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 0.425 kg ball that is traveling horizontally at 12.0 m/s . Your mass is 68.5 kg . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Astronaut rescue. Part A If you catch the ball, with what speed do you and the ball move afterwards? Express your answer in centimeters per second.

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cjmejiab cjmejiab · Answer 1 · 2019-11-09T04:58:32+01:00

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]

Where,

[tex]m_1[/tex] = mass of ball

[tex]m_2[/tex] = mass of the person

[tex]u_1[/tex] = Velocity of ball before collision

[tex]u_2[/tex] = Velocity of the person before collision

[tex]v_1[/tex] = velocity of ball afer collision

[tex]v_2[/tex]= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

[tex]v_1 = v_2[/tex]

[tex]m_1u_1 + m_2u_2 = (m_1+m_2)v_2[/tex]

Re-arrenge to find [tex]v_2[/tex],

[tex]v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}[/tex]

Our values are,

[tex]m_1[/tex]= 0.425kg

[tex]u_1[/tex]= 12m/s

[tex]m_2[/tex]= 68.5kg

[tex]u_2[/tex]= 0m/s

Substituting,

[tex]v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}[/tex]

[tex]v_2 = 0.074m/s[/tex]

The speed of the person would be 0.074m/s after the collision between him/her and the ball