Projectile motion of two bullets on inclined plane

AI Thread Summary
Two bullets shot from the same gun at different velocities will strike an inclined plane simultaneously, regardless of their speed. The discussion emphasizes the importance of analyzing the motion using kinematic equations and transforming the coordinate system to account for the incline. By focusing on the vertical and horizontal components of motion, it becomes clear that the time taken to hit the plane is independent of the initial velocity. The conversation highlights the utility of coordinate transformations in simplifying complex projectile motion problems. Understanding these principles is crucial for solving similar physics problems effectively.
drawar
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Homework Statement


Two bullets are shot with different velocity from the same gun placed parallel to an inclined plane. Which one will strike the plane first?
1. the faster
2. the slower
3. neither, both strike the plane at the same time

Homework Equations



Kinematics equations...

The Attempt at a Solution


It can be seen from here http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html that both fall onto the plane at the same time but I don't know how to explain it mathematically. I think we should find the time it takes each bullet to reach its maximum height (t1), then deduce the time they travel until hitting the plane (t2), the sum of them should be an expression independent of v. I'm able to write down t1, but I'm stuck at determining t2. Any help would be appreciated.
 
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How would you go about solving the problem if there wasn't an inclined plane, if the bulets were shot parallel to the x axis?
 
genericusrnme said:
How would you go about solving the problem if there wasn't an inclined plane, if the bulets were shot parallel to the x axis?

In this case the vertical component of initial velocity is zero. Since they are shot at the same height, the vertical distance they've traveled until they hit the ground is equal, that is, they land at the same time (h=1/2*g*t^2).
Why are you talking about this?
 
drawar said:
In this case the vertical component of initial velocity is zero. Since they are shot at the same height, the vertical distance they've traveled until they hit the ground is equal, that is, they land at the same time (h=1/2*g*t^2).
Why are you talking about this?

How can you transform the system so that you're dealing with nearly the exact same situation?
 
You mean choosing a coordinate system with reference to the incline?
 
Yes, continue!
What coordinate system are we going to choose?
 
Try to combine the equations for the x- and y-range for a situation without incline, by eliminating the time. Then use trigonometry to take account for incline, for instace

x=L*cos(a)

where l is the distance alonf the incline, xthe distance along the horizontal, and a the angle between the incline and the horizontal.
 
That would be errm... the x-axis is along the incline and the y-axis is perpendicular to it, right?
 
drawar said:
That would be errm... the x-axis is along the incline and the y-axis is perpendicular to it, right?

Yes, so what's going to happen to the components of acceleration on the bullets in this rotated coordinate system?
 
  • #10
Let alpha be the angle of inclination wrt the ground.
gx=gsin(alpha), gy=gcos(alpha), ux=u, uy=0
Motion along the x-axis: x=ut-1/2*gsin(alpha)*t^2
Motion along the y-axis: y=-1/2*gcos(alpha)*t^2
 
  • #11
Yes!
So you can now conclude that the initial velocity along the x-axis (that is, parallel to the incline) has no effect on the time it takes for the bullet to hit the plane!
 
  • #12
Yeah, got it now.
Thank you for your clear explanation, it does help a lot!
 
  • #13
No problem :biggrin:

It's always handy to bear in mind that a lot of the problems you will come across can be transformed into problems that you already know about by making use of clever coordinate transformations.
 
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