Physics: Force Question...

• May 18th 2008, 02:24 AM
looi76
Physics: Force Question...
Question:
http://img100.imageshack.us/img100/1...sicsp25wz8.png

Use data from Figure to determine,
(a) the distance travelled by the ball during the first 0.40s,
(b) the change in momentum of the ball, of mass 45g, during contact of the ball with the surface,
(c) the average force acting on the ball during contact with the surface.

Attempt:
(a) 0.8m
(b) $\displaystyle Change \ in \ momentum = mv - mu$
$\displaystyle = (45 \times 10^{-3} \times 4.2) - (45 \times 10^{-3} \times 0)$
$\displaystyle = 0.19Ns$

Answer in text book is $\displaystyle 0.35Ns$, where did I go wrong?
• May 18th 2008, 03:42 AM
mr fantastic
Quote:

Originally Posted by looi76
Question:
http://img100.imageshack.us/img100/1...sicsp25wz8.png

Use data from Figure to determine,
(a) the distance travelled by the ball during the first 0.40s,
(b) the change in momentum of the ball, of mass 45g, during contact of the ball with the surface,
(c) the average force acting on the ball during contact with the surface.

Attempt:
(a) 0.8m
(b) $\displaystyle Change \ in \ momentum = mv - mu$
$\displaystyle = (45 \times 10^{-3} \times 4.2) - (45 \times 10^{-3} \times 0)$
$\displaystyle = 0.19Ns$

Answer in text book is $\displaystyle 0.35Ns$, where did I go wrong?

It's hard to read the scale .....

(b) I get the time of intial contact to be 0.43 s (when v = 4 m/s) and the time when the ball leaves the surface (speed starts to decrease) to be 0.58 s (when v = -3.6 m/s).

Then $\displaystyle \Delta p = m \Delta v = .......$ where $\displaystyle \Delta v = v_{final} - v_{initial} = -3.6 - 4 = -7.6$ m/s.

(c) $\displaystyle F_{average} = \frac{\Delta p}{\Delta t}$. Get $\displaystyle \Delta p$ from (b). $\displaystyle \Delta t = 0.58 - 0.43 = 0.15$ s.

Scale reading uncertainty: You should assume all my values for t are $\displaystyle \pm 0.01$ and my values for v are $\displaystyle \pm 0.1$ ........
• May 18th 2008, 07:13 AM
topsquark
Quote:

Originally Posted by mr fantastic
Then $\displaystyle \Delta p = m \Delta v = .......$ where $\displaystyle \Delta v = v_{final} - v_{initial} = -3.6 - 4 = -7.6$ m/s.

I haven't looked at any of the numbers, but note that $\displaystyle v_{initial}$ is negative and $\displaystyle v_{final}$ is positive. So $\displaystyle \Delta v = +7.6~m/s$ according to your working.

-Dan