A hammer of mass M strikes a nail of mass m w | vedclass

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(B) Step $1$: Use the principle of conservation of linear momentum during the collision between the hammer and the nail.$M 	imes 20 + m 	imes 0 = (M

Vedclass · Accepted Answer

$\frac{2 M^2}{M+m} 	imes 10^4$

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(B) Step $1$: Use the principle of conservation of linear momentum during the collision between the hammer and the nail.$M 	imes 20 + m 	imes 0 = (M + m) 	imes v'$$v' = \frac{20M}{M+m}$Step $2$: Use the work-energy theorem to find the average resistance force $F$ of the wall.The work done by the resistance force equals the change in kinetic energy of the system $(M+m)$.$F 	imes d = \frac{1}{2} (M+m) (v')^2$Here,$d = 1 \ cm = 10^{-2} \ m$.Step $3$: Substitute the values into the equation.$F = \frac{(M+m) 	imes (v')^2}{2d}$$F = \frac{(M+m)}{2 	imes 10^{-2}} 	imes \left( \frac{20M}{M+m} ight)^2$$F = \frac{(M+m)}{2 	imes 10^{-2}} 	imes \frac{400M^2}{(M+m)^2}$$F = \frac{200M^2}{(M+m) 	imes 10^{-2}} = \frac{2M^2}{M+m} 	imes 10^4$Thus,the average resistance is $\frac{2M^2}{M+m} 	imes 10^4$.

$A$ hammer of mass $M$ strikes a nail of mass $m$ with a velocity $20 \ m/s$ into a fixed wall. The nail penetrates into the wall to a depth of $1 \ cm$. The average resistance of the wall to the penetration of the nail is

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