Physics Conservation of Matter Question

AI Thread Summary
To calculate the distance where the ball bounces after being launched from a ramp, the conservation of energy principle is applied, using the equations Eg = Ek and mgh = (1/2)mv^2. The height of the ramp "h" and the height of the bench "H" are known, allowing for the calculation of the ball's velocity without needing its mass. The discussion highlights the importance of considering both horizontal and vertical forces acting on the ball, suggesting the use of energy components. Ultimately, the user successfully resolves the problem with assistance from peers. Understanding the energy conservation approach is crucial for solving similar physics problems.
TigerLilly
Messages
5
Reaction score
0

Homework Statement



Here is the set-up:
-there is a lab bench that is a certain height above the floor, "H"
-on the bench is a ramp that is a certain height above the bench top, "h"
-a ball is dropped on the ramp and then is "launched" in the air and eventually falls to bounch on the ground. How can I calculate the distance from the bench where the ball bounces first?

I have "h" and "H".



Homework Equations



To calculate the speed of the ball I did:

Eg=Ek
mgh=(1/2)mv^2


I'm unsure how to proceed from here to find the distance. I don't have the mass of the ball, the acceleration or any angle for the ramp.

Other possible equations I could use are:
W = ∆ K Ek
W = F d cos θ


The Attempt at a Solution



I know I have to take into account that there are 2 forces acting on the ball. The first is the horizontal force propelling the ball forward and the force of gravity pulling on the ball.

I was thinking of finding Ek and Eg again and add them together for Et to use in the work formula. But I don't have the mass of the ball.

I'm just unsure what to do from here.
 
Physics news on Phys.org
From what I understand, you don't need the m of the ball because:

mgh=1/2mv^2

therefore gh=1/2v^2
 
I think you have to break it up into components (x and y)

Sorry if this doesn't help
 
No it did actually. I was working on it with a friend and it helped a lot. I figured it out now. Thanks so much!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Conservation of Energy when lifting a box up off the floor
Replies
5
Views
2K
Conservation of energy problem: Ball rolling down inclined plane and then through a loop-the-loop
Replies
10
Views
2K
A ball climbing a ramp while "rolling the wrong way"
Replies
32
Views
3K
Conservation of momentum (wrecking ball hits a stationary object)
Replies
10
Views
3K
Conservation of Energy with heat
Replies
5
Views
2K
Two balls, dropped with a delay of ##\Delta t##, meet after rebound
Replies
34
Views
2K
Horizontal impulse on a ball at rest on a plane (with friction)
Replies
14
Views
3K
Projectile Motion Using Vectors
Replies
39
Views
4K
Conservation of energy/Angular momentum for elastic collison
Replies
5
Views
2K
Conservation of energy interpretation of percentage conserved
Replies
20
Views
2K

Hot Threads

Recent Insights

Back
Top