A 25 gram Bullet is fired with a velocity of 400 metre per second into a sandbag of mass 3.975 kilogram suspended by a rope the bullet gets embedded into the bag the loss of kinetic energy is
Answers
Explanation:
Initially, only the bullet is moving, and so the total momentum of the bullet+bag system is mv. Once the bullet is embedded in the bag, the bullet and bag are moving together, and therefore have the same velocity (call it u). So, by conservation of momentum, we have
(M+m)u=mv
u=
M+m
mv
We now have what we need to compare the initial and final kinetic energies. Probably the most useful way to do this is as a ratio: the final kinetic energy, as a fraction of the initial kinetic energy, is
2
mv
2
m+M
2
u
2
=
m
M+m
(
v
u
)
2
=
m
m+M
(
m+M
m
)
2
=
m+M
m
We could also see this by noting that the kinetic energy of a single moving mass is equal to
2m
p
2
, and that p isn’t changing during the collision, so the kinetic energy will vary inversely with (uniformly moving) mass.
KE
loss
=KE
ini
(1−
m+M
m
)=
2
1
mv
2
(
m+M
m
)
At first find your first momentum then second momentum.
after finding equal the both momentum and yu will get the answer.
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