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

Answer:

0.074m/s

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 $v_2$,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$= 0.425kg

$u_1$= 12m/s

$m_2$= 68.5kg

$u_2$= 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

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