A 63-kg student stands on a 34-kg cart that is free to move

In summary, a 63-kg student standing on a 34-kg cart with a velocity of 3.2 m/s [W] starts to walk with a velocity of 1.8 m/s [25° W of N] relative to the floor. The new velocity of the cart can be calculated using the equation p=mv. The direction of 25 degrees E of S is likely due to a change in reference frame.
  • #1
jasonbans
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Homework Statement


A 63-kg student stands on a 34-kg cart that is free to move in any direction. The cart and student are moving together with a velocity of 3.2 m/s [W]. The student then starts to walk so that her velocity is 1.8 m/s [25° W of N] relative to the floor. Calculate the new velocity of the cart.


Homework Equations


p=mv


The Attempt at a Solution


i don't know what is going on
 

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  • #2


jasonbans said:

Homework Statement


A 63-kg student stands on a 34-kg cart that is free to move in any direction. The cart and student are moving together with a velocity of 3.2 m/s [W]. The student then starts to walk so that her velocity is 1.8 m/s [25° W of N] relative to the floor. Calculate the new velocity of the cart.


Homework Equations


p=mv


The Attempt at a Solution


i don't know what is going on

why is the direction point to 25 degrees E of S
 

1. How does the mass of the student and cart affect their motion?

The mass of the student and cart affects their motion by determining their combined inertia, or resistance to changes in motion. The greater the mass, the greater the inertia, making it more difficult to change their state of motion.

2. What happens to the motion of the cart when the student jumps off?

When the student jumps off the cart, the cart will experience a change in motion due to the conservation of momentum. The cart will move in the opposite direction of the student's jump, with a speed dependent on the mass and velocity of the student.

3. How does the force exerted by the student on the cart compare to the force exerted by the cart on the student?

According to Newton's Third Law of Motion, for every action there is an equal and opposite reaction. This means that the force exerted by the student on the cart is equal in magnitude, but opposite in direction, to the force exerted by the cart on the student.

4. Can the student and cart move with a constant velocity?

No, the student and cart cannot move with a constant velocity as there is a net force acting on the system due to the weight of the student and cart. In order to have a constant velocity, the net force on the system must be zero.

5. How does the acceleration of the cart change as the student moves around on it?

The acceleration of the cart will change as the student moves around on it. The direction of the acceleration will depend on the direction of the net force on the system, which is determined by the distribution of the student's weight on the cart. As the student moves around, the distribution of weight will change, resulting in a change in the direction and magnitude of the net force and therefore the acceleration of the cart.

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