Why Does Calculating Bullet Transit Time Yield Different Results?

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SUMMARY

The discussion focuses on calculating the time a bullet spends in a target during a ballistics test. A 25.0g bullet traveling at 1100m/s penetrates a 40.0cm thick target and exits at 900m/s. The initial calculation of time using the final speed was incorrect due to not accounting for the target's movement. The correct approach involves using the average speed formula, resulting in a time of 4.0 * 10^-4 seconds.

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  • Understanding of basic physics concepts such as velocity and acceleration
  • Familiarity with the equations of motion, specifically s = (u+v)/2 * t
  • Knowledge of ballistics and projectile motion
  • Ability to perform unit conversions (e.g., cm to m)
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  • Study the principles of momentum conservation in collisions
  • Learn about the effects of target mass on bullet penetration
  • Explore advanced ballistics calculations using software tools
  • Investigate the impact of different bullet weights and speeds on transit time
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This discussion is beneficial for physics students, ballistics researchers, and anyone interested in the dynamics of projectile motion and impact analysis.

ZzZerozZ
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In a ballistics test, a 25.0g bullet traveling horizontally at 1100m/s goes through a 40.0cm -thick 250kg stationary target and emerges with a speed of 900m/s . The target is free to slide on a smooth horizontal surface.

the 1st question of this problem ask: How long is the bullet in the target?

I simply do:

40.0cm=0.4m ; 0.4m/(900m/s)=4.4*10^-4

The answer shows 4.0*10^-4

Did I make a mistake or it is just a rounding problem?
 
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Hi ZzZerozZ! Welcome to PF! :smile:

i] in your distance/speed, you've used the width of the block as the distance, but the block is moving, so the distance is longer

ii] your speed is the final speed, so it's too small :redface:
 
ZzZerozZ said:
In a ballistics test, a 25.0g bullet traveling horizontally at 1100m/s goes through a 40.0cm -thick 250kg stationary target and emerges with a speed of 900m/s . The target is free to slide on a smooth horizontal surface.

the 1st question of this problem ask: How long is the bullet in the target?

I simply do:

40.0cm=0.4m ; 0.4m/(900m/s)=4.4*10^-4

The answer shows 4.0*10^-4

Did I make a mistake or it is just a rounding problem?

Using s = (u+v)/2 *t the answer is 4 *10^-4. s= distance, u = inital vel, v= final vel.
 

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