Bounce Trajectory

Classical Mechanics Level 3

In the $xy$ coordinate system, a massive particle is initially at rest at $(x,y) = (\frac{1}{\sqrt{2}}m,2m)$ . Under the influence of gravity $(10 \frac{m}{s^{2}}$ in the -y direction $)$ , the particle falls and then bounces off of a unit circle situated at the origin. It then continues on through space until its path intersects the $x$ axis.

How far is the particle (in $meters$ ) from the origin when it intersects the $x$ axis (to three decimal places)?

Note: For our purposes, define a "bounce" as an elastic impact with the surface of an object, such that the normal component of the velocity is reversed and the tangential component is preserved. "Normal" and "tangential" are defined in relation to the surface geometry at the point of impact.

The answer is 2.619.

1 solution

Steven Chase
Oct 6, 2016

Solution outline:

1) The ball impacts the circle at $(x,y) = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
2) The kinetic energy upon impact is $mg(2-\frac{1}{\sqrt{2}}) = \frac{1}{2} m v^{2}$
3) Upon impact, $v$ is thus equal to $\sqrt{g(4-\sqrt{2}})$

4) Because the surface normal vector at the impact point is at 45 degrees with respect to the horizontal, and the velocity is initially vertical, the velocity after the bounce is purely horizontal. This is possible to prove in a rigorous way, but I will leave that as an exercise for the reader.

5) We need to determine the flight time for the ball after the bounce, which is the time taken to fall to $(y=0)$ . The relevant equation is:

$\frac{1}{\sqrt{2}} = \frac{1}{2} g\, t_f^{2}$

6) The flight time after the bounce is thus $\sqrt{\frac{\sqrt{2}}{g}}$

7) The x intercept position is therefore as shown below:

$x_{int} = \frac{1}{\sqrt{2}} + \sqrt{g(4-\sqrt{2}})\sqrt{\frac{\sqrt{2}}{g}} = \frac{1}{\sqrt{2}} + \sqrt{4\sqrt{2} - 2} \approx 2.619$

yes, great solution !

A Former Brilliant Member - 4 years, 5 months ago

Bounce Trajectory

The answer is 2.619.

1 solution

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