Which Material Exerts Greater Impulse When Jumping Off a Dock?

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SUMMARY

The discussion centers on the impulse experienced when jumping off a dock and landing on different materials, specifically sand and water. Using the impulse-momentum theorem, it was determined that while the impulse calculated for landing on sand is greater (1000 Ns) compared to water (800 Ns), the correct conclusion is that both materials exert the same impulse, but sand exerts a greater average force. The confusion arises from the interpretation of impulse and average force in the context of stopping time.

PREREQUISITES
  • Understanding of the impulse-momentum theorem
  • Basic knowledge of physics equations, specifically J = m*(Vf-V0)
  • Familiarity with concepts of impulse and average force
  • Ability to perform calculations involving mass and velocity
NEXT STEPS
  • Review the impulse-momentum theorem in detail
  • Study the relationship between impulse and average force
  • Explore examples of impulse in different materials and their effects on stopping time
  • Investigate real-world applications of impulse in sports and safety engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of jumping and landing on various surfaces.

Maxo
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Homework Statement


Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, determine which one of the following statements is correct:

a) In bringing you to a halt, the sand exerts a greater impulse on you than does the water.

b) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

c) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.

Homework Equations


J = m*(Vf-V0)

The Attempt at a Solution


Let's say we have mass = 100 kg, v0 = -10 m/s and vfw = -2 m/s for water and vfs = 0 m/s for sand. Then the impulse in each case would be J = m(Vf-V0) = 100*(-2-(-10))=800 Ns for water and J = 100*(0-(-10))=1000 Ns for sand.

So this seems to give a greater impulse for sand. However my book says the answer is b). Is there something wrong with my calculation? Isn't J=m*(Vf-V0) the equation for impulse?
 
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Maxo said:
In bringing you to a halt

Is the above consistent with

Maxo said:
vfw = -2 m/s for water
 

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