Question 198: Work, Energy and Power & Circular Motion

Question

A bullet fired into a fixed target loses half of its velocity after penetrating $3$ $cm.$ How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Accepted Answer

Let $K$ be the initial kinetic energy and $F$ be the resistive force. Then according to work-energy theorem,  
$$W = \Delta K$$
i.e., $3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} ight)^2}...\left( 1 ight)$
Let the bullet will penetrate x cm more before coming to rest.
$	herefore$ $Fx = {1 \over 2}m{\left( {{v \over 2}} ight)^2} - {1 \over 2}m{\left( 0 ight)^2}...\left( 2 ight)$
Dividing eq. $(1)$ and $(2)$ we get,
${x \over 3} = {1 \over 3}$ or x = 1 cm

Answer

$2.0$ $cm$

Answer

$3.0$ $cm$

Answer

$1.0$ $cm$

Answer

$1.5$ $cm$

Work, Energy and Power & Circular Motion

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

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