Question : A bullet is fired into a fixed target. It loses 1/3 rd of its velocity after travelling 4cm. It penetrates further p×10–3 m before coming to rest. Find p.
Doubt by Nisha
Solution :
Case I :
Initial speed = u
Final speed, v=2u/3
[The bullet losed 1/3 of its initial velocity so the remaining velocity would be 2/3 of u and not 1/3 of u]
Distance Covered, s1= 4 cm = 0.04 m
Let constant retardation = a
Using
v²-u²=2as
(2u/3)²-u²=2as1
4u²/9-u²=2as1
(4u²-9u²)/9=2as1
-5u²/9=2as1
-5u²/18s1=a
a=-5u²/18s1 — (1)
Case I :
Initial speed = u
Final speed, v=2u/3
[The bullet losed 1/3 of its initial velocity so the remaining velocity would be 2/3 of u and not 1/3 of u]
Distance Covered, s1= 4 cm = 0.04 m
Let constant retardation = a
Using
v²-u²=2as
(2u/3)²-u²=2as1
4u²/9-u²=2as1
(4u²-9u²)/9=2as1
-5u²/9=2as1
-5u²/18s1=a
a=-5u²/18s1 — (1)
Case II :
Initial speed, u = 2u/3
Final speed, v = 0 m/s
Initial speed, u = 2u/3
Final speed, v = 0 m/s
Distance Covered = s2 = p×10-2 m
retardation = a
v²-u²=2as
(0)²-(2u/3)²=2as2
-4u²/9=2as2
-4u²/9=2[-5u²/18s1]s2
-4u²/9=-10u²s2/18s1
4=10s2/2s1
4=5s2/s1
4s1=5s2
4s1/5=s2
s2=4s1/5
s2=4(0.04)/5
s2=0.16/5
s2=0.32
s2=32×10-2 m
p×10-2 m=32×10-2 m
p=32
retardation = a
v²-u²=2as
(0)²-(2u/3)²=2as2
-4u²/9=2as2
-4u²/9=2[-5u²/18s1]s2
-4u²/9=-10u²s2/18s1
4=10s2/2s1
4=5s2/s1
4s1=5s2
4s1/5=s2
s2=4s1/5
s2=4(0.04)/5
s2=0.16/5
s2=0.32
s2=32×10-2 m
p×10-2 m=32×10-2 m
p=32
Hence, p=32