Answer to Question #83913 in Mechanics | Relativity for similoluwa fashola

Question

A girl kicks a ball along the ground at a wall 2.0 m away. The ball strikes the wall normally at a velocity of 8.0m/s and rebounds in the opposite direction with an initial velocity of 6.0m/s . The girl, who has not moved, stops the ball a short time later.
(a) The ball has a mass of 0.45kg and is in contact with the wall for 0.10s. For the period of time the ball is in contact with the wall,
(I)calculate the average acceleration of the ball
(II)calculate the average force acting on the ball.
(III)state the direction of the average force acting on the ball.
(b)Determine the amount of kinetic energy lost by the ball during the collision. What happens to this energy.

Answer 1

(I) The average acceleration is

$$a=(v1-v0)/t=(6-(-8))/0.10=140 m/s^2$$

(II) The average force is

$$F=ma=0.45*140=63 N$$

(III) The direction of the average force relative to the wall is normally aside the wall since the ball moves towards the girl.

(b) The amount of KE lost in this experiment with the wall and the ball is

$$Q=KE1-KE0=mv1^2/2-mv2^2/2=m/2*(v1^2-v2^2),$$ $$Q=0.45/2*(6^2-8^2)=-6.3J$$

This energy wastes for deformation of the ball, for the increase in temperature for the ball and the wall, and maybe for deformation of the wall.