Answer to Question #20662 in Mechanics | Relativity for krishna

Question #20662
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?
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&#039;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;

Answer: 10 cm further

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