Answer to Question #20662 in Mechanics | Relativity for krishna

Question

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?

Answer 1

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;

Answer: 10 cm further