2.Two children play a game of catch, running back and forth to catch each throw from the other. Assume the children are standing 5m apart with one child standing at the origin before the game begins. The ball makes the following motions:

5 meters right, 8 meters left, 3 meters right, 6 meters left, 15 meters right, and 4 meters left.

Part A:
Calculate the total distance traveled by the child on the left.

Part B:
Calculate the total displacement of the child on the right.

Part A: 24 m

Part B: 9 m

Explanation:

Part A:

The children takes turns so the child on the left moves with when the ball moves 5 meters right, 3 meters right, 15 meters right, while the child on the right moves with when the ball moves 8 meters left, 6 meters left, 4 meters left

The total distance traveled by the child on the left is given by the relation;

Initial position of the child on the left = 5 m

First motion of the child on the left = Distance to origin + first distance ball moved

∴ First motion of the child on the left = 5 + 5 = 10 m location of child = 5 m left

Second motion of the child on the left = location of child - position of the ball = 5 - (5 - 8 + 3) = 5 m left location of the child = 0 m

Third motion of the child on the left = location of child - position of the ball = 0 - (6 - 15) = 9 m right

Total distance traveled by the child on the left = 10 + 5 + 9 = 24 m

Part B:

First motion of the child on the right = Distance of child to origin + second position of ball

Initial location of the child on the right = 0 m

∴ First motion of the child on the right = 0 + (5 - 8) = -3 m location of child = 3 m left

Second motion of the child on the right = location of child - current position of the ball = 3 - 6 = -3 m left location of the child = -6 m

Third motion of the child on the right = location of child - position of the ball = -6 - 5 = 11 m right

Final location of the child on the right = 9 m right

Total displacement of the child on the right = Final location of the child on the right - Initial location of the child on the right 9 - 0 = 9 m.

K = 25351. 69 N / m

Explanation:

Given : Fk = 515 N , v = 1.8 m /s  , d = 5.0 m , β = 22.0 ° , m = 150 kg

Using the work done by all forces at initial and the end can determine the constant of the spring

Ws + We - Fk = Em - Ef

- ¹/₂ * K * x²  + m*g*h - F*d = 0 - ¹/₂ * m * v²

Also the round motion part

K* x = F + We

K * x = F + m*g*h

Replacing numeric to equal the equations and find the constant

¹/₂ * K * x² = 150*9.8* 5* sin (22°) - 5150* 5 + ¹/₂*150*(1.8m/s)²

K * x² = 421.358

Now use the other equation

K * x = 515 + 150*9.8* sin(22°)

K * x = 3268.35

Both equation give x' as a

x = 0.1289 m now using in any equation can find K

K = 25351. 69 N / m