homework-and-exercisesnewtonian-mechanics
Could someone please explain the trajectory of the ball that is bouncing in this picture…
The vertical component of the velocity of a bouncing ball is shown in the graph below. The positive Y direction is vertically up. The ball deforms slightly when it is in contact with the ground.
I'm not sure what the ball is doing and when, what happens at 1s?
You already have all the points needed. I'll quote you:
The y-component of the objects velocity will be identical for both reference frames, but the horizontal velocity is going to differ.
and then you say
The x-compoents of the plane and the object must be equal, the plane however has no vertical velocity.
Now what you should do is to think of a famous theorem that connects the sum of the squares of the components of a vector to its magnitude. The answer will pop out then..!
First of all $mv^2/r $ is a pseudo force that acts in frame of rotating body itself. So, let's work it like that.
The centrifugal force acts outwards radially from the circle of rotation (not along the thread).
First let's see the the first image.
$$T\cos\theta=mg$$ $$T\sin\theta=\dfrac{mv^2}r$$
As we move the body faster. $T\sin\theta$ must increase keeping the $T\cos\theta$ component constant(equals $mg$.)
So, both $T,\theta$ must increase to balance the centrifugal force and keep vertical component constant.
Hence , body moves upwards while speed of rotation increases.
Also, see that the motion with $\theta\ge90^0$ are not possible due to the vertical component are not balanced. The max. tension ($T\to\infty$) comes when $\lim \theta\to 90$
Best Answer
You need to remember that this isn't a plot of position vs. time, it's velocity vs. time. Now that that's emphasised, let's analyse the plot.
Clearly, the ball starts out at zero velocity and the velocity is increasing in the negative direction linearly with time. But we also know that the acceleration on a body is given by $$ a = \frac{dv}{dt}$$ i.e., the rate of change of velocity. I don't know familiar with calculus you are, but for a straight line the slope ($dv/dt$) is constant. Therefore we can infer that the acceleration is also constant.
Do we know a physical situation where this happens? Sure... a ball released from a height above ground will follow exactly that motion, with the constant acceleration being provided by gravity.
For what happens at $t = 1$ $s$ - The velocity of the ball goes from $-9$ $m/s$ to $9$ $m/s$ really quickly. This just means that the direction of the velocity changed, without changing the magnitude. This happens when a ball bounces off the ground, neglecting friction and other losses. So that's your system - A ball bouncing off the ground.
P.S - There's another way to see that this is a ball falling - Acceleration is the rate of change of velocity. And we know that the acceleration due to Earth's gravity is ~ $9.8$ $m/s^2$ so in $1$ $s$ the velocity will be ~ $9.8$ $m/s$, which is roughly what the graph is showing here.