Ethan, whose mass is 67 kg, stands at one end of a very long, stationary wheeled cart that has a mass of 500 kg. He then starts sprinting toward the other end of the cart. He soon reaches his top speed of 8.0 m/s, measured relative to the cart. What is the speed of the cart when Ethan has reached his top speed?

v = - 0.938 m/s (the negative sign indicating the direction of the cart is opposite the motion of Ethan)

Explanation:

mass of Ethan (Me) = 67 kg

mass of cart (Mc) = 500 kg

Ethan's top speed (Ts) = 8 m/s

let the velocity of the cart be v

relative to the earth, Ethan will be moving forward while the cart will be moving backward, therefore

Ethan, s velocity relative to the earth = 8 - v,

apply the conservation of momentum equation

MeUe + McUc = MeVe + McVc

where Ue and Uc are the initial velocities of Ethan and the cart and are all zero, the left side of the equation is equal to zero

0 = MeVe + McVc

0 = (67 x (8-v)) + (500 x v)

0 = 536v - 67v + 500v

0 = 469 + 500v

v = - 469/500

v = - 0.938 m/s (the negative sign indicating the direction of the cart is opposite the motion of Ethan)