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?

Expert's answer

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

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

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