a machine gun fires n bullets per second with speed U and mass of each Bullet is M let`s hit a wall and rebounds with same speed the force acting on the wall is

							Force is rate of change of momentumPer second the n bullets hit the wall with velocity v and return at same velocity So change in momentum of 1 ball is { mv-m(-v)} =2mv So rate of change if momentum for n balls is 2mvn/1 so force is 2nmv
						

							Here we are given with n bullets and the speed with which they are being fired is u and each having mass m. So, my dear friends what you have got to do is that first consider 1 bullet`s momentum I.e.,mu. Now when it rebounds then momentum is (-mu). Since force is the rate of change of momentum I.e., mu- (-mu)=2mu therefore for n bullets = 2nmu. Now 2nmu/1sec gives us the force acting on the wall by n no. Of bullets. = 2nmu.
						

							As force=vdm/dr hence n bullets hits the wall per second so we write the dm/dt is n*m so force required is f=U(nm) =nmU
						

							
 
Here we are given with n bullets and the speed with which they are being fired is u and each having mass m. So, my dear friends what you have got to do is that first consider 1 bullet`s momentum I.e.,mu. Now when it rebounds then momentum is (-mu). Since force is the rate of change of momentum I.e., mu- (-mu)=2mu therefore for n bullets = 2nmu. Now 2nmu/1sec gives us the force acting on the wall by n no. Of bullets. = 2nmu.
						

							
orce is rate of change of momentumPer second the n bullets hit the wall with velocity v and return at same velocity So change in momentum of 1 ball is { mv-m(-v)} =2mv So rate of change if momentum for n balls is 2mvn/1 so force is 2nmv