# Falling hailstones on windscreen

Two motor cars have their wind screens at $\theta_1=15^{\circ}$ and $\theta_2=30^{\circ}$ respectively. While moving in a hailstorm the drivers see the hailstones bounced by the wind screen of their cars in the vertical direction. What is the ratio $\dfrac{v_1}{v_2}$ of the velocities of the cars $?$ Assume the hailstorms fall vertically.

My solution:-

Let the initial velocity of hailstone w.r.t Earth be represented by $\vec{u}_{H/E}$, following the same analogy we define the following:-

1. $\vec{v}_{H/E}$ - Final velocity of hailstone w.r.t Earth
2. $\vec{v}_{1/E}$ - Velocity of car $1$ w.r.t Earth
3. $\vec{v}_{2/E}$ - Velocity of car $2$ w.r.t Earth
4. $\vec{v}_{H/1}$ - Velocity of hailstone w.r.t car $1$
5. $\vec{v}_{H/2}$ - Velocity of hailstone w.r.t car $2$

Now consider the following vector diagrams  From the above vector diagram ,we can conclude the following:-

$$|\vec{v}_{H/E}|\cos{\alpha}=|\vec{v}_{1/E}| \tag {1}$$ $$|\vec{v}_{H/E}|\cos{\beta}=|\vec{v}_{1/E}| \tag {2}$$

$\therefore$ The ratio $\dfrac{v_1}{v_2}$ can be obtained by $(1)/(2)$, so we get

$$\dfrac{|\vec{v_1}|}{|\vec{v_2}|}=\dfrac{\cos{\alpha}}{\cos{\beta}}=\dfrac{\cos\left(\dfrac{\pi}{2}-2\theta_1\right)}{\cos\left(\dfrac{\pi}{2}-2\theta_2\right)}=\dfrac{\sin{2\theta_1}}{\sin{2\theta_2}}=\dfrac{\sin{60^{\circ}}}{\sin{30^{\circ}}}=\sqrt3$$

Book's Solution:-

According to the observers in the cars, hailstones bounce in the vertical direction which implies that the angle of reflection is $\theta_1$ as shown in figure below, which is same as angle of incidence in the cars' reference frame. Velocity of hailstones relative to the first car is $\vec{v}-\vec{v}_1$ as shown in the figure. From figure,

$\alpha + 2\theta_1=\frac{\pi}{2}$ and $\tan{\alpha}-\dfrac{v}{v_1}$

Hence, $\tan{\alpha}=\tan{\left(\dfrac{\pi}{2}-2\theta_1\right)}-\cot{2\theta_1}$. So, $\dfrac{v}{v_1}=\cot(2\theta_1)$

Similarly, for second car, $\dfrac{v}{v_2}=\cot{2\theta_2}$

Thererfore ratio of velocities of the two cars, $\dfrac{v_1}{v_2}=\dfrac{\cot{2\theta_2}}{\cot{2\theta_1}}=3$

My deal with the question:-

My main question:- Why isn't in the frame of reference of Earth is the reflection property applicable.

Now, what I thought regarding this is as follows:-

When the hailstone collides with the windscreen the impulsive force acting on the hailstone is only in the direction to their common normal. So, this only changes the component of the velocity of the stone perpendicular to the windscreen, but the velocity along the surface of the screen remains unchanged. Now, if this collision is drawn as a vector diagram(which I have shown) then we find that $\angle i=\angle r$($\angle i$-angle of incidence, $\angle r$-angle of reflection). Also the cars are not moving with an acceleration hence the hailstones do not experience any pseudo force(though pseudo force doesn't come into play here but still I just thought while trying to figure it out myself, that it could be a possibility so just mentioned it).

So, to conclude, my deal with the question is why is the reflection property only applicable in the frame of reference of car and not Earth.

• I congratulate you on the careful diagram and showing all your work. I wish more people with problems like this would put in so much effort. – Floris Aug 1 '16 at 12:36
• Please note that this is not a homework help site. The site policy is that questions should ask about a specific concept of physics, so that questions are of benefit to the broader community of users. If you only need someone to check your work, you can ask your teacher or classmates, or use another site such as Physics Forums. see : meta.physics.stackexchange.com/questions/714/…. – sammy gerbil Aug 1 '16 at 14:17
• @sammygerbil - As you said I added my particular doubt about the concept related to relative velocity should that suffice. – user350331 Aug 1 '16 at 15:00
• Thanks. Yes, I think your final sentence is good enough (for me, at least). However, the previous sentence stretches across 9 lines and has multiple clauses! If that sentence is saying something relevant, please would you consider splitting it up to make it more easily readable. If the meaning is essentially the same as the last sentence (which is clear enough on its own) I suggest you consider deleting it. – sammy gerbil Aug 1 '16 at 15:15
• @sammygerbil - Yeah let me do it, I just wrote it without stopping so that I would not forget some of the arguments to support my question :D. Now, let me sort it and make some sense out of all that. – user350331 Aug 1 '16 at 15:32

I think you are getting confused, in your diagram, about what velocity is in what frame of reference. The diagram that you drew shows the same angle $$\theta_1$$ for the incoming and outgoing hail stone - but with velocities in the Earth frame of reference. Clearly, the angle is the same only in the vehicle frame of reference, and that is why you are getting the wrong answer.

This diagram might help. It shows a collision in the frame of reference of a stationary wall, and relative to an observer that is moving with respect to the wall: It should be pretty clear that all the observed velocities must be transformed in the same way (adding the red arrow). But this means immediately that, in the frame moving to the right, the incident angle is no longer equal to the reflected angle (velocities in the new frame of reference in green).

• I updated my post so as to include what I thought on the issue that I am having could you please refer it, clarify some more why $\theta$ doesn't remain same in the reference of Earth. – user350331 Aug 1 '16 at 15:05
• Imagine you are dropping a ball, and hit it with a tennis racket you hold perfectly vertical. The angle of incidence (in the earth frame) is 90 degrees, but as you are imparting velocity of the ball, the angle of reflection will not be. This should tell you intuitively why "angle in = angle out" only in the frame where the surface is stationary. – Floris Aug 1 '16 at 15:54
• Okay, the example that you provided did tell me why the reflection property works in that frame which makes the collision surface stationary, but I am still not comfortable with it. If you can provide a somewhat more general example. – user350331 Aug 1 '16 at 17:11
• Okay, the figure that you included summed it up pretty nicely. – user350331 Aug 2 '16 at 2:10
• Glad it helped. – Floris Aug 2 '16 at 2:15