Question

A ball of mass m is moving towards a batsman at a speed v. The batsman strikes the ball and deflects it by an angle $\theta$ without changing its speed. The impulse imparted to the ball is given by

• A. $mvcos\left(\theta \right)$
• B. $mvsin\left(\theta \right)$
• C. $2mvcos\left(\frac{\theta }{2}\right)$
• D. $2mvsin\left(\frac{\theta }{2}\right)$

Answer 1

The correct option is C

$2mvcos\left(\frac{\theta }{2}\right)$

The ball moving along AB with speed v strikes the bat at point B and is deflected along BC with the same speed v. Vector AB can be resolved into two rectangular components $AD=vcos\left(\frac{\theta }{2}\right)$ and $AF=vsin\left(\frac{\theta }{2}\right)$. Similarly, vector BC can be resolved into two rectangular components $BE=vsin\left(\frac{\theta }{2}\right)$ and $BF=vcos\left(\frac{\theta }{2}\right)$. As we can see, velocity components AF and BE are equal in magnitude and are along the same direction. There is no momentum change along this direction. However, along the normal direction, the component of velocity has reversed. Hence, change in momentum of the ball is given by $\Delta P=mvcos\left(\frac{\theta }{2}\right)-\{-mvcos\left(\frac{\theta }{2}\right)\}=2mvcos\left(\frac{\theta }{2}\right)$
Hence, the correct choice is (c).