Answer:
-0.01052 m/s
Explanation:
M = mass of ship = 
m = mass of shell = 1100 kg
v = velocity of shell = 550 m/s
u = recoil velocity of ship
As linear momentum is conserved

The recoil velocity of the ship taking the firing direction to be the positive direction is -0.01052 m/s
Answer:
0.488 m
Explanation:
If θ be the angle ladder makes with the plane
cos θ = 1.2 / 5
Tan θ = 4.04
Let the height a person of weight 600 N can climb be h from the ground .
Distance from the base point where ladder touches the floor = h / tanθ
= h / 4.04
Total reaction force = total downward force
R = 200 + 600
800 N
Frictional force = μ R
= .2 x 800
= 160 N
Taking moment of force about the point on the ladder where it touches the floor and balancing them
200 x 1.2 x .5 + 600 x h / tanθ = μ R x 1.2 / tanθ ( reaction at the top point of ladder where it touches the wall is R₁ and
R₁ =μ R )
= 200 x 1.2 x .5 + 600 x h / tanθ = 160 x 1.2 / tanθ
120 - 600 h / 4.04 = 47.52
120 - 47.52 = 600 h / 4.04
72.48= 148.51 h
h = 0.488 m
=
Here in the above situation when boy pushes the girl there will be equal and opposite force on boy as per Newton's III law.
Due to this action reaction law boy will also go back with some speed.
Since there is no external force on this girl + boy system so we can use momentum conservation principle here.
As per momentum conservation




So boy will go back with speed 0.2 m/s
Part b)
Since the boy and girl will always exert same force on each other by Newton's III law so it has no it matter whether the boy pushed the girl, the girl pushed the boy, or they put their hands together and pushed each other away.
As in all above cases the as per Newton's III law the force on them is always equal and opposite.