Dynamics Question #3

  • Thread starter looi76
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[SOLVED] Dynamics Question #3

Homework Statement


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.

Homework Equations


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

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?
 

Answers and Replies

  • #2
dx
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Yes, they look good.
 
  • #3
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Thank you dx
 

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