Will the Magnet Stick to the Car or Fall to the Ground?

[AirForceOne]

Homework Statement

A person throws out the back of a car a strong magnet that is attracted to the metal of the car. Will the magnet return and stick to the back of the car or fall to the ground? The initial velocity of the magnet is v=sqrt(5)i m/s. Treat the magnet/car interaction as an acceleration that act on the magnet equal to -a_mi, where a_m=10m/s^2 and L=1m. If the magnet hits the car at a point more than L below its starting height, it will encounter the plastic bumper and thus fail to stick.

The Attempt at a Solution

So I modeled the magnet's position then its acceleration, and then its position, getting:

However, I got y=1 for the magnet for all times t, which is not correct. If I understand correctly, the acceleration vector should be:

right?

Thanks.

 

[Spinnor]

Does the following look like a good start to solving the problem?

 

[RTW69]

Spinnor's diagram is very helpful. Hint: Where is the Vx=0 and at what time does that occur

 

[AirForceOne]

Thank you very much everyone. I understand it now.

 

[rwisz]

Hello,

Thank you for providing the context and your attempt at a solution. I would approach this problem by first clarifying the variables and assumptions being made. The homework statement mentions a magnet being thrown out of a car and being attracted to the metal of the car. However, there is no mention of any other forces acting on the magnet, such as air resistance or gravity. Are we assuming that the car is moving at a constant speed and there are no external forces acting on the magnet? Additionally, the statement mentions an initial velocity of the magnet, but it is not clear in which direction the magnet is being thrown and at what angle.

Assuming that the car is moving at a constant speed and there are no external forces acting on the magnet, we can use the equations of motion to calculate the distance the magnet falls. The initial velocity of the magnet can be resolved into its x and y components, and the acceleration due to the car's metal can be represented as a negative acceleration in the y direction. The equation for the position of the magnet at any time t would be y(t) = y0 + v0y t + (1/2) a_m t^2, where y0 is the initial height of the magnet, v0y is the initial velocity in the y direction, and a_m is the acceleration due to the car's metal. Plugging in the given values, we would get y(t) = (1/2) (-10 m/s^2) t^2 + sqrt(5) m/s t. This equation can be solved for t when y(t) = 0, which would give us the time at which the magnet falls to the ground. From there, we can calculate the distance the magnet falls using the equation d = v0y t + (1/2) a_m t^2.

However, if we consider the plastic bumper on the car, then the magnet may not fall to the ground. In this case, we would need to consider the additional forces acting on the magnet when it hits the bumper. This would require more information about the properties of the bumper and how it affects the motion of the magnet.

I hope this helps. If you have any further questions or clarifications, please let me know.