A bullet fired from a target loses half of its velocity after penetrating 12cm. How much further will it penetrate before coming to rest?
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Answered by
1
Explanation:
Case I :
[
2
u
]
2
−u
2
=2.a.3
or −
4
3u
2
=2.a.3⇒a=−
8
u
2
Case II :
0−[
2
u
]
2
=2.a.x or −
4
u
2
=2[−
8
u
2
]×x
⇒x=1cm
Alternative method : Let K be the initial energy and F be the resistive force. Then according to work-energy theorem,
W=ΔK
i.e., 3F=
2
1
mv
2
−
2
1
m[
2
v
]
2
3F=
2
1
mv
2
[1−
4
1
]
3F=
4
3
[
2
1
mv
2
] ............(1)
and Fx=
2
1
m[
2
v
]
2
−
2
1
m(0)
2
i.e.,
4
1
[
2
1
mv
2
]=Fx .........(2)
Comparing eqns. (1) and (2)
F=Fx
or x=1cm
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