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!
To solve this problem, let's use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before and after the mass is thrown off should be the same.
Let's denote the total mass of the cart and luggage as M, the initial velocity of the cart as vA, the velocity of the mass that is thrown off as v, and the new velocity of the cart as vB.
Step 1: Calculate the initial momentum of the cart and luggage system.
The initial momentum (p_initial) is given by the product of the total mass and the initial velocity:
p_initial = M * vA
Step 2: Calculate the final momentum of the cart and luggage system after mass is thrown off.
After throwing off 18% of the total mass, the remaining mass is 0.82M (82% of the total mass). Since the mass is thrown off in the forward direction, the momentum is given by the product of the remaining mass and vA:
p_final = (0.82M) * vA
Step 3: Calculate the momentum of the mass that is thrown off.
The momentum of the mass that is thrown off (p_thrown) is given by the product of the mass and the velocity:
p_thrown = (0.18M) * v
Step 4: Apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the sum of the initial momentum and the momentum of the mass that is thrown off should be equal to the final momentum:
p_initial + p_thrown = p_final
Substituting the corresponding expressions from the previous steps, we get:
M * vA + (0.18M) * v = (0.82M) * vA
Simplifying the equation will result in:
vB/vA = (0.18M) / (0.82M)
= 0.18 / 0.82
≈ 0.2195
Therefore, the ratio vB/vA is approximately 0.2195.
To solve this problem, we can use the conservation of momentum principle. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces acting on the system.
Let's break down the problem into two parts:
Part 1: When 18 percent of the mass is thrown off the cart:
Let's assume the initial mass of the cart and luggage is M, and the initial velocity of the cart is vA.
After 18 percent of the mass is thrown off the cart, the remaining mass is 0.82M. Let's call this mass M'. Since the mass is thrown in the forward direction, the momentum lost by the mass is given by 0.18MvA.
As the cart comes to a halt, the momentum gained by the remaining mass is 0.82M * v_f, where v_f is the final velocity of the cart.
Using the conservation of momentum, we have:
Initial momentum = Final momentum
(Initial momentum of mass thrown off) + (Initial momentum of remaining mass) = (Final momentum of remaining mass)
0.18MvA + M' * vA = M' * v_f
Substituting the mass values, we get:
0.18MvA + 0.82M * vA = 0.82M * v_f
Simplifying:
0.18vA + 0.82vA = 0.82v_f
1vA = 0.82v_f
v_f = vA / 0.82
So, the final velocity after throwing off 18 percent of the mass is v_f = vA / 0.82.
Part 2: Reversing the thrown mass with the same relative speed:
In this part, the thrown mass (0.18M) is reversed (opposite direction) but with the same speed relative to the cart, meaning the momentum lost by the mass is the same as before (0.18MvA).
As there are no external forces acting on the system, the final momentum should be the same as in Part 1. Therefore, the momentum gained by the remaining mass should be equal to 0.82M * vA.
Using the conservation of momentum again, we have:
Initial momentum = Final momentum
(Initial momentum of mass thrown off) + (Initial momentum of remaining mass) = (Final momentum of remaining mass)
0.18MvA + M' * vA = M' * v_B
Substituting the mass values, we get:
0.18MvA + 0.82M * vA = 0.82M * v_B
Simplifying:
0.18vA + 0.82vA = 0.82v_B
1vA = 0.82v_B
v_B = vA / 0.82
So, the new velocity of the cart after reversing the thrown mass is v_B = vA / 0.82.
Finally, to find the ratio vB/vA, we can substitute the values we found:
vB/vA = (vA / 0.82) / vA
vB/vA = 1 / 0.82
vB/vA ≈ 1.2195
Therefore, the ratio vB/vA is approximately 1.2195.
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