a bullet after entering a wooden block at rest covers 30 cm inside it and loses half of its velocity after how much further distance will it come to rest?
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Expert's answer
2012-12-18T09:01:06-0500
Let the velocity of the bulletwhen it hits the surface be v. We know that on the distance l = 30 cm it performed work against the dissipation forces A = F*l; It converted a part of its kinetic energy for this purpose, losing half of the velocity, thus A = 0.5*m*v^2 - 0.5*m*(v/2)^2;
Let's assume the resistance force stays the same. The bullet will come to a stop after moving by additional distance x (further than l = 30 cm). Then 0.5*m*(v/2)^2 = F*x;
From here, x = 0.5*m*(v/2)^2 * (1/F) = 0.5*m*(v/2)^2 * (l/A) = l * 0.5*m*(v/2)^2 / [0.5*m*v^2 - 0.5*m*(v/2)^2] = l * (1/3); x = 30 cm * (1/3) = 10 cm;
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