Physical lifts, high school physical lifts problem

The moment the rope is interrupted, the pull of the rope to the ball disappears, and the ball is only subject to gravity. At this point, the ball has acceleration g downwards and the lift has acceleration 5 upwards, and the instantaneous velocity of both is equal. The problem translates into the following form, find the time it takes for the ball to meet the ground of the lift.

Cut the rope, is the height of the ball relative to the floor h? The ball should be thrown upright before it is free-falling! H has changed! ”

That's true, at this time relative to the ground. If it is relative to the lift, the muzzle velocity can be ignored.

We can do it again and again, and when you are done, you will find that the initial speed is useless.

Let the initial velocity at shear be v0, then the displacement of the lift floor is:

s1=v0t+1/2at^2

The motion of the ball is:

s2=v0t-1/2gt^2

At the beginning, the distance between the ball and the lift floor is h

When the ball is in contact with the floor, the lift floor goes up more than the ball, that is, s1=s2+h

v0t+1/2at^=v0t-1/2gt^2+h1/2（g+a)t^2=h

t^=2h/(g+a)

t = sub-root [2h (g+a)] = sub-root (2h 15) is an algorithm that considers the initial velocity with the earth as a reference.

If the elevator is taken as a reference, there is no initial velocity problem, and the direct equation 1 2(g+a)t 2=h

The solution gives t = infra-root [2h (g+a)] = sub-root (2h 15).

Using the ground as the reference frame, the acceleration of the ball is the acceleration of gravity (downwards) and the acceleration of the lift is 5m s 2 (upwards). Using the lift as the reference frame, the acceleration and velocity of the lift are both zero. Whereas, the acceleration of the ball is g-(-5ms2) (downward).

The initial velocity of the ball is zero. Therefore, in the upstairs solution, the symbol of acceleration is wrong, even more so:

The distance between the small ball and the floor of the lift before cutting the rope is h

Because h=1 2(g-a)t 2

So: t 2 = 2h (g-a) = 2h 15 so time t = under the root (2h 15).

Cut the rope, is the height of the ball relative to the floor h? The ball should be thrown upright before it is free-falling! H has changed!

For terrestrial reference frames, this is true. But for the lift, it is a falling motion with zero initial velocity and g+5 acceleration! The moment the lift and ball are sheared, there is no relative movement.

Thank you for your understanding. Also, you might as well use the ground frame of reference to solve the problem and see if the conclusion is the same, but the latter solution is more troublesome (first assume the upward velocity of the ball at the beginning, and also consider the movement of the lift floor).

How high is the ball from the floor when it falls?

The elevator rises in an overweight state, and the spring scale indicator must be greater than 5N, and the two sides of the balance, that is, the weight and the object, are overweight, so the indicator remains unchanged, so choose C

The answer is c, the balance is constant.