A man with a rifle shoots bullets at a speed 
Hello all
So I was doing this question.
I had a doubt that, initially as both the (man and rifle ) system and the bullet were at rest we put 0 instead of m1u1 + m2u2 in equation of conservation of momentum and hence get the speed of the system as 0.08 m/s after the first shot.
But , after the first shot we can't do the same for the subsequent shots as both the  system and the bullet will have some velocity ( they both will have the same velocity in the same direction. ) and hence for subsequent shots the velocities gained by system would be different than 0.08 m/s. Then how can we just give the answer as 10 × (the velocity gained in the first shot.) .
ie:  0.8 m/s
This was the answer given in the book.
 A: We can do the same for each subsequent shot, because the momentum change with each subsequent shot is the same as the momentum change with the first shot.  Look at the impact = momentum relationship.
$F\ \Delta t =m\ \Delta v$
Each shot produces the same $F\ \Delta t$ and therefore produces the same momentum change in the man and his muzzle.
Another way to look at it is that the bullet exits with the same muzzle velocity with each shot, so the fact that the muzzle is moving backwards after the first shot is irrelevant.  Even though the bullet moves less than 800 m/s with respect to the ground after the first shot, it started with a negative velocity, so that its change in velocity and therefore, its momentum change is the same as it was with the first shot.  So it imparts the same momentum change in the man and his muzzle as did the first shot.
Since each shot produces the same momentum change in the man as the first shot, it is correct to multiply the effect of the first shot by the total number of shots.
A: Intrigued by @Bill Watts answer I decided to do this answer step by step and found out that Bill was indeed right. 
But I also observed that we all were also right but the thing was that we weren't actually calculating the values on pen and paper and hence believed that multiplying the initial velocity by 10 would be wrong.
Anyway here is the solution

A: Imagine only one bullet is being fired. If (after the firing) the velocity of this bullet is $v_b$ and the velocity of the system (man + remaining bullets) is $v_s$, then we apply momentum conservation as follows:
$m_Sv_s + m_bv_b = 0$ or $\frac{v_s}{v_b} = -\frac{m_b}{m_s} $ 
Therefore, after firing one shot, the velocities of the man and the bullet, will in general not be the same. 
We can iterate this for 10 bullets, by changing our 'system' variables for each shot fired. 
Alternatively we need only apply momentum conservation once if we look at the system before the first shot and after the last shot since all the bullets are fired independently and the momentum of the man + rifle + ten bullets does not change (since no forces act on this larger system as a whole)
Therefore if $m_r$ and $v_r$ denote the mass and the velocity of the man + rifle and $v_b$ and $m_b$ represent the velocity and mass of each bullet (velocity of a bullet does not change once fired) the total momentum of this system will be 
$m_rv_r + 10m_bv_b = 0$ 
Since we know the total mass and the mass of each bullet, we can solve for $v_r$ here which is the required answer. 
The reason this can be done directly, is because through the firing of the 10 shots, no external force acts on the bullets + rifle + man system, so its total linear momentum is what it was before the first shot is fired, which is zero. 
EDIT: Okay so if muzzle velocity here refers to the velocity of the bullet wrt the muzzle alone, then momentum conservation must be applied ten times separately. We can do this from the ground frame by calculating the velocity of each bullet wrt to the ground since we know the muzzle velocity and the velocity of the gun after each shot. In the ground state, any force on the rifle due to the bullet (and vice versa) will still be internal forces as long as we re calculate the right velocities after each shot. 
A: Let us start with basic what question actually want since man has mass of 100kg (including gun mass) ,the question is all about momentum conservation since no impulsive force is present momentum relationship.
F Δt=0 so initial momentum =final momentum so lets start when bullet one is fired initial momentum was zero so the final momentum should also.equate to zero 1000 multiply x (x:velocity of man experience in opposite direction of bullet travel )=800 multiply 0.01 so you get the velocity x now for firing the shot 2 man has already momentum of mx so equate and you will yourself able to analyze the pattern please comment if you not able to solve for part b momentum relationship.
F Δt=10F delta t=10 initial.momentum of man was zero final you calculate in part 1
