798 J
Change in kinetic energy
= 1931.6825 - 1834.798 = 96.845 J
(c) When they are moving in opposite direction
Initial momentum of the system
= mpVP + mBVB = 117.5*(5.5)+ 0.44*(-26.5) = 634.59 kg-m/s
Final momentum of the system
=  (mP + mB)V = (117.5+0.44)V
Applying momentum conservation
(117.5+0.44)V = 634.59
V = 5.381 m/s
(d) Initial energy of the system
=(1/2)mPVP2 + (1/2)mBVB2
= (1/2)*(117.5)*(5.5)2 + (1/2)*0.44*(26.5)2 = 1777.1875 + 154.495 = 1931.6825 J
Final energy of the system
=(1/2)(mP +mB)V2 = (1/2)(117.5+0.44)*(5.381)2 = 1707.486 J
Change in kinetic energy
= 1931.6825 - 1707.486 = 224.1965 J

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