Solved Dynamics Question #3: Calculate Velocity, Momentum, and Force

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SUMMARY

The discussion focuses on solving a dynamics problem involving a ball with a mass of 0.20 kg dropped from a height of 2.5 m. The calculations yield a velocity of 7.0 m/s upon impact with the floor and a velocity of 5.6 m/s as it leaves the floor. The change in momentum is calculated to be 2.5 kg·m/s, and the average force exerted by the floor on the ball during a 40 ms impact time is determined to be 63 N. The solutions are confirmed as correct by participants in the discussion.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = u² + 2as
  • Knowledge of momentum concepts, including change in momentum = final momentum - initial momentum
  • Familiarity with force calculations, specifically Force = Change in momentum / Time
  • Basic grasp of gravitational acceleration, specifically 9.81 m/s²
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  • Explore the concept of impulse and its relationship with momentum
  • Learn about energy conservation principles in dynamics
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Students studying physics, particularly those focusing on dynamics, as well as educators and tutors looking for clear examples of momentum and force calculations.

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[SOLVED] Dynamics Question #3

Homework Statement


A ball of mass 0.20kg is dropped from a height of 2.5m and bounce back up to 1.6m. Taking the acceleration due to gravity as 9.81 ms^{-2}, 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


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

The Attempt at a Solution



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

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

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

(d) Force = Change in momentum / Time

F = \frac{2.5}{40 \times 10^{-3}}

F = 63N

Are my answers correct?
 
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Yes, they look good.
 
Thank you dx
 

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