Inclined plane projectile problem from dynamics course

[CaityAnn]

Homework Statement

the 2 kg ball is fired from point a with an Vo of 10 m/s up the smooth inclined plane. The angle is 37 degrees, it gives me the planes dimentions the ball rolls on, it is 1.5 m high and 2 m wide. It asks the distance in the x it lands. Also the velocity with which it strikes.

I know what the answer is supposed to be, 8.53m+2m=10.53m and velocity of 10 m/s- same as original.

The Attempt at a Solution

I found Vy=10*sin37= 6 m/s and Vx=10*cos37=8 m/s

We are in the energy section of the chapter so I used KE=PE to find the max hieight= 1.83m, don't know why, just did. Then I used the speed as a function of time kinematic equation, setting Vfy=0, found the time to reach max height= .6116 s. So now that I know the time to get to the top of the plane and the max height, how can I find the distance it will go?
I used the position as a function of time, using yo as 1.83' and vy=6 a=-9.81 and solved for the time from the max height to the ground and found .649 s. I am wondering if I should find the acceleration in the X using the speed as a function of time eq? HELP ME, I am going in circles...

 

[Mindscrape]

I'm sort of confused. The ball goes on a plane for a while, and then jumps off the plane? If so, you need to consider that the motion of the ball on the plane is going to be different from if it were just fired at 37 degrees.

 

[CaityAnn]

The ball is fired with an intial velocity of 10 m/s up a smooth inclined plane. The planes dimensions are 1.5 m high and 2 m long on the bottom so the hypotenuse of the plane is sqrt (1.5^2+2^2). It has enough velocity to make a parabolic shape as it is coming off the plane. .. Confusing I know.

 

[CaityAnn]

mindscape, thanks, could I use energy?? I am really stumped.

 

[learningphysics]

Do you see why the final velocity must be the same as the initial?

For the first part, calculate the total velocity at the top of the incline (using conservation of energy)... then calculate the time the object takes to hit the ground (from the top of the incline)... you can use that to get the distance travelled...

 

[Mindscrape]

Yeah, you can use conservation of energy because the ball should be assumed frictionless (probably assumed, but use point A as zero potential for convenience). If you know the velocity at the top of the plane, and the plane's angle then you can basically just treat that as some starting point with the velocity you calculated, like those "cannon on a cliff" problems that intro physics classes usually have.

 

[CaityAnn]

Learning physics: The velocity is the same at impact because the vx component does not change, but what of the Vy component? It makes sense that it doesn't actually increase more than its given velocity.. but not sure why its the EXACT same.

So here's my plan- Ill set up the energy equation to find velocity at top of the hill. 1/2mv^2=mgh+1/2mv^2, because there is PE and KE at the top of the hill, right?? Also, do I add the angle to my intial and final velocities? Seems resonable to at least add an angle to my intial velocity.
Then I will use 0=1.5+Vsintheta*t-1/2*9.8*t^2, to find t and then use
x=Vox*t to find the x displacement... sound good?

Thanks ahead

 

[learningphysics]

Learning physics: The velocity is the same at impact because the vx component does not change, but what of the Vy component? It makes sense that it doesn't actually increase more than its given velocity.. but not sure why its the EXACT same.

So here's my plan- Ill set up the energy equation to find velocity at top of the hill. 1/2mv^2=mgh+1/2mv^2, because there is PE and KE at the top of the hill, right?? Also, do I add the angle to my intial and final velocities? Seems resonable to at least add an angle to my intial velocity.
Then I will use 0=1.5+Vsintheta*t-1/2*9.8*t^2, to find t and then use
x=Vox*t to find the x displacement... sound good?

Thanks ahead

That should do it. in the energy equation don't worry about angles... you're trying to find the magnitude of the velocity there.

Ah... the final velocity isn't the same as the initial, because the direction is different. But the magnitude of the final velocity is the same as the initial... are you asked for the speed or velocity? If it asks for velocity, then you should find the angle...

The reason the final speed is the same as the initial is by conservation of energy...
Initial Energy = Final Energy
KEinitial + PEinitial = KEfinal + PEfinal

but the heights at the beginning and the end are exactly the same (they're both 0)... so they cancel.

KEinitial = KEfinal

that means the speeds are the same. The directions are different. The vertical component of the final velocity is downward... whereas initially it was upward.

 

[Mindscrape]

Yes, everything looks good, though be careful because the displacement you find from the ball being launched is not the net displacement!

As far as the velocity at the very start and very end being the same, just think about the energy of the system. More pedagogically, how would simply shooting the ball out of a cannon compare to the ball going through a roller coaster type track (starting with the same velocity and same height as the cannon) that extends for about a mile? Should both end with the same velocity given that they both return to the original height? Will both have gone the same distance?

 

[CaityAnn]

thank you both very much.

Mindscrape: The difference is that the ball has tangental and normal acceleration components, however using energy they should both come out to be the same speed, because neither of them ever stops. but because their acceleration and velocity vectors are so different I would imagine they would not have gone the same distance. ??