.Calculating Time for .22 Rifle Bullet to Stop in Soft Wood

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SUMMARY

The discussion focuses on calculating the time it takes for a .22 rifle bullet, traveling at 350 m/s and penetrating 0.130 m into soft wood, to stop. The bullet has a mass of 1.80 g, and the retarding force exerted by the wood is 848 N. The correct approach involves using the equation Vf = Vi + a*t, where the acceleration (a) is derived from F = ma. The initial calculations were incorrect due to a miscalculation of time, which was off by a factor of 2.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with kinematic equations
  • Basic knowledge of unit conversions (grams to kilograms)
  • Concept of constant retarding force
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  • Practice solving kinematic equations with different initial conditions
  • Explore the effects of varying penetration depths on stopping time
  • Investigate real-world applications of bullet penetration in soft materials
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theonerealazn
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Ok, so here's my question:

A .22 rifle bullet, traveling at 350 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.130 m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.80 g. Assume a constant retarding force.

I know this has already been posted in another thread, but i have an additional question that i can't get.
1. How much time is required for the bullet to stop?
i tried 0.130m/350m/s and got .00037143 seconds as an answer. This was not correct. Any suggestions on what to do?

I already know that the wood exerts 848 N on the bullet
 
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F=ma where F=net force on the bullet=848N and m=mass of the bullet=.0018kg. Find a

Vf=Vi+a*t
Vf, Vi and a are all known at this point. Solve for time.

Your answer is close. You are off by a factor of 2.
 

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