A bullet of mass 20 ,g moving with a velocity | vedclass

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Question

(A) Given: Mass of the bullet $m = 20 \,g = 0.02 \,kg$. Initial velocity $u = 200 \,m/s$. Final velocity $v = 0 \,m/s$. Time taken $\Delta t = \frac{1

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$4 \,Ns, 200 \,N$

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(A) Given: Mass of the bullet $m = 20 \,g = 0.02 \,kg$. Initial velocity $u = 200 \,m/s$. Final velocity $v = 0 \,m/s$. Time taken $\Delta t = \frac{1}{50} \,s = 0.02 \,s$.Impulse $(J)$ is equal to the change in momentum:$J = \Delta p = m(v - u)$$J = 0.02 \,kg 	imes (0 - 200) \,m/s = -4 \,kg \cdot m/s$.The magnitude of impulse is $4 \,Ns$.Average force $(F_{avg})$ is given by the rate of change of momentum:$F_{avg} = \frac{J}{\Delta t} = \frac{4 \,Ns}{0.02 \,s} = 200 \,N$.Thus,the impulse is $4 \,Ns$ and the average force is $200 \,N$.

$A$ bullet of mass $20 \,g$ moving with a velocity of $200 \,m/s$ strikes a target and is brought to rest in $\left(\frac{1}{50}\right) \,s$. The impulse and average force of impact are respectively:

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