# Colliding particles, level 3 difficulty

1. Sep 17, 2006

### mbrmbrg

Halliday 7e chapt 4 #16 (with web generated values)

In the figure (attatched), particle A moves along the line y = 33 m with a constant velocity of magnitude 3 m/s and 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 of magnitude 0.47 m/s2. What angle θ between and the positive direction of the y axis would result in a collision?

I got theta = 74 degrees, which is wrong. Here's my work:

PARTICLE A:
x_0=0m
y_0=y=33m
v_0=v=3 m/s
a=0m/s^2

PARTICLE B:
x_0=0m
y_0=0m
v_0=0m/s
a=0.47m/s^2
a_x=0.57sin theta
a_y=0.47cos theta
(NOTE: theta is not conventional. It is measured from the y-axis)

GENERAL:
collision occurs when A and B have the same x and y co-ordinates.

I did some simplifying of $x=x_{0}+v_{0}t+\frac{1}{2}at^2$ and arranging what I know and got the following 6 equations:

$y_{A}=y_{B}$
$x_{A}=x_{B}$
$y_{A}=y_{0A}$
$x_{A}=v_{0A}t$
$y_{B}=\frac{a\cos\theta\t^2}{2}$
$x_{B}=\frac{a\sin\theta\t^2}{2}$

and from these I get the following two eqations:

1) $y_{0A}=\frac{a\cos\theta\t^2}{2}$
2) $v_{0A}t={a\sin\theta\t^2}{2}$

solve equation 2 for t, and subsitute that value into eqution 1 to get:

3) $y_{0A}=\frac{a\cos\theta}{2}\(\frac{2v_{0A}}{a\sin\theta})^2$

distribute the squared and kill all fractions in equation 3 to get:
$y_{A}a(\sin^2\theta)=2(v_{0A})^2\cos\theta$

Replacing sin^2(theta) with 1-cos^2(theta) and moving everything to one side gives a quadratic equation in the form ax^2+bx+c=0. Plug it into the quadratic equation to get:

$\frac{-2(v_{0A})^2\pm\sqrt{4v_{0A}+(4)(y_{A}^2)(a^2)}}{2y_{A}}$

Sove in values from the table, and voila, either 144.699 degrees or74.296 degrees. Both of which are wrong. Help?

#### Attached Files:

• ###### diagram chapt 4.doc
File size:
19.5 KB
Views:
83
Last edited: Sep 17, 2006
2. Sep 17, 2006

### mbrmbrg

Actually, when taking the arctan of an angle, sometimes you have to subtract your answer from 180 degrees. Is this a similar situation? If so, what do you add/subtract when using arccos?

EDIT: I would think to subtract the angle from 360 degrees, but that would give me an angle poingting below the x-axis, and in this case, that would prevent the two particles from ever colliding. So it can't be that.

Last edited: Sep 17, 2006
3. Sep 18, 2006

### mbrmbrg

4. Sep 18, 2006

### 6Stang7

is the answer around 61-62 degrees?

5. Sep 18, 2006

### mbrmbrg

Beats me; it's a WebAssign problem. I type in the answer and the computer says yes or no. It's not 74 degrees; I think I may have punched in fifty-something degrees at one point, also.

6. Sep 18, 2006

### 6Stang7

hmmm. cause i worked out the problem and got down to a tan(x)sec(x)=3.9 and had to use a graphing calculator to solve for x. it's been a while since i have done trig-ID's, so a implicit way to solve for x is escaping me.....or i might hae just done the problem wrong.

7. Sep 18, 2006

### mbrmbrg

shame... :sigh:. Well, the due date has passed to hand in the assignment, so I'll just get the answer in recitation... I wish I could figure this out, though!

8. Sep 18, 2006

### 6Stang7

well, here is how i worked it. maybe this will help.

(.47)sin(x)=a_y (acceleration in the y direction).
(.47)cos(x)=a_x (acceleration in the x direction).

(position of partical A)X_a=(3m/s)t
(position of partical B in the y direction)33=(1/2)(a_y)t^2
(position of partical b in the x direction)X_b=(1/2)(a_x)t^2

X_a=X_b
(3m/s)t=(1/2)(a_x)t^2
t=(6/a_x)

33=(1/2)(a_y)(6/a_x)^2
(a_y/(a_x)^2)=66/36

this resulted in tan(x)sec(x)=3.9

however, i'm thinking this might be wrong since the answer is explicit as of now.

9. Sep 18, 2006

### mbrmbrg

But the angle we're looking for is with respect to the y-axis?

Woof. I am too tired to understand this now(12:41 my time), I'll look it over tomorrow. Thanks, though!