Momentum of carts of different masses

In summary, the drawing shows two carts, one with a mass of 1.0 kg and the other with a mass of 2.84 kg, in contact with a compressed spring. The question asks for the mass needed on the first cart for both carts to hit the ends of the track at the same time after the explosion. Using the equations for final momentum and time, it is determined that the mass needed is 1.54 kg. For the second part of the question, the correct statement is that px = py. The incorrect statements are that Ex = 2/3 Ey, px > py, Ex < Ey, and px = py/2. The correct equation for Ex is (1/2)(2.
  • #1
cyclonefb3
7
0

Homework Statement



The drawing below shows two laboratory carts (each has a mass of 1.0 kg) X and Y in contact with a compressed exploder spring between them. The mass on cart Y is 2.84 kg, distance A is 12 cm, distance B is 18 cm. What mass must be placed on cart X, such that after the explosion both carts will hit the ends of the track at the same time?

After the explosion has taken place, which of the following statements regarding the kinetic energies, EK, and the magnitudes of the momenta, pi, are correct or incorrect? (Note: The subscripts x and y refer to the carts, not the direction along x or y.)
Ex = 2/3 Ey
px > py
px = py
Ex < Ey
px = py/2


Homework Equations



Final momentum = inital momentum = 0

MxVx + MyVy = 0

T = d/v

The Attempt at a Solution



z = unknown mass on cart x
(1 + z)Vx + (1.00 + 2.84)Vy = 0

T= 12/Vy = -18/Vy so Vx = -18/12 Vy

(1+z) *-18/12 Vy + 3.84 Vy = 0

Vy[(1+z)*-18/12 + 3.84] = 0

-18/12 + -18/12 z + 3.84 = 0

-1.5Z = -2.34

Z = 1.54 kg


After the explosion has taken place, which of the following statements regarding the kinetic energies, EK, and the magnitudes of the momenta, pi, are correct or incorrect? (Note: The subscripts x and y refer to the carts, not the direction along x or y.)
Correct Incorrect Ex = 2/3 Ey
incorrect energy should be the same
Correct Incorrect px > py
not sure: depends on answer in part 1 because p = mv
Correct Incorrect px = py
Incorrect unless Mx = 2.84
Correct Incorrect Ex < Ey
Incorrect energy should be the same
Correct Incorrect px = py/2
 
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  • #2
confused by this part:

"T= 12/Vy = -18/Vy so Vx = -18/12 Vy"

Do you mean 12/Vy = -18/Vx?
 
  • #3
yea.. and from that you can derive that Vx = (-18/12)Vy
 
  • #4
ttk3 said:
yea.. and from that you can derive that Vx = (-18/12)Vy

ok... So Ex = (1/2)(1+1.54)vx^2 = 1.27(1.5vy)^2 = 2.8575vy^2

Ey = (1/2)(1+2.84)vy^2 = 1.92vy^2

so it seems like out of all the choices the only right one is px = py.

BTW Ex is not the same energy as Ey.
 

FAQ: Momentum of carts of different masses

What is momentum?

Momentum is a physical quantity that describes the amount of motion an object has. It is calculated by multiplying an object's mass by its velocity. The unit for momentum is kilogram-meters per second (kg*m/s).

How does the mass of a cart affect its momentum?

The mass of an object is directly proportional to its momentum. This means that the greater the mass of a cart, the greater its momentum will be, as long as the velocity remains constant.

What happens to the momentum when the velocity of a cart changes?

When the velocity of a cart changes, the momentum also changes. This is because momentum is directly proportional to velocity. If the velocity increases, the momentum will also increase, and vice versa.

What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. This means that momentum cannot be created or destroyed, only transferred between objects.

How can the momentum of carts of different masses be equal?

In a collision between carts of different masses, the momentum can be equal if the velocities are different. This is because momentum takes into account both mass and velocity, so as long as the values are adjusted accordingly, the overall momentum can be equal.

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