1. Sep 12, 2004

### justagirl

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.

2. Sep 12, 2004

### Staff: Mentor

Good.
Good.

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

3. Sep 12, 2004

### justagirl

and...

thanks doc, but I still don't see how I can eliminate Vob...

4. Sep 12, 2004

### Staff: Mentor

Combine these two equations:
(1) gt = m(v0b -gt)
(2) H = V0b(t)

Eliminate V0b, solve for t.

5. Sep 28, 2005

### zenity

mhmm if m=2... i got Xa = (2/3)H...

is that right?

Last edited: Sep 28, 2005