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

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The discussion focuses on calculating the collision height of two balls with different initial velocities, where one ball is dropped from a height H and the other is thrown upward. Key equations include the positions of the balls as functions of time, leading to a condition for their collision. Participants explore how to express the collision height as a fraction of H based on the speed ratio m, with some confusion about eliminating the initial velocity of ball B. The conversation also touches on the implications of negative values for m, suggesting that certain values may yield unphysical results. Overall, the thread emphasizes the mathematical relationships governing the collision dynamics of the two balls.
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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?
 
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