Projectile Motion Collision Problem

[TheRedDevil18]

Homework Statement

A particle P is projected from the origin with initial speed 5 m/s and projection angle theta. At time t = 0, when P is at the origin, another particle Q begins to move on the
(horizontal) x-axis with initial speed 15/4 m/s and constant acceleration a = 3g/4 m/s^2.
Determine the time t when P collides with Q, and the distance from the origin where this
occurs.

Homework Equations

The Attempt at a Solution

I know the times are the same but I can't solve for the equations because the angle is unknown. Any advice on how to go about solving this ?

 

[ehild]

What are the equations? Show them, please. The angle is one of the unknowns.

ehild

 

[TheRedDevil18]

What are the equations? Show them, please. The angle is one of the unknowns.

ehild

Δy = V0SinΘ - 1/2 gt^2

Knowing that delta y = 0,

0 = 5SinΘ - 1/2 (-9.8)t^2.....1

x = V0t - 1/2at^2, this is for particle q

x = (15/4)t - 1/2(-9.8)t^2......2

If the angle was given this could have been easily solved because the times are the same for P and Q

 

[Nathanael]

If I'm not misunderstanding the problem, then there is only one angle in which the particles collide (i.e. there's only one angle in which the paths intercept at the same time)
This is how you know which angle to use.

P.S.
Are we supposed to assume the position of Q at t=0 is at the origin?



Edit:

x = V0t - 1/2at^2, this is for particle q

Don't you mean plus 1/2at^2?

x = (15/4)t - 1/2(-9.8)t^2

That is not the acceleration that was given for particle Q

 

[ehild]

Δy = V0SinΘ - 1/2 gt^2

Knowing that delta y = 0,

0 = 5SinΘ - 1/2 (-9.8)t^2.....1

x = V0t - 1/2at^2, this is for particle q

x = (15/4)t - 1/2(-9.8)t^2......2

If the angle was given this could have been easily solved because the times are the same for P and Q

What about the x coordinate of particle P?

As Nathanael pointed out, particle Q does not accelerate with -9.8 m/s^2. And in general, x= v_ot+a/2t^2 is the formula for the x displacement in case of uniformly accelerating motion.

ehild

 

[TheRedDevil18]

What about the x coordinate of particle P?

As Nathanael pointed out, particle Q does not accelerate with -9.8 m/s^2. And in general, x= v_ot+a/2t^2 is the formula for the x displacement in case of uniformly accelerating motion.

ehild

Yes, my mistake

For particle Q
x = (15/4)t + 1/2(7.35)t^2...,

@Nathanael, why is it plus 1/2 at^2 ?, I thought the formula had a minus sign in it

The X co-ordinate of particle P is,
x = 5cosΘt

I don't know if we can assume that the Q started at the origin else their displacements should be the same. Still don't know how to get the angle though

 

[Nathanael]

I don't know if we can assume that the Q started at the origin else their displacements should be the same.

I think that if we don't assume that Q started at the origin, then the problem is unsolvable (as far as I can tell).

@Nathanael, why is it plus 1/2 at^2 ?, I thought the formula had a minus sign in it

It only has a minus sign if the acceleration is "negative" (which is just a direction).
Consider the case of zero initial velocity and a positive acceleration. The formula becomes $x=\frac{1}{2}at^2$.
If the acceleration is (in the) positive (direction), wouldn't you expect the displacement to be (in the) positive (direction)?

Still don't know how to get the angle though

Have you thought about it? Can you elaborate on why you're stuck? Contemplate it a bit.

I haven't gone through a specific solution, but I can tell you that each angle corresponds with one-and-only-one final distance, and one-and-only-one "time in the air"

The other particle must be at that same distance at that same time. So you will need to find the angle which works.
(It's like a game where you're tring to land one particle on the other particle by altering the launch angle)
(Except you have physics/math, so you will beat the game first try )

 

[ehild]

Yes, my mistake

For particle Q
x = (15/4)t + 1/2(7.35)t^2...,
The X co-ordinate of particle P is,
x = 5cosΘt

I don't know if we can assume that the Q started at the origin else their displacements should be the same. Still don't know how to get the angle though

You have three equations and three unknowns: the angle θ, the time t and x.

x=5cos(θ) t

y=0=5sin(θ) t -g/2 t²

x = (15/4)t + 1/2(7.35)t^2

I assume you can solve a system of equations.
You need t and x in the final result. It would be nice to eliminate the angle θ. It is a good method to use that sin²(θ)+cos²(θ)=1

ehild

 

[TheRedDevil18]

Thanks guys, I finally solved it

I got my time to be 0.29's and displacement to be 1.39m

Thanks again to ehild and Nathanael :)

 

[dean barry]

By trial and error, i agree, with g @ 9.8 (m/s)/s, the result is :
launch angle = 16.25 °
x distance to collision = 1.3707 metres
elapsed time = 0.2855 seconds

 

[ehild]

Thanks guys, I finally solved it

I got my time to be 0.29's and displacement to be 1.39m

Thanks again to ehild and Nathanael :)

Well done!
I have just noticed that a =3/4 g was given. So the evaluation can be much simpler.

x=5cos(θ) t

y=0=5sin(θ) t -g/2 t²

x = (15/4)t + 1/2(3g/4)t^2


You could have done it on a more straightforward way:

Isolate t from the second equation: $t=\frac{10}{g}\sin(θ)$

Equate the first and third equations and divide by t.

$5\cos(θ)=\frac{15}{4}+\frac{3g}{8}t$

Substitute t=10/g sin(θ) for t:

$5\cos(θ)=\frac{15}{4}+\frac{3g}{8}\frac{10}{g}\sin(θ)\rightarrow 5\cos(θ)=\frac{15}{4}+\frac{15}{4}\sin(θ)$

Multiply by 4/5:

$4\cos(θ)=3+3\sin(θ)$

take the square, substitute cos²(θ)=1-sin²(θ) and solve for sin(θ).

Spoiler

sin(θ)=0.28, θ=16.26°

ehild