Question

A 370 g ball with a speed v of 3.9 m/s strikes a wall at an angle θ of 22° and then rebounds with the same speed and angle. It is in contact with the wall for 13 ms. In unit-vector notation, what are (a) the impulse on the ball from the wall and ?

Answer 1

Let the ball of mass `m=0.37kg` strikes the vertical wall with velocity v at an angle of incidence `theta=22^@` with the positive directon of x-axis and it rebounds with same speed and angle.

Vectorially the velocity of striking the wall may be epressed as `vec(v_s)=vcosthetahati+vsinthetahatj`

and velocity of rebounce will be `vec(v_b)=-vcosthetahati+vsinthetahatj`

Here the horizontal component is reversed and vertical component is remaining same and no action of gravity is considered.

The impulse or change in momentum for the ball of mass `m=370g=0.37kg` is

`=m(vec(v_b)-vec(v_s))=-2mvcosthetahati`
`=-2xx0.37xx3.9xxcos22hati" kgm/s"`

`=-2.675hati" kgm/s"`

(a) So magnitude of change in momentum or impulse

`=abs(m(vec(v_b)-vec(v_s)))= 2.675" kgm/s"`

(b) If the duration of contact with the wall be `t=13ms=13xx10^-3s` then the average force on the wall from the ball will be
`"Force "(F)="impulse"/"time"=2.675/(13xx10^-3)N~~205.8N`