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Q4. A machine gun fitted on a trolley of mass 2000 kg on a horizontal frictionless surface. If the gun fires bullet each of mass 10 gm with a velocity of 500 m/s with respect to the trolley and if the number of bullet fired per second is 10 then calculate the average thrust exerted by ejected bullet on the system.

Asked by sbidipak | 05 Jul, 2017, 07:20: PM
answered-by-expert Expert Answer
- The average thrust exerted by ejected bullet is nothing but the momentum gained by that bullet when fired.
- Momentum can be understood as a force or an energy gained by a moving object.
- This force is given by F = mv (mass and velocity of bullet). 
- But the number of bullets per second also amplifies this momentum, therefore, the force in this case is expressed as F = n x mv ('n' is   the number of bullets)
- Converting the mass of a bullet in kilograms we get m=0.010 kgs.
- Check if all the units involved either belong to the SI (prefered) or CGS system.
- Substituting the values of m,v and n in the above formula for 'F' we get,
  F = 10 x 0.010 x 500 = 50 N
Answered by Abhijeet Mishra | 06 Jul, 2017, 01:57: PM

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