Answer:
The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.
Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.

If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.
The wall exerts a force of equal magnitude but in the opposite direction. So the force by the wall is 10 N to the right.
Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :


So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Answer:
rolling ball down a hill
Explanation:
A rolling ball has kinetic energy
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle