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rjblayz rjblayz
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11 years ago
A bullet of mass 15g travelling with a velocity of 150m/s strikes a tree and penetrates a distance of 250mm into the timber before coming to rest. What was the average acceleration of the buleet and what was the average force causing the accleration?

I really appreciate any help!!!
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wrote...
11 years ago
use (work done by the force=kinetic energy)
v0^2=2*a*s=2*F/m*s
then
a=v0^2/(2*s)=4.5*10^(4) m/s^2 Leftwards Arrow---------
F=m*a=675 N Leftwards Arrow---------------

ps- all moduli, of course.. but it's simple to understand..
not so simple...!!!!
wrote...
11 years ago
u^2+2as = v^2
where u = the initial velocity of the bullet.
a = the acceleration(actually deceleration) of the bullet.
s = the distance that the bullet travels in the timber.
v = the final velocity of the bullet.
i.e.,
150^2+2a*(250/1000)=0^2
22500+0.25*2*a = 0
22500+0.5a = 0
0.5a = -22500
a = -45000
The average acceleration(deceleration) of the bullet was 45000ms^(-2).

So, F = ma
= [(15/1000)*(-45000)]N
= 675N in the opposite direction that the bullet was travelling.
The average force that was causing the acceleration(deceleration) was 675N in the opposite direction that the bullet was travelling.
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