A projectile of mass 3m explodes at highest pt of its path.It breaks into three equal parts.One part retraces its path, the second one comes to rest.The range of the projectile was 100m if no explosion would have taken place.The distance of the third part from the pt of projection when it finally lands on the ground is
Asked by m.nilu
18th September 2018,
10:28 AM
Answered by Expert
Answer:
When the projectile reaches the highest point, its vertical component of velocity is completely vanished.
Only it has horizontal component of velocity remains.
Hence Momentum before explosion = 3m×u×cosα .........................(1)
It is broken into three equal parts. One part of mass m retrace its path.
Hence its velocity after explosion is u×cosα in horizontal direction.
Second part of mass comes to rest. Hence its velocity is zero.
let U be the velocity of remaining third part.
As per conservation of momentum, 3m×u×cosα = m×u×cosα + m×0 + m×U ..............(2)
from eqn.(2), we get U = 2×u×cosα .............................(3)
Hence the third part has velocity 2×u×cosα in horizontal direction just after explosion, at same time it starts falling due to gravity.
time taken for the original mass 3m to reach the highest point = u×sinα/g ................(4)
time taken for the third part of mass m to reach ground staring from highest point also same as given in eqn.(4)
Hence horizontal distance travelled by the third part of mass m = 2×u×cosα × u×sinα/g = u2×sin2α/g ..................(5)
RHS of eqn.(5) is same as horizontal range of projectile which is given as 100 m.
Hence the third part of mass will travel a distance 100 m after explosion.
Before explosion distance travelled by the original mass is half of range i.e. 50 m.
Hence the required landing distance of third part of mass from point of projection is 100+50 = 150 m.
Answered by Expert
18th September 2018,
11:53 AM
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