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.

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

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

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

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

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

Explanation / Answer

given

m = 50 g = 50*10^-3 kg

hi = 1.5 m

hf = 1 m

P_before = -m*v_before

= -m*sqrt(2*g*hi)

= -50*10^-3*sqrt(2*9.8*1.5)

= -0.271 kg.m/s

P_after = +m*v_after

= m*sqrt(2*g*hf)

= 50*10^-3*sqrt(2*9.8*1)

= +0.221 kg.m/s

Jy = p_after - p_before

= 0.221 - (-0.271)

= 0.492 kg.m/s or N.s

we know, Jy = Favg*t

==> F_avg = Jy/t

= 0.492/(15*10^-3)

= 32.8 N

K_after - K_before = 0.5*m*(v_after^2 - v_before^2)

= 0.5*m*(2*g*hf - 2*g*hi)

= m*g*(hf - hi)

= 50*10^-3*9.8*(1 - 1.5)

= -0.245 J

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