Question

A baggage cart full of luggage is coasting at a speed vA across the airport taxiway. When eighteen 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 vB. Calculate the ratio vB/vA.

Any help would be nice!

Answer 1

three masses
m
.18 m
and
m - .18 m = .82 m
-----------------------
momentum = m Va forever
so
m Va = .82 m (0) + .18 m (v)
v is relative to ground
or Va = .18 v
throw speed relative to moving cart = v - Va = .82 v
v = Va/.18 = 5.56 Va

new speed of baggage relative to ground = Va - .82 v
m Va = .82 m Vb + .18 m (Va - .82 v)
Va = .82 Vb + .18 (Va - 4.55 Va)
Va (1 - .18 + .82) = Vb
Vb/Va = 1.64