A question

F=(mv-mu)/t <---momentum

F=[17m-(-10)m]/0.8 <---注意:係-10, 因為個波向反方向移動

F=27m/0.8

F=3.375m N

By F=ma

3.375m=ma

a=3.375ms^-2

因為你TAKE向牆果邊係POSITIVE, 所以當個波向番反方向行GE時侯要加個MINUS响前面

因為VELOCITY係VECTOR~

in this case
we use force = change of momentum/time taken
where change of momentum = final momentum - inital momentum

let m be the mass ,v be the final velocity ,u be the inital velocity t be time taken for the hiting process

therefore force = (mv-mu)/t
and by f=ma
(mv-mu)/t=ma
(v-u)/t=a

IMPORTANT : we should choose one direction as +ve then the other direction is -ve
in this case if i choose the direction that the ball travel after hitting as =+ve,then the original direction of the ball will be -ve

here we take final velocity = +10 ms^-1 the inital velocity= -17 ms^-1
then a =(v-u)/t=(10-(-17))/0.8
this is the same as a = (17-(-10))/0.8= +33.75 ms^-2
since the answer is positive , this mean the acclearation is in the rebound direction
as i assigned +ve as the rebound direction

The important things is u should assign one direction as positive and the other be -ve before u write down the formula
This is nesscary since velocity , momentum , force and acceleration are all vectors which contain "direction information".