A cricket ball of mass 0.25 ,kg with speed 10 | vedclass

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(D) Given: Mass $m = 0.25 \,kg$, initial velocity $u = 10 \,m/s$, final velocity $v = -10 \,m/s$ (since it returns in the opposite direction), and tim

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) Given: Mass $m = 0.25 \,kg$, initial velocity $u = 10 \,m/s$, final velocity $v = -10 \,m/s$ (since it returns in the opposite direction), and time interval $\Delta t = 0.01 \,s$.Change in momentum $\Delta P = m(v - u) = 0.25 	imes (-10 - 10) = 0.25 	imes (-20) = -5 \,kg \cdot m/s$.The magnitude of change in momentum is $|\Delta P| = 5 \,kg \cdot m/s$.The force exerted on the ball by the bat is $F = \frac{\Delta P}{\Delta t} = \frac{5}{0.01} = 500 \,N$.According to Newton's third law, the force exerted on the bat by the ball is equal in magnitude, which is $500 \,N$.

$A$ cricket ball of mass $0.25 \,kg$ with speed $10 \,m/s$ collides with a bat and returns with the same speed within $0.01 \,s$. The force exerted on the bat is: (in $\,N$)

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