Physics Momentum

Question

A baggage cart full of luggage is coasting at a speed v_A across the airport taxiway. When forty-two percent of the mass of the cart and luggage is thrown off the cart, parallel to the ground and in the forward direction, the cart is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the cart remains the same, the cart accelerates to a new speed v_B. Calculate the ratio v_B/v_A.

Answer 1

Let m be the mass of the cart + luggage. Let v' be the speed of the mass thrown off the cart. The sum of momenta before separation must be equal the sum of momenta after by conservation of momentum.

In the first scenario,

P₀ = m•v_A = P_f = (0.42 m)•v' + (1 – 0.42)m • (0 m/s)

^ the momentum of the now-stopped cart = 0

By canceling the ´m´s on both sides of the equation we find that

v' = v_A / 0.42

In the second scenario, the initial momentum is once again P₀ = m•v_A

The final is = –(0.42 m)*v' + (1 – 0.42)m • v_B

Note the minus sign appeared because the luggage is being thrown in the opposite direction of the cart's motion.

Now we can substitute v' = v_A / 0.42 in and Cancel the 'm' s again

v_A = –v_A + (1 – 0.42)v_B

2v_A = (1 – 0.42)v_B

v_B / v_A = (2 / 0.58)

Hope this helps! (^_^)