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mbrmbrg
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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 [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 equation 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?
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 equation 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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