Question

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.

Answer 1

0.074m/s

Step-by-step explanation:

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

`m_1u_1+m_2u_2 = m_1v_1+m_2v_2`

Where,

`m_1` = mass of ball

`m_2` = mass of the person

`u_1` = Velocity of ball before collision

`u_2` = Velocity of the person before collision

`v_1` = velocity of ball afer collision

`v_2`= velocity of the person after collision

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

`v_1 = v_2`

`m_1u_1 + m_2u_2 = (m_1+m_2)v_2`

Re-arrenge to find
`v_2`,

`v_2 = (m_1u_1+m_2u_2)/(m_1+m_2)`

Our values are,

`m_1`= 0.425kg

`u_1`= 12m/s

`m_2`= 68.5kg

`u_2`= 0m/s

Substituting,

`v_2 = ((0.425)(12)+(68.5)(0))/(0.425+68.5)`

`v_2 = 0.074m/s`

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