As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is

Question

As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height of hi=1.5m . It collides with a table, then bounces up to a height of hf=1.0m . The duration of the collision (the time during which the superball is in contact with the table) is tc=15ms . In this problem, take the positive y direction to be upward, and use g=9.8m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance.

A. Find the y component of the momentum, pbefore,y, of the ball immediately before the collision.

B. Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.

C. Find Jy, the y component of the impulse imparted to the ball during the collision.

D. Find the y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball.

E. Find KafterKbefore, the change in the kinetic energy of the ball during the collision, in joules.

Explanation / Answer

Mass of the ball m = 50 g = 50 x10 -3 kg

Initial height hi = 1.5 m

Final height hf =1 m

Time of contact tc= 15 ms= 15 x10 -3 s

Acceleration due to gravity g = 9.8 m/s 2

(A).Velocity of the ball just before hit the table v = [2g hi] 1/2

                                                                      = [2 x9.8 x 1.5 ] 1/2

                                                                      = 5.422 m/s

So,the y component of the momentum, pbefore,y, of the ball immediately before the collision = -mv

                                                                                                                                      = - 0.2711 kgm/s

Here negative indicates that it is along downward direction.

(B).Velovity of the ball just after collision V = [ 2ghf] 1/2

                                                               = [2 x9.8 x1] 1/2

                                                               = 4.427 m/s

So,the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table = mV

        = 0.2213 kg m/s

(C). The y component of the impulse imparted to the ball during the collision J = mV - mv

                                                                                                                = 0.4924 kg m/s

(D).The y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball

     is F = J /tc

            = 0.4924 / (15 x10 -3 )

            = 32.83 N

(E).The change in the kinetic energy of the ball during the collision = ( 1/2) mV 2 - (1/2) mv 2

                                                                                                = ( 1/2) m[V 2 -v 2]

                                                                                                = -0.24499 J

Here negative sign indicates that kinetic energy is lost                                                                                        

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