How does a mass moving up a ramp change its velocity?

[phantomvommand]

Homework Statement

A point mass is moving at speed v, on a horizontal plane, until it reaches an incline. Immediately, its direction changes, but its speed remains at v. How does this happen?

Relevant Equations

Understanding of forces

A point mass is moving at speed v, on a horizontal plane, until it reaches an incline. Immediately after just climbing up the incline, its speed remains at v, but its direction changes. How does this happen?

Q2: Now, I drop a point mass such that it falls vertically downward onto a fixed ramp.
Is only velocity parallel to the slope of the ramp preserved? I think this is because the ramp exerts an impulse on the ball such that the momentum perpendicular to the slope is removed.

 

[BvU]

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[berkeman]

I think this is because the ramp exerts an impulse on the ball such that the momentum perpendicular to the slope is removed.

I think the concept of impulse is useful for this problem. Can you show us some equations involving impulse, momentum, mass and velocity? Can you then write the equations for the two parts of the question?

 

[phantomvommand]

I think the concept of impulse is useful for this problem. Can you show us some equations involving impulse, momentum, mass and velocity? Can you then write the equations for the two parts of the question?

@BvU
For Q1:
I am trying to see the transition onto the incline as a round curve. I believe it collides with the ramp, and so there is some force between the mass and the ramp, which leads to an impulse on the mass. Let's say the angle of the ramp is $\theta$.

It seems, that for the velocity vector v to change direction but retain its magnitude, the impulse must be directed at an angle $90 - \frac {\theta} {2}$ from the ground. This would give a resultant velocity with magnitude same as before and angled at $\theta$. May I know why the angle of the impulse is $90 - \frac {\theta} {2}$, instead of perpendicular to the slope (ie $90 - \theta$)

 

[haruspex]

I am trying to see the transition onto the incline as a round curve

Yes, that is one way it is possible for speed to be maintained (initially). But that is not a single impact, more like an infinite sequence of tiny impacts.
If we assume a point particle and a smooth curve of radius r then the only loss of speed results from the climb of $r(1-\cos(\theta)$. As we make r vanishingly small, that tends to zero.
Alternatively, it could be, say, a short ramp at angle $\theta/2$ followed immediately by the main ramp at angle $\theta$. An elastic bounce off the short ramp does the trick.

 

[ergospherical]

It seems, that for the velocity vector v to change direction but retain its magnitude, the impulse must be directed at an angle $90 - \frac {\theta} {2}$ from the ground. This would give a resultant velocity with magnitude same as before and angled at $\theta$. May I know why the angle of the impulse is $90 - \frac {\theta} {2}$, instead of perpendicular to the slope (ie $90 - \theta$)

Say a particle is traveling along the $x$ axis at a speed $v$, until it reaches the origin at $t=0$ and is subject to a kick $F(t) = I\delta(t)$ parallel to the $y$ axis. It's momentum in the $y$ direction changes as $\Delta p_y = I \int \delta(t) dt = I$ and it obtains a speed $v_y = I / m$ in the $y$ direction. Its direction has changed by an angle $\theta = \mathrm{arctan}\left( I/ mv \right)$. Except, its speed is now $v' = \sqrt{v^2 + v_y^2} \geq v$.

What if you fixed this by kicking it with an additional force $G(t) = -J \delta(t)$ parallel to the $x$ axis? It's momentum in the $x$ direction changes as $\Delta p_x = -J$ and its speed in the $x$ direction reduces to $v_x = v - J/m$. The total speed of the particle is then\begin{align*}
v' = \sqrt{v_x^2 + v_y^2} = \sqrt{v^2 - 2vJ/m + J^2/m^2 + I^2 / m^2}
\end{align*}If the speed has not changed, $v' = v$, then the condition relating $I$ and $J$ is $J^2 -2mvJ + I^2 = 0$,\begin{align*}\implies J &= \dfrac{2mv - \sqrt{4 m^2 v^2 - 4I^2}}{2} = mv - \sqrt{m^2 v^2 - I^2}
\end{align*}The smaller root is taken since $I, J \leq mv$. The speed in the $x$ direction, in terms of $I$, would be\begin{align*}
v_x = \sqrt{v^2 - I^2/m^2}
\end{align*}The particle travels at an angle $\theta$ which satisfies\begin{align*}
\tan \theta = \dfrac{1}{\sqrt{m^2 v^2/I^2 - 1}} \equiv \dfrac{1}{\sqrt{\kappa^2 -1}} \\
\end{align*}where $\kappa \equiv mv/I$. Meanwhile, the total impulse $(-J, I)$ applied to the ball was at an angle of $\theta_\mathrm{im}$ to the vertical, with\begin{align*}
\tan \theta_\mathrm{im} = J/I = \kappa - \sqrt{\kappa^2 - 1}
\end{align*}What can you say about the angles $\theta$ and $\theta_{\mathrm{im}}$? Is this what you expect, by symmetry?