Answer:
The average acceleration of the ball during the collision with the wall is [tex]a=2,800m/s^{2}[/tex]
Explanation:
Known Data
We will asume initial speed has a negative direction, [tex]v_{i}=-30m/s[/tex], final speed has a positive direction, [tex]v_{f}=26m/s[/tex], [tex]\Delta t=20ms=0.020s[/tex] and mass [tex]m_{b}[/tex].
Initial momentum
[tex]p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s[/tex]
final momentum
[tex]p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s[/tex]
Impulse
[tex]I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s[/tex]
Average Force
[tex]F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}[/tex]
Average acceleration
[tex]F=ma[/tex], so [tex]a=\frac{F}{m_{b}}[/tex].
Therefore, [tex]a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}[/tex]
