Halliday 7e chapt 4 #16 (with web generated values)(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]x=x_{0}+v_{0}t+\frac{1}{2}at^2[/itex] and arranging what I know and got the following 6 equations:

[itex]y_{A}=y_{B}[/itex]

[itex]x_{A}=x_{B}[/itex]

[itex]y_{A}=y_{0A}[/itex]

[itex]x_{A}=v_{0A}t[/itex]

[itex]y_{B}=\frac{a\cos\theta\t^2}{2}[/itex]

[itex]x_{B}=\frac{a\sin\theta\t^2}{2}[/itex]

and from these I get the following two eqations:

1) [itex]y_{0A}=\frac{a\cos\theta\t^2}{2}[/itex]

2) [itex]v_{0A}t={a\sin\theta\t^2}{2}[/itex]

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

3) [itex]y_{0A}=\frac{a\cos\theta}{2}\(\frac{2v_{0A}}{a\sin\theta})^2[/itex]

distribute the squared and kill all fractions in equation 3 to get:

[itex]y_{A}a(\sin^2\theta)=2(v_{0A})^2\cos\theta[/itex]

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:

[itex]\frac{-2(v_{0A})^2\pm\sqrt{4v_{0A}+(4)(y_{A}^2)(a^2)}}{2y_{A}}[/itex]

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

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# Homework Help: Colliding particles, level 3 difficulty

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