question_answer

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

A)  \[15\text{ }Ns\]         

B)  \[25\text{ }Ns\]

C)  \[30\text{ }Ns\]         

D)  \[10\text{ }Ns\]

Correct Answer: A

Solution :

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