Answer on Question #44306 – Physics - Mechanics | Kinematics | Dynamics
The initial velocity u of a bullet in penetrating a distance 's' through a target is reduced by u/n. how far will the bullet proceed through the target before coming to rest?
Solution:
u – initial velocity of the bullet;
S – first traveled distance;
nu – final velocity after travelling first distance;
D – traveled distance before coming to rest;
a – deceleration of the bullet;
t1 – time of travelling distance S;
t2 – time of travelling distance D;
Equation of motion of the bullet for the travelled distance S:
S=ut1−2at12
Rate equation for the bullet:
nu=u−at1a=(t1u−nt1u)=t1u(1−n1)
(2) in(1):
S=ut1−t1u(1−n1)⋅2t12S=ut1−2u(1−n1)t1t1=u−2u(1−n1)S=2u+2u⋅n1S=u(n+1)2S
(3) in(2):
a=u(n+1)2Su(1−n1)=2Su2(n+1)(1−n1)=2Su2(n+1)⋅(nn−1)=2Su2(n+1)⋅(nn−1)=2nSu2(n2−1)
Rate equation for the bullet (final velocity of the bullet is zero):
0=u−at2t2=au
Equation of motion of the bullet for the travelled distance D:
D=ut2−2at22
(5) in(6):
D=u⋅au−2a(au)2=2au2D=2⋅2nSu2(n2−1)u2=u2(n2−1)u2nS=n2−1nS
Answer: Distance that bullet proceed through the target before coming to rest is equal to n2−1nS
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