Answer to Question #20662 in Mechanics | Relativity for krishna
We know that on the distance l = 30 cm it performed work against the
A = F*l;
It converted a part of its kinetic energy for this purpose, losing half of the
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;
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;
Answer: 10 cm further
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