Given:
The bullet loses one-third of its velocity after traveling s=4cm=4×10−2m, with final velocity v=32u.
Using the kinematic equation:
v2−u2=2a⋅s
Substituting the values:
(32u)2−u2=2a⋅(4×10−2)
Simplifying:
94u2−u2=2a⋅(4×10−2)
94u2−99u2=2a⋅(4×10−2)
−95u2=2a⋅(4×10−2)
a=72×10−2−5u2
Now, for the bullet to come to rest:
v2−u2=2a⋅D
Substitute v=0, a=72×10−2−5u2, and solve for D:
0−u2=2⋅(72×10−2−5u2)⋅D
u2=72×10−210u2⋅D
D=1072×10−2
D=32×10−3m=32mm.
The Correct answer is: 32mm