JEE MAIN - Physics (2006 - No. 44)
A player caught a cricket ball of mass $$150$$ $$g$$ moving at a rate of $$20$$ $$m/s.$$ If the catching process is completed in $$0.1s,$$ the force of the blow exerted by the ball on the hand of the player is equal to
$$150$$ $$N$$
$$3$$ $$N$$
$$30$$ $$N$$
$$300$$ $$N$$
Explanation
We know, Force$$ \times $$ time = Impulse = Change in momentum
$$\therefore$$ $$F \times t = m\left( {v - u} \right)$$
$$ \Rightarrow $$ $$F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$$
$$\therefore$$ $$F \times t = m\left( {v - u} \right)$$
$$ \Rightarrow $$ $$F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$$
Comments (0)
