Question

a) A ball of mass 20 g moving at horizontal speed of 20 m s' hits a vertical wall. The ball bounces back on the same path with the same speed after touching the wall for 0.01 s. i) Calculate the total changes of momentum of the ball. ii) Calculate the average force exerted by the ball on the wall. iii) Is this collision elastic or inelastic? Prove your answer. b) From (a), if the ball bounces back with half its initial speed on the same path, calculate i) the change of momentum of the ball. ii) the change of kinetic energy of the ball.​

Answer 1

Step-by-step explanation:

i) the change total change in momentum of an object is p = m(vf - vi). we know that the ball bounced back with the same initial speed in the opposite direction.

20 gram = 20/1000 = 0.02kg.

p = 0.02( -20 -20) = 0.02x-40 = -0.8N

ii) Average force = m (vf - vi)/t ; 0.02x(-20-20)/0.01 = -80N

iii) collision is elastic because the kinectic energy before and after collision are the same.

b. i) if the ball bounces back with half its initial speed, then the change in momentum will be, p = m( 0.5v - v)

p = 0.02( -0.5x20 - 20 ) = 0.02x -30 = -0.6N

ii) kinetic energy = 6J

`(1)/(2)m(vf-vi)^(2) = (1)/(2) m(-10-20)^(2) ;\\(1)/(2) 0.02(-30^(2) ) == 300*0.02 = 6J`