Their combined momentum after they meet is 0 .
The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.

= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
Answer:
Explanation:19,2 or 0/4 or 5 or 40,4
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s
To find force we use formula:
F = m * a
a is acceleration. To find it we use formula:
a = Δv / Δt
a = 25.2 / 0.00195
a = 12923.1 m/s^2
Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N
To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N
We can see that weight is much smaller than the applied force so it's influence in negligible.