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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a ball of mass 0.400 kg that is traveling horizontally at 11.5 m/s. Your mass is 75.0 kg.
(A) If you catch the ball, with what speed do you and the ball move afterwards?
(B) If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 7.80m/s in the opposite direction, what is your speed after the collision?

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opudodennis

Answer:

(a) 0.061 m/s

(b) 0.103 m/s

Explanation:

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

[tex]m_1v_1=(m_1+m_2)v_c[/tex] where [tex]v_c[/tex] is the common velocity, [tex]v_1[/tex] is the velocity of the ball [tex]m_1[/tex] and [tex]m_2[/tex] are masses of the ball and person respectively

Substituting the given values then

[tex]0.400\times 11.5 = (0.400+75)v_c\\v_c=0.061007958\approx 0.061 m/s[/tex]

(b)

Momentum, p=mv where m is the mass and v is the velocity

[tex]m_1v_1=m_1v_2+m_2v_3[/tex]

[tex]v_1[/tex] is the velocity of the ball , [tex]v_2[/tex] is the velocity of ball afterwards and [tex]v_3[/tex] is your speed, [tex]m_1[/tex] and [tex]m_2[/tex] are masses of the ball and person respectively. Since it bounces back, we give it a negative value hence

[tex]0.400\times 11.5= 0.4\times -7.8+75v_3\\v_3=0.102933333\approx 0.103 m/s[/tex]

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