Question

A cricket ball of mass $0.5\, kg$ strikes a cricket bat normally with a velocity of $20\, ms ^{-1}$ and rebounds with velocity of $10\, ms ^{-1}$. The impulse of the force exerted by the ball on the bat is

• A. $15\,Ns$
• B. $25\,Ns$
• C. $30\,Ns$
• D. $10\,Ns$

Answer 1

Answer: (A)

$15\,Ns$

Answer 2

During collision of ball with the wall horizontal momentum changes (vertical momentum remains constant)
$\therefore F=\frac{\text { change in horizontal momentum }}{\text { time of contact }}$
$=\frac{m(v+u) \cos \theta}{t}$
$=0.5(20+10) \cos \theta=15 \,N s$