When will the particles collide?

[rudransh verma]

Homework Statement

In fig, particle A moves along the line y=30m with a constant v= 3m/s parallel to x. At the instant particle A leaves the y axis particle B leaves origin with zero initial speed and constant acceleration a =0.4 m/s^2. Find theta

Relevant Equations

$$x=vt$$
$$s=ut+1/2at^2$$

I am stuck. Please ignore my handwriting. I am working on latex.
All I am taking is x and y coordinates same of both particles.
Yes they will meet at some time t.

 

[phystro]

It will help, if you could please create a drawing that shows both particles on the x-y axis, or at least include it here if it is given. I will assume that θ is the angle of the particle B with respect to the x or y axis. However, this angle could really take any value based only on your provided data. I will assume that the particles will meet at some time t, but you will need to specify they will meet, otherwise it is impossible to say what is θ.

 

[phystro]

1)Please write down the length of each side of the triangle.
2)Use Pythagorean theorem to find t (the time particles will meet)
3)Finally, you can then use the definition of either cosθ or sinθ to find the angle theta.

So to start with, what would be the length of the hypotenuse of the triangle?

 

[rudransh verma]

1)Please write down the length of each side of the triangle.
2)Use Pythagorean theorem to find t (the time particles will meet)
3)Finally, you can then use the definition of either cosθ or sinθ to find the angle theta.

So to start with, what would be the length of the hypotenuse of the triangle?

$$\sqrt{(vt)^2 + 30^2}$$
I can simply use the definition of tan but the problem is what is t?

 

[phystro]

That's correct! You need to equate this with x=(0.5)(a)(t²), and set t= t_m.
So you will have: 30²+ (3t_m²)²=((0.5)(0.4)(t_m²))²

You can copy and paste this to something like wolframalpha to find t_m.

 

[Steve4Physics]

So you will have: 30²+ (3t_m²)²=((0.5)(0.4)(t_m²))²

You can copy and paste this to something like wolframalpha to find t_m.

Hi @phystro . I think
30²+ (3tₘ²)²=((0.5)(0.4)(tₘ²))²
should be
30²+ (3tₘ)²=((0.5)(0.4)(tₘ²))²

However, in the spirit of the PF homework forum, note that it's better to let @rudransh verma attempt to derive the equation rather than simply supply it.

Note, the equation can be quite easily solved without any special software.

 

[rudransh verma]

However, in the spirit of the PF homework forum, note that it's better to let @rudransh verma attempt to derive the equation rather than simply supply it.

Note, the equation can be quite easily solved without any special software.

$$30^2+3t^2=(1/2(.4)(t^2))^2$$
$$900+9t^2=.04t^4$$
After solving and taking $$t^2=x$$ I got $$x= 262.9, -37.9$$
So $$t=16.2 sec$$
$$\tan (\theta)=(3*16.2)/30$$
$$\theta = 58.3$$

 

[Steve4Physics]

$$30^2+3t^2=(1/2(.4)(t^2))^2$$

Note ##3t^2## should be ##(3t)^2## but the line after that is correct.

$$900+9t^2=.04t^4$$
After solving and taking $$t^2=x$$ I got $$x= 262.9, -37.9$$

I don't get the same answer as you. Check your quadratic equation solutions by back-substitution. You can also check with a tool such as this: https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

Good work with the Latex. Note you can get fractions by using, for example:
\frac 1 2 which gives ##\frac 1 2## or
\frac {rudranch}{verma} which gives ##\frac {rudranch}{verma}##

Also, if you want your Latex in-line (not on a separate line) use ## instead of $$

 

[kuruman]

I get 60°. I used the horizontal kinematic equation to find an expression for the collision time and substituted that into the vertical kinematic equation.

The discrepancy between your answer and mine is probably due to round-off errors. If you substitute ##a=\frac{4}{10}## m/s² (instead of 0.4 m/s²) and simplify, you end up with a quadratic in ##\cos\theta## that can be solved quite easily as @Steve4Physics suggested.

 

[rudransh verma]

I get 60°. I used the horizontal kinematic equation to find an expression for the collision time and substituted that into the vertical kinematic equation.

The way you told me got 30 degrees.
$$vt=\frac 12 \frac {4}{10}cos(theta) t^2$$
$$30=\frac 12\frac {4 }{10} sin(theta) t^2$$
$$6= \frac {sin(theta) v^2} {cos^2 (theta)}$$
$$\frac 23=\frac {sin theta}{1-sin^2theta}$$
$$theta=30 degree$$

 

[Steve4Physics]

The way you told me got 30 degrees.
$$vt=\frac 12 \frac {4}{10}cos(theta) t^2$$
$$30=\frac 12\frac {4 }{10} sin(theta) t^2$$
$$6= \frac {sin(theta) v^2} {cos^2 (theta)}$$
$$\frac 23=\frac {sin theta}{1-sin^2theta}$$
$$theta=30 degree$$

If you refer to the diagram in Post#1, you will see that ##\theta## is defined as the angle relative to the y-axis. I think you have incorrectly treated it as the angle relative to the x-axis.

I can’t understand how you got your 3rd equation. Where does ##v^2## come from?
[EDIT. I'm being silly. I understand.]

By the way, in Latex, \theta gives you ##\theta##.

 

[haruspex]

By the way, in Latex, \theta gives you θ.

And \sin avoids "sin" appearing in italics. Similarly tan, cos, and maybe atan etc., but not sec, cosec.

 

[rudransh verma]

If you refer to the diagram in Post#1, you will see that θ is defined as the angle relative to the y-axis. I think you have incorrectly treated it as the angle relative to the x-axis.

The two eqns will be ##vt=\frac12\frac{4}{10}\sin\theta t^2##
$$30=\frac12 0.4 \cos\theta t^2$$
After substituting the value of t we get $$60=\frac{v^2\cos\theta}{1-\cos^2\theta}$$
$$20\cos^2\theta + 3\cos\theta-20=0$$
$$\cos\theta=0.928, -1.078$$
$$\cos\theta=1,-1$$
-1 not possible so 1 is the value
$$\theta= 0$$
I didn’t got 60 degrees.

 

[haruspex]

After substituting the value of t we get $$60=\frac{v^2\cos\theta}{1-\cos^2\theta}$$

That's not what I get. Factor of 10 problem.

 

[rudransh verma]

That's not what I get. Factor of 10 problem.

What did you got?

 

[haruspex]

What did you got?

6 where you have 60.

 

[rudransh verma]

6 where you have 60.

I got it. slight calculation error.

 

[kuruman]

I understand that in post #13 you made an algebraic mistake and you got 60 where you should have gotten 6 as @haruspex pointed out. You also made a mistake in logic which I will point out so that you will be aware of it. You wrote,

$$\cos\theta=0.928, -1.078$$
$$\cos\theta=1,-1$$
-1 not possible so 1 is the value
$$\theta= 0$$

I agree that the value of the cosine can be no larger than +1 and no less than -1. Given your solutions, why do you conclude that the cosine must be equal to 1? The proper way to interpret the two solutions is to say

The second solution for the cosine, -1.078, is not acceptable because it is less than -1. Therefore we must accept the first solution ##\cos\theta=0.928## which gives ##\theta=\arccos(0.928)=21.9^{\text{o}}.##​

It looks like you rejected both solutions and, for some unknown reason, you picked the value zero for the angle. You cannot do that. You have to base your conclusions on the evidence that you have. If you're wrong, you're wrong, but you cannot second guess yourself.

 

[rudransh verma]

It looks like you rejected both solutions and, for some unknown reason, you picked the value zero for the angle. You cannot do that. You have to base your conclusions on the evidence that you have. If you're wrong, you're wrong, but you cannot second guess yourself.

Yes! i was confused about the solutions of quadratic eqns. I know cos value can be 1 but i don't know how it can be -1. I googled it and found cos is -1 at Pi. I guessed that -1 is not possible only 1. Hence I neglected -1 after rounding off both values to 1 and -1. I was wrong.