A bomb initially at rest at a height of 40 metres above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m/s², what is the separation between the fragments 2 s after the explosion?
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# Answer - 40 m.
## Explaination-
Consider, bomb with mass M splits into two pieces of mass m moving with velocities v1 & v2.
Initially bomb was at rest, hence initial momentum will be zero.
By law of conservation of momentum,
0 = mv1 + mv2
v1 = -v2
Now,
s = vt
s = 10×2
s = 20 m
But two fragments travel in opposite directions, hence,
seperation = 2×s = 40 m.
After 2 s, the separation between fragments will be 40 m.
Hope that cleared your doubt...
# Answer - 40 m.
## Explaination-
Consider, bomb with mass M splits into two pieces of mass m moving with velocities v1 & v2.
Initially bomb was at rest, hence initial momentum will be zero.
By law of conservation of momentum,
0 = mv1 + mv2
v1 = -v2
Now,
s = vt
s = 10×2
s = 20 m
But two fragments travel in opposite directions, hence,
seperation = 2×s = 40 m.
After 2 s, the separation between fragments will be 40 m.
Hope that cleared your doubt...
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