The average force imparted on the wall when a 0.75 kg ball moving with a speed of 18 m/s strikes it at an angle of 60' for a contact time of 0.15 s and gets reflected with the same speed and at the same angle is P newtons. Then, the value of P/10 is: (Take `sqrt3 = 1.73` )

Answer

`15.57`

Explanation

Components of momentum parallel to the wall add each other and components of momentum perpendicular to the wall are equal and opposite. Thus, the change in momentum is only due to the change in direction of the perpendicular components.
`:.` Change in momentum
`= ON sin 60^@ - (-OM sin 60^@)`
`= mv sin 60^@ - (-mv sin 60^@)`
`= 2 mv sin 60^@`
Now, `Fxxt` = change in momentum
`:. F = (2 mv sin 60^@)/t`
`:. F =(2xx0.75xx18xxsin 60^@)/0.15`
`:. F = (2xx0.75xx18xx1.73)/(0.15 xx 2)`
`:. F = 155.7 N`