# Thread: Physics: Force Question...

1. ## Physics: Force Question...

Question:

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?

2. Originally Posted by looi76
Question:

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 $\displaystyle \pm 0.1$ ........

3. 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

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### The change in momentum of the ball, of mass 45 g,during contact of the ball with the surface

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