Velocity vs Time Bounce

Question

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?

Answer 1

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.

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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.

Answer 2

This is a graph of velocity versus time, so the slope of the graph at any point gives the change in velocity over time, or dv/dt which is acceleration. Acceleration occurs due to the application of force to a mass (f=ma, or a=f/m).

Any change in the slope of the graph must occur due to a change in (dv/dt), which is a change in acceleration, and hence a change in the applied force. The only forces at work here are gravity and the "spring" like force caused by deformation of the ball. The changes at 1s and 1.25s are due to the addition and removal of the "spring" force from the deforming ball as it strikes and then leaves the ground. Gravity applies throughout the graph. Both these forces are "constant" where applied, giving uniform acceleration and hence a straight line dv/dt graph with some slope.

Velocity-time graphs can be slightly confusing in that a change in direction of motion, something we would see as a major event, is indicated only where the line crosses zero. If this change of direction is due to constant acceleration, like a ball at the top of its arc or at the transition of a spring-like deformation from compression to expansion, this occurs in the middle of a line of constant slope (2.25s and 1.125s above), and seems unremarkable. Crossing zero is always a point of potential interest in a graph.

Answer 3

At $t_0$=1s, the ball bounces in a perfectly elastic manner and there is no loss of momentum. Therefore, $v(t=t_0+dt) = - v(t=t_0)$. If the recording had a finer resolution in time, this would show up as a vertical jump but here you have a milder slope since it is recording only every 0.1 s. Between the bounces, gravity takes over and the ball accelerates downwards. If you compute the slope of the decreasing part of the curve, you should find a well-known constant.