Dynamics Question #3

[SOLVED] Dynamics Question #3

1. The problem statement, all variables and given/known data
A ball of mass [tex]0.20kg[/tex] is dropped from a height of [tex]2.5m[/tex] and bounce back up to [tex]1.6m[/tex]. Taking the acceleration due to gravity as [tex]9.81 ms^{-2}[/tex], calculate:
(a) the velocity of the ball as it hits the floor.
(b) the velocity of the ball as it leaves the floor.
(c) the change in momentum caused by the impact
(d) the average force of the floor on the ball if the impact time is 40ms.

2. Relevant equations
[tex]v^2 = u^2 + 2as[/tex]
Change in momentum = Final momentum - Initial momentum
Force = Change in momentum / Time

3. The attempt at a solution

(a) [tex]v^2 = u^2 + 2as[/tex]
[tex]v^2 = 0^2 + 2 \times 9.81 \times 2.5[/tex]
[tex]v = \sqrt{2 \times 9.81 \times 2.5}[/tex]
[tex]v = 7.0ms^{-1}[/tex]

(b) [tex]v^2 = u^2 + 2as[/tex]
[tex]v^2 = 0^2 + 2 \times 9.81 \times 1.6[/tex]
[tex]v = \sqrt{2 \times 9.81 \times 1.6}[/tex]
[tex]v = 5.6 ms^{-1}[/tex]

(c) Change in momentum = Final momentum - Initial momentum
[tex]= mv - mu[/tex]
[tex]= (0.2 \times 5.6) - (0.20 \times (-7.0))[/tex]
[tex]= 2.5 kgms^{-1}[/tex]

(d) Force = Change in momentum / Time

[tex]F = \frac{2.5}{40 \times 10^{-3}}[/tex]

[tex]F = 63N[/tex]

Are my answers correct?

 

Thank you dx