Okay.
If the box were attached to the spring, it would go on forever. That makes sense. It also makes sense that there is a point in the cycle where the unattached box and the attached box are the same. It was pretty much established that equilibrium in not that point.
This leaves me with...
Would that fraction be 1/2? Here's my perspective, it lands, compress, and let's go once it returns to equilibrium. If it were represented graphically, the entire process would be on one side of the x-axis; it's one half of a total period.
Technically, this is problem has multiple parts, and the block lands (as in dropped) on the spring at 5 m/s after falling some unmentioned number of meters. I did not mention that because I thought it it meant nothing (velocity wasn't in the period formula, so I left it out).
Under normal circumstances, would the box fly off one the spring hits equilibrium again, or does it fly off when the spring is extended? That probably play a big role in this.
T = 2π * √(2/300), T = .513 seconds.
If I divide it by 4/3, I get a final answer of .385 seconds of touch.
I know the box isn't attached for the entire oscillation, so T has to be divided. To me, it makes sense to divide it by 4/3 (when the box falls, the spring is compressed, hits...
Okay. In other words, the wheels essentially exist so we can ignore friction, and my car's energy should be based on the body of the car rather than the wheels. That actually makes a lot of sense. Thank you.
That's true. Does the car's mass make a difference? I know mass is there, but I just figure it could be ignored (we weren't given masses). I might sound like a fool, but I don't even know how an external mass such as the car would affect the wheels of their velocity.
Hello,
I tried to put it in an equation, but it didn't really work out. In this situation, the car was about the size of a model, and, while not exact, the radius of each wheel couldn't have been more than like a centimeter. Conversely, the ball was like twice the size of the car and had a...