# Collision of two particles: constant velocity, constant acceleration

In the figure, particle $A$ moves along the line $y = 25~\text{m}$ with a constant velocity $v$ of magnitude $3.0~\text{m}/\text{s}$ and directed parallel to the $x$ axis. At the instant particle A passes the $y$-axis, particle $B$ leaves the origin with zero initial speed and constant acceleration $a$ of magnitude $0.44~\text{m}/\text{s}^2$. What angle $\theta$ between $a$ and the positive direction of the $y$-axis would result in a collision?

I did the problem but I'm not getting the right answer? How would this be solved and/or what am I doing wrong?

You don't have to read this part but to solve it, Both have a same final displacement in the x direction so if you use an equation $3t = 0.5 \cdot 0.44 \cdot t^2$ with $t = 13.64~\text{s}$.

Since it has the same $y$ point ($25$ m height) the angle can be found with $\arctan(13.64\cdot 3/25)$. Then you subtract that from $90$ to get the angle with reference to the $y$-axis.

What am I doing wrong here?

• I don't think I have to include the picture of the figure (it didn't give an option to) but the A vector is a straight line at y = 25, and the B vector is like a line flowing at 60 degrees, going upwards towards the A vector, and then crossing it.
– Sal
Feb 14, 2016 at 20:37

Mathematical Model

It is convenient to use vectors here. For particle $A$ we have: $$r_A(t) = r_A(0) + v_A t = (x_A(0), 25) + (3, 0) t = (x_A(0) + 3t, 25)$$ where the text gives $x_A(0) = 0$.

For particle $B$ we have $$a_B(t) = a \\ v_B(t) = v_B(0) + a t = at \\ r_B(t) = r_B(0) + \frac{1}{2}a t^2 = \frac{1}{2}a t^2 = ((1/2)\lVert a \rVert \sin(\theta) t^2,(1/2)\lVert a \rVert \cos(\theta) t^2)$$ The above image shows the trajectory of $A$ (green) and of $B$ (blue) and where the trajectories intersect at point $X$. Also the angle is given (named $\alpha$ instead of $\theta$).

You can fiddle with a live version here.

Calculating the Collision Point

For a collision we need $r_A(t) = r_B(t) = X$ for some $t$.

The intersection point $X$ is $$X = (\xi, 25)$$ with $\tan(\theta) = \xi / 25 \iff \xi = 25 \tan(\theta)$.

$A$ arrives at $X$ if $$t = (25/3) \tan(\theta) \quad (1)$$ $B$ arrives at $X$ if $$\frac{1}{2} \lVert a \rVert \sin(\theta) t^2 = 25 \tan(\theta) \quad (2) \\ \frac{1}{2} \lVert a \rVert \cos(\theta) t^2 = 25 \quad (3)$$ Alas equation $(2)$ and $(3)$ are equivalent, division of both sides of $(2)$ by $\tan(\theta)$ gives $(3)$.

Inserting $(1)$ into $(3)$ gives $$\frac{1}{2} \lVert a \rVert \cos(\theta) \left(\frac{25}{3}\right)^2 \tan^2(\theta) = 25 \iff \\ 25 \lVert a \rVert \sin^2(\theta) = 18 \cos(\theta) \quad (4)$$ To limit the solutions we should note that $\theta \in (0, \pi/2)$.

Algebraic Solution

Substituting (see Tangent half-angle formulas) $$t = \tan(\theta/2) \\ \sin(t) = \frac{2t}{1+t^2} \\ \cos(t) = \frac{1-t^2}{1+t^2} \\$$ (note: this $t$ is not the time $t$ from further above, but just used for this intermediate calculation of $\theta$ from equation $(4)$) we get $$25 \lVert a \rVert \left(\frac{2t}{1+t^2} \right)^2 = 18 \frac{1-t^2}{1+t^2} \iff \\ 25 \lVert a \rVert 4t^2 = 18(1-t^2)(1+t^2) = 18(1-t^4) \\ t^4 + \frac{50}{9} \lVert a \rVert t^2 - 1 = 0 \quad (5)$$ Introducing $s = t^2$ we get $$s^2 + \frac{50}{9} \lVert a \rVert s - 1 = 0 \quad (6)$$ which has the solutions $$s = \frac{\pm \sqrt{81+25^2 \lVert a \rVert^2}-25\lVert a \rVert}{9}$$ we discard the negative solution and apply $25 \lVert a \rVert = 11$ and get $$s = \frac{\sqrt{202} - 11}{9} \\ t = \frac{\sqrt{\sqrt{202}-11}}{3} \\ \theta = 2 \arctan{\frac{\sqrt{\sqrt{202}-11}}{3}} \approx 1.0771 \approx 61.714^\circ$$ Inserting in equation $(1)$ gives $$t \approx (25/3) \tan(1.0771) \approx 15.485 \, \text{s}$$

• The t I calculated above using the 3t = 1/2at^2, since the set distance is equal. So I have to solve for θ but I don't see how I would work to solve it, using tangent wouldn't work even though the tangent vector refers to the angle complementary to the angle of reference, so subtracting it from 90 should work, but for some reason it doesn't.
– Sal
Feb 14, 2016 at 21:21
• Simplifying that down you get cos θ/sin^2 θ = 50a / 36. Since a = 0.44 multiple 50 by that, divide by 36, and you get Is that correct? 0.611.
– Sal
Feb 14, 2016 at 21:37
• @Sal I guess I am done.
– mvw
Feb 15, 2016 at 0:02

How can there be an angle between an acceleration and the $y$-axis? I suppose you mean there is an angle between the line along which particle $B$ moves and the $y$-axis.

In order for $B$ to collide with $A$, it has to move farther than $A$ does, because $B$ travels the entire hypotenuse of a right triangle while $A$ travels just one leg of that triangle. So your first equation for $t$ is wrong: the time of collision, $0.5 \cdot 0.44 \cdot t^2 > 3t$; the two are not equal.

The ratio of the distance traveled by $A$ to the distance traveled by $B$ is just the sine of the angle $\theta$ (in fact it is practically the definition of $\sin(\theta)$). The correct equation is $$3t = (0.5 \cdot 0.44 \cdot t^2) \sin(\theta).$$ This is not enough information to solve the problem. But you can also use the fact that the other leg of the right triangle has length $25$, that is, $$25 = (0.5 \cdot 0.44 \cdot t^2) \cos(\theta).$$

This gives you two equations in two unknowns. They are not the easiest equations to solve simultaneously, but they are solvable. Wolfram Alpha gives four solutions, only one of which has a positive real value of $t$; in that solution, $t > 15$, and the angle is about $62$ degrees.