A ball of mass m strikes a rigid walJ with speed u and rebounds with the same speed. The impulse imparted to the ball by the wall is

Options

A. `("mu sin"30^(@))/(t)`
B. `("2mu sin"30^(@))/(t)`
C. `("mu cos"30^(@))/(t)`
D. `("2mu cos"30^(@))/(t)`

Answer

D

Explanation

Initial momentum of the ball is `bar(p)_(i)=-"mu"cos30^(@)hati-"mu"sin30hatj`
Final momentum of the ball is
`bar(p)_(f)=-"mu"cos30^(@)hati-"mu"sin30hatj`
Change in momentum `Deltabar(p)=bar(p)_(f)-bar(p)_(i)=-2"mu cos"30^(@)hati`
Impusle=Change in momentum=`-2"mu cos" 30^(@)hati`. As impulse and force are in the same direction, therefore, force on the ball due to the wall is normal to hte wall along the negative x-axis. Using Newton's `3^(rd)` law of motion the force on the wall due to the ball is normal to the wall along the positive x-direction
`therefore F=(2"mu cos"30^(@))/(t)`