What Is the Collision Height of Two Balls with Different Initial Velocities?

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Homework Help Overview

The discussion revolves around a physics problem involving two balls: one dropped from a height and the other thrown upward. The goal is to determine the height at which they collide, considering their initial velocities and the ratio of their speeds. The problem also explores different scenarios, including cases where the balls move in the same direction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for both balls and the conditions for collision, including position and velocity relationships. There are attempts to express the collision height in terms of the building height and the speed ratio. Questions arise about eliminating variables and the implications of negative speed ratios.

Discussion Status

Some participants have provided partial equations and insights into the relationships between the variables involved. There is ongoing exploration of how to manipulate the equations to isolate specific variables, and some guidance has been offered on combining equations to facilitate this process. Multiple interpretations of the problem are being considered.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the requirement to express the collision height as a fraction of the building height and the implications of varying the speed ratio. There is also mention of potential unphysical scenarios when considering negative values for the speed ratio.

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Problem 1

Ball A is dropped from the top of a building of height H at the same instant ball B is thrown vertically upward from the ground. First consider the situation where the balls are moving in opposite directions when the collide. If the speed of ball A is m times the speed of ball B when they collide, find the height at which they collide in terms of H and m. Take x = 0 at the ground, positive upward.

1.1 With two equations, describe the conditions at the collision (position and velocities of the balls).

- I got Xa = H - 1/2gt2
Xb = Vob(t) - 1/2gt2

At collision: Xa = Xb
H = Vob(t)

Va = -gt
Vb = Vob - gt

At collision, Va = Vb

t = mVob/2g

so Va = -g(mVob)/2g
Vb = Vob - g(mVob)/2g

1.2 Write the expressions for position and velocity of the balls as a function of time.

See above

1.3 Solve the above equations to find the height at which the balls collide. Your answer should be expressed as a fraction of the height of the building H and it should depend on the speed ratio m.

I tried solving it but couldn't eliminate the Vob...help!

1.4 Now suppose that m can be negative (i.e. balls A and B are moving in the same direction when they collide). Use the formula derived above to graph the height of the collision (again expressed as a fraction of the building height H) as a function of m for -5 < m < 5. Are there values of m for which the answer is unphysical?

Problem 2

Suppose, for a change, the acceleration of an object is a function of x, where a(x) = bx and b is a constant with a value of 2 seconds-2. In order to solve this problem you should use the chain rule: for arbitrary variables u, v and t, remember that dr/dt = (dr/ds) * (ds/dt).

2.1 If the speed at x = 1 m is zero, what is the speed at x = 3 m? Be sure to show your work.

By integration, should V = x squared?

2.2 How long does it take to travel from x = 1 to x = 3 m?

Problem 3

A small rock sinking through water experiences an exponentially decreasing acceleration as a function of time, given by a(t) = ge-bt, where b is a positive constant that depends on the shape and size of the rock and the physical properties of water.

3.1 Derive an expression for the position of the rock as a function of time, assuming the initial speed of the rock is zero.

3.2 Show that the rock's acceleration can be written in a simple form involving its speed v: a = g - bv (still assuming that its initial speed is zero). This is, perhaps, a more common form of expressing acceleration in the presence of drag.
 
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comments on Prob 1

justagirl said:
1.1 With two equations, describe the conditions at the collision (position and velocities of the balls).

- I got Xa = H - 1/2gt2
Xb = Vob(t) - 1/2gt2

At collision: Xa = Xb
H = Vob(t)
Good.
Va = -gt
Vb = Vob - gt
Good.

At collision, Va = Vb
No. Speed(a) = m x speed(b). Note: speed, not velocity. So that means:
gt = m(v0b -gt)
 
and...

thanks doc, but I still don't see how I can eliminate Vob...
 
justagirl said:
thanks doc, but I still don't see how I can eliminate Vob...
Combine these two equations:
(1) gt = m(v0b -gt)
(2) H = V0b(t)

Eliminate V0b, solve for t.
 
mhmm if m=2... i got Xa = (2/3)H...

is that right?
 
Last edited:

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