- #1

Physiophysio

- 7

- 0

## Homework Statement

Assume a ping pong of mass 3g is falling, and it has fallen 1m and is now traveling at 5 m/s. If the air resistance force is half its weight, how far will it travel in the next 0.1 seconds. [Hint: you may want to sketch a velocity-time graph)

## Homework Equations

Understanding that area under v-t graph is distance travelled.

F=ma

a=v/t

v=d/t

## The Attempt at a Solution

So I thought:

If mass=3g, then weight is 0.03N

And as air resistance is half that, then the

**resultant force**on the ping pong ball is 0.03-(0.03/2)=0.015N

Therefore, as F=ma,

0.015=m * a

But we know m=0.003 kg, so,

0.015=0.003 * a

a=5 metres per second per second.

Hence, in 0.1 seconds,

5 * 0.1 = 0.5 m/s

So the velocity of the ping pong ball

**increases by 0.5 m/s**

*in the next 0.1 seconds*.

And we know initial velocity was 5 m/s, so it increased to 5.5 m/s in 0.1 seconds.

So on a velocity-time graph, this forms a trapezium shape, and you find the area of that trapezium to find the distance the ping pong ball has traveled in 0.1 seconds:

[ (5+5.5)/2 ] * 0.1 =

**0.525 m**.

__So the ping pong ball travels 525 centimetres (or 0.525 metres) in the next 0.1 seconds.__

__Is that right?__Thanks :)