- #1

brotherbobby

- 699

- 163

- Homework Statement
- Shown in the picture below, two particles ##1## and ##2## are set into motion simultaneously from the edge of a table with velocities ##v_1## and ##v_2## at angles ##\theta_1## and ##\theta_2##. If their initial distance of separation is ##\boldsymbol{s}##, answer the following questions :

1. What is the necessary condition for ##1## and ##2## to collide?

2. Prove that for collision to be, the relative velocity is also the velocity of approach.

3. Find the time after which collision occurs.

- Relevant Equations
- 1. For two particles of the given type to ##\textbf{collide}##, there should be ##\textbf{no motion}## perpendicular to the line joining them. This implies that one particle would "see" the other move along the line joining them.

2. Relative velocity between the two particles ##v_{R} = (v_1^2+v_2^2+2v_1v_2\cos({\theta_1+\theta_2}))^{1/2}##. (You may need to draw the diagram to convince yourself of this - it's not trivial).

3. For the two given particles, the velocity of approach ##v_A = v_1\cos\theta_1+v_2\cos\theta_2 = -\frac{ds}{dt}##

**The statement is given above. Here I attach the image of the problem to the right.
**

__Problem__:

__Attempt__:(1)

**Condition for collision :**For the two particles to collide, there should be no relative velocity perpendicular to their "separation vector" ##\boldsymbol{s}##. Hence, we must have : ##\boxed{v_1 \sin\theta_1=v_2\sin\theta_2}## ##\huge{\checkmark}## (agrees with book)

(2)

**For collision, ##\boldsymbol{v_R = v_A}## :**While I understand that the author's demand is a mathematical one, it can be show intuitively too. If the two particles possesses no relative velocity perpendicular to their separation vector ##\boldsymbol{s}##, the only other component of relative velocity is along their separation vector. But that is also the velocity of approach - proving the supposition. ##\\[10pt]##

Now for the same result mathematically. We have the following three relations :

\begin{equation}

v_1\sin\theta_1 = v_2\sin\theta_2

\end{equation}

\begin{equation}

v_R^2 = v_1^2+v_2^2+2v_1v_2\cos(\theta_1+\theta_2)

\end{equation}

\begin{equation}

v_A = v_1\cos\theta_1+v_2\cos\theta_2

\end{equation}

Squaring the last equation, we get

\begin{equation}

v_A^2=v_1^2\cos^2\theta_1+v_2^2\cos^2\theta_2+2v_1v_2\cos\theta_1\cos\theta_2

\end{equation}

Squaring (1), we get $$v_1^2\sin^2\theta_1=v_2^2\sin^2\theta_2\Rightarrow v_1^2-v_1^2\cos^2\theta_1 = v_2^2-v_2^2\cos^2\theta_2\Rightarrow v_1^2\cos^2\theta_1 = v_1^2-v_2^2+v_2^2\cos^2\theta_2$$

Substituting the value of ##v_1^2\cos^2\theta_1## into (4), we get

$${\scriptstyle v_A^2 = v_1^2-v_2^2+2v_2^2\cos^2\theta_2+2v_1v_2\cos\theta_1\cos\theta_2\Rightarrow v_A^2=\underbrace{v_1^2+v_2^2+2v_1v_2\cos(\theta_1+\theta_2)}_{v_R^2}-2v_2^2+2v_2^2\cos^2\theta_2+2v_1v_2\sin\theta_1\sin\theta_2,}$$

where I have "forced" ##v_R^2## to appear on the right and had to subtract and add terms accordingly to make up for it.

Taking ##-2v_2^2## out as common and remembering that ##v_1 \sin\theta_1=v_2\sin\theta_2##, the last equation simplifies to

$$v_A^2 = v_R^2 -\cancel{2v_2^2\sin^2\theta_2}+\cancel{2v_2\sin^2\theta_2}= v_R^2\Rightarrow v_A = \pm v_R.$$

The relative velocity and the velocity of approach ##v_R, v_A\ge 0##. If two particles are moving "away" from one another, we would have ##v_A = +\dfrac{ds}{dt}##. Hence, our answer is ##\boxed{v_R=v_A}##; additionally, ##\boxed{v_A = -\dfrac{ds}{dt}}##, as particles 1 and 2 move "towards" one another, implying that their distance of separation decreases with time.

**(3) Time of collision :**The timeat which collision would occur is : ##\boxed{t = \dfrac{s}{v_1\cos\theta_1+v_2\cos\theta_2}}## ##\huge{\checkmark}## (agrees with book)

**Though I have been correct I think, I would appreciate if point no. 2 above, where ##v_A = v_R## could have been done better. Or any other comment when it comes to the "velocity of approach" as against the "relative velocity" between two vectors.**

Thank you for your time.

Thank you for your time.

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