A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?
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* Physics - bobpursley, Sunday, February 25, 2007 at 5:25pm
Use conservation of momentum. I will be happy to critique your thinking.
* Re: Physics - COFFEE, Sunday, February 25, 2007 at 11:19pm
Ok, so I use mv = m1v1 + m2v2??? And I would get m1 and m2 by dividing the weights by 9.8 m/s^2? Please explain.
The original momentum is..
(915+2805)g18
That is equal to the final momentum
915g*(-40+18) + 2805g(V+18)
solve for V.
0.95
To solve this problem, you can use the principle of conservation of momentum. The initial momentum of the system (man + flatcar) is equal to the final momentum of the system.
The initial momentum is the sum of the momentum of the man and the momentum of the flatcar. The momentum of an object is given by the product of its mass and velocity.
The mass of the man can be found by dividing his weight by the acceleration due to gravity (9.8 m/s^2):
mass of man = weight of man / acceleration due to gravity
mass of man = 915 N / 9.8 m/s^2
mass of man = 93.37 kg
Similarly, the mass of the flatcar can be found by dividing its weight by the acceleration due to gravity:
mass of flatcar = weight of flatcar / acceleration due to gravity
mass of flatcar = 2805 N / 9.8 m/s^2
mass of flatcar = 286.22 kg
Now we can calculate the initial momentum of the system:
initial momentum = (mass of man + mass of flatcar) * initial velocity
initial momentum = (93.37 kg + 286.22 kg) * 18.0 m/s
initial momentum = 37923.96 kg⋅m/s
The final momentum of the system is equal to the momentum of the man running in the opposite direction plus the momentum of the flatcar with an increased velocity:
final momentum = mass of man * (final velocity of man) + mass of flatcar * (final velocity of flatcar)
final momentum = 93.37 kg * (-40.00 m/s) + 286.22 kg * (V + 18.0 m/s)
Solving for V, we equate the initial momentum to the final momentum:
37923.96 kg⋅m/s = 93.37 kg * (-40.00 m/s) + 286.22 kg * (V + 18.0 m/s)
Solving the above equation for V will give us the resulting increase in the speed of the flatcar.
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the man starts running is equal to the total momentum after he starts running.
The initial momentum is the sum of the momentum of the man and the momentum of the flatcar. The momentum is given by the product of mass and velocity.
The mass of the man can be found by dividing his weight (915 N) by the acceleration due to gravity (9.8 m/s^2), therefore the mass of the man is 93.47 kg.
The mass of the flatcar can be found by dividing its weight (2805 N) by the acceleration due to gravity, therefore the mass of the flatcar is 286.22 kg.
The initial momentum is then given by:
Initial momentum = (Mass of the man * Velocity of the man) + (Mass of the flatcar * Velocity of the flatcar)
Initial momentum = (93.47 kg * 0 m/s) + (286.22 kg * 18.0 m/s) (The man is stationary, so his velocity is 0 m/s)
Initial momentum = 0 + 5151.96 kg*m/s
The final momentum is the sum of the momentum of the man and the momentum of the flatcar after the man starts running. The velocity of the man is given as -40.00 m/s relative to the flatcar.
The final momentum is then given by:
Final momentum = (Mass of the man * Velocity of the man) + (Mass of the flatcar * Velocity of the flatcar)
Final momentum = (93.47 kg * -40.00 m/s) + (286.22 kg * V) (V is the increase in velocity of the flatcar)
Now, according to the principle of conservation of momentum, the initial momentum is equal to the final momentum.
Initial momentum = Final momentum
5151.96 kg*m/s = (93.47 kg * -40.00 m/s) + (286.22 kg * V)
Now we can solve for V, the increase in velocity of the flatcar.
5151.96 kg*m/s = -3738.8 kg*m/s + 286.22 kg * V
Add 3738.8 kg*m/s to both sides of the equation:
5151.96 kg*m/s + 3738.8 kg*m/s = 286.22 kg * V
8889.76 kg*m/s = 286.22 kg * V
Divide both sides of the equation by 286.22 kg:
V = 8889.76 kg*m/s / 286.22 kg
V ≈ 31.04 m/s
Therefore, the resulting increase in the speed of the flatcar is approximately 31.04 m/s.