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# If a body loses half its velocity on penetrating 3cm in a wooden block, them how much will it penetrate more before coming to rest?

Vikas TU
14149 Points
4 years ago
With penetrating 3 cm block,
v^2/4 = v^2 – 2a*3
3v^2/4 = 6a
v^2 = 8a.........(1)
Now,
0^2 = v^2 – 2ax.......(2)
put (1) in (2)
x = 1 cm
net distance would be => 1 + 3 = 4cm.
Samuel Garry
61 Points
3 years ago
First case
initial velocity(u)=v , final velocity(v)=v/2 ;distance covered(s)=3cm
we know 2as=v2-u2
2*a*3=(v/2)2 - v2
a= -v2 / 8
in second case
initial velocity(u)=v , final velocity(v)=o ;distance covered(s)=we hav to find out
here we know acceleration=-v2 / 8
2as=v2-u2
2*(v2 / 8)*s=(0)2 - v2
on solving we get
s=4cm
but it had already pnetrated 3 cm thrfore it will pertrate 1 cm more to come to rest
ankit singh
one year ago
Let the initial velocity of the body be u. Its velocity becomes u/2 after penetrating 3 cm in the block. We can calculate the deceleration of the body using these.
v²= u² +2as
u²/4 = u² + 2a×3
-3u²/4 = 6a
a = -u²/8
Now, we can calculate the distance the body travels till it comes to rest.
v² = u² + 2as
0 = u² +2×(-u²/8)×s
u² = u²s/4
s = 4 cm
Therefore, the body penetrates 1 cm (4-3=1cm) more before coming to rest.
I hope you understand the concept.
All the best!