What are the different methods for solving this bouncing ball problem?

[crazyman]

Homework Statement

A rubber ball is dropped onto a ramp that is tilted at 20°. A bouncing ball obeys the "law of reflection" which says that the ball leaves the surface at the angle it approached the surface. The ball's next bounce is 3 m to the right of its first bounce. What is the ball's rebound speed on its first bounce?

Homework Equations

All kinematics equations. Maybe projectile motion equations
http://www.tvdsb.ca/webpages/pcowling/files/equation%20sheet.pdf"

The Attempt at a Solution

I thought of this question like a projectile motion question. Cause the ball is dropped, it travels a distance in the air in a curved path, then lands at a point away from the starting point. So, we know that Δx=3 m, Δy=-1.09 m, and the ball's launch angle is 50° above the horizontal (The ball's angle with the ramp is 70° and since the ramp is 20°, the angle with the horizontal becomes 50°). We also know the vertical acceleration due to gravity is -9.8 m/s. So basically, I've got a bunch of random pieces of information, but I don't know what to do with them...

Any help would be most appreciated :)

 

[gneill]

Set up the equations for the standard projectile problem. You've got the x-distance and y-distance values to plug into them (what would normally be what you'd be looking to solve for), along with the launch angle.

 

[crazyman]

Ok I do that. So I get

Vix=Δx/t
Vix=3/t

Δy=ViyΔt + 1/2aΔt
-1.1=ViyΔt -4.9Δt^2

...And then I don't know what to do.

EDIT: OH! If I let Vix=Viysin15°, I now get

Viysin15=3/t

And now I have two equations and two unknowns. I can now solve for Viy right, find Vix, use pythagorean threorem, and I'm good?

 

[gneill]

I'm not sure where your 15° is coming from. As far as I can tell, the launch angle is 50°, so that tan(50°) = Viy/Vix .

 

[crazyman]

Oops, sorry I meant Vix=Viy/tan50°

Ok my final answer is 6.48 m/s (6 m/s with sig figs). Can anyone confirm this?

 

[gneill]

Try checking your results by plugging that velocity back into your equations of motion to see if you get the right Y-displacement when the X-displacement is 3m.

 

[crazyman]

I don't see what you mean. Plug it back into what equations? The final answer is the final velocity of the ball, but all the equations deal with initial velocity.

 

[gneill]

I don't see what you mean. Plug it back into what equations? The final answer is the final velocity of the ball, but all the equations deal with initial velocity.

Are you sure that you were solving for the final velocity? What does the problem statement say?

 

[crazyman]

It tells me to find the ball's rebound speed. But since we're assuming that the collisions are perfectly elastic, isn't the speed of the the ball right before it lands the same as the rebound speed?

 

[gneill]

It tells me to find the ball's rebound speed. But since we're assuming that the collisions are perfectly elastic, isn't the speed of the the ball right before it lands the same as the rebound speed?

What about the potential difference between where the ball is launched (bounced) and where it finally lands? They're at different heights, yes?

Anyways, the problem was to determine the launch speed, and thus the vertical and horizontal components of the launch velocity. If you've got those, then you can enter them into the original equations of motion in order check to see that they do indeed produce the expected landing point.

 

[crazyman]

What about the potential difference between where the ball is launched (bounced) and where it finally lands? They're at different heights, yes?

Anyways, the problem was to determine the launch speed, and thus the vertical and horizontal components of the launch velocity. If you've got those, then you can enter them into the original equations of motion in order check to see that they do indeed produce the expected landing point.

Wait what? When do we have to determine the launch speed? We have to determine the rebound speed after it travels 3 m. Here, I'll show you my solution. Can you look over it and see if everything's in order?

V_ix=Δx/Δt
V_iy/tan50°=3/Δt
V_iy=3tan50°/Δt (1)

Δy=V_iyΔt + 1/2aΔt²
-1.1=V_iyΔt - 4.9Δt² (2)

*Sub (1) into (2)

-1.1=(3tan50°/Δt)(Δt) - 4.9Δt²
-4.67=4.9Δt²
Δt=0.97 s

V_iy=3tan50°/Δt
V_iy=3tan50°/0.97
V_iy=3.66 m/s

V_ix=Δx/Δt
V_ix=3/0.97
V_ix=3.07 m/s

V_fy²=V_iy² + 2aΔy
V_fy²=3.66² - 19.6(-1.1)
V_fy²=32.5
V_fy²=5.7 m/s

V_f²=3.07² + 5.7²
V_f²=6.48 m/s

Since all collisions are perfectly elastic, the ball's speed at the instant before it lands is the same at the ball's rebound speed. Therefore, 6 m/s is the rebound speed of the ball.

 

[gneill]

Wait what? When do we have to determine the launch speed? We have to determine the rebound speed after it travels 3 m. Here, I'll show you my solution. Can you look over it and see if everything's in order?

V_ix=Δx/Δt
V_iy/tan50°=3/Δt
V_iy=3tan50°/Δt (1)

Δy=V_iyΔt + 1/2aΔt²
-1.1=V_iyΔt - 4.9Δt² (2)

*Sub (1) into (2)

-1.1=(3tan50°/Δt)(Δt) - 4.9Δt²
-4.67=4.9Δt²
Δt=0.97 s

V_iy=3tan50°/Δt
V_iy=3tan50°/0.97
V_iy=3.66 m/s

V_ix=Δx/Δt
V_ix=3/0.97
V_ix=3.07 m/s

V_fy²=V_iy² + 2aΔy
V_fy²=3.66² - 19.6(-1.1)
V_fy²=32.5
V_fy²=5.7 m/s

V_f²=3.07² + 5.7²
V_f²=6.48 m/s

Since all collisions are perfectly elastic, the ball's speed at the instant before it lands is the same at the ball's rebound speed. Therefore, 6 m/s is the rebound speed of the ball.

From the original question:

The ball's next bounce is 3 m to the right of its first bounce. What is the ball's rebound speed on its first bounce?

I take that to mean that the first bounce is where it first struck the ramp, i.e., its first hit after being dropped. This is what I have been referring to as the launch speed.

Regardless, your workings look fine. You're losing some precision on your results by substituting values for symbols early on and rounding as you go through multiple steps. The value I obtain for the final speed, for example, is 6.66 m/s as opposed to your 6.48 m/s.

 

[crazyman]

Yeah numbers can be fixed easily though. The mechanics, the logic, that all looks good then?

 

[gneill]

Yeah numbers can be fixed easily though. The mechanics, the logic, that all looks good then?

Sure, the methodology works.

 

[crazyman]

Are there any different ways to solve this problem? I don't see any, but I'm sure that there are. And I think knowing a bunch of different methods for the same problem would only be beneficial.

 

[willem2]

Are there any different ways to solve this problem? I don't see any, but I'm sure that there are. And I think knowing a bunch of different methods for the same problem would only be beneficial.

You could try to do it in a coordinate system with the x-axis parallel to the slope and
the y-axis perpendicular to the slope.

Initial x and y speeds are $v_i cos(70)$ and $v_i sin(70)$. The magnitude of the initial velocity doesn't change. You get x and y components of gravity that are different tough.

It will be easy to compute the bounce time if you know the initial $v_y$, and if you know the bounce time it's easy to compute the horizontal displacement in that time, which should then be equal to distance along the slope to the landing point (a bit larger than 3m)