Question

.
In fig, Disc of mass 2kg is kept
floating horizontally in air by firing
bullets of mass 0.1kg each vertically
at it at rate of 5 per sec. If
bullets rebound with same speed,
Find speed with which these are fired.​

Answer 1

Answer:

weight of disc =mg=5kg×10m/S2=50N

force exerted by 10 bullets much be equal to 50N

F=∆P/t

F=nm×(v-(v))1t

F=2nmv/t

from above

v=Ft/(2nm)

=(50n×1s)/(2×10/50×10^-3kg)

=50m/s

speed of each bullet is 50m/s

Step-by-step explanation:

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Answer 2

Answer:

v=19.6 m/s

Step-by-step explanation:

as there is force Mg on plate on downward side.......
So, mass of plate = 2 , g=-9.8
Mg=2*9.8=19.6 N....

Now....... P initial=mv=0.1v
P final= mv=0.1v
  change in momentum=P final - P initial=0.1(2v)

Now, we need to find v that pushes the plate upward side
So, v=F*change in momentum
  where F= rate of firing
v=5*(0.1*2v)
=by calculating we get
v
so we should need v taht will balance the plate in the air so
v=19.6m/s.........