 #1
brotherbobby
 421
 107
 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}##
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 $$v_1^2\sin^2\theta_1=v_2^2\sin^2\theta_2\Rightarrow v_1^2v_1^2\cos^2\theta_1 = v_2^2v_2^2\cos^2\theta_2\Rightarrow v_1^2\cos^2\theta_1 = v_1^2v_2^2+v_2^2\cos^2\theta_2$$
Substituting the value of ##v_1^2\cos^2\theta_1## into (4), we get
$${\scriptstyle v_A^2 = v_1^2v_2^2+2v_2^2\cos^2\theta_2+2v_1v_2\cos\theta_1\cos\theta_2\Rightarrow v_A^2=\underbrace{v_1^2+v_2^2+2v_1v_2\cos(\theta_1+\theta_2)}_{v_R^2}2v_2^2+2v_2^2\cos^2\theta_2+2v_1v_2\sin\theta_1\sin\theta_2,}$$
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.
Last edited: