JEE MAIN - Physics (2005 - No. 57)
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?
$$2.0$$ $$cm$$
$$3.0$$ $$cm$$
$$1.0$$ $$cm$$
$$1.5$$ $$cm$$
Explication
Let $$K$$ be the initial kinetic energy and $$F$$ be the resistive force. Then according to work-energy theorem,
$$$W = \Delta K$$$
i.e., $$3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2}...\left( 1 \right)$$
Let the bullet will penetrate x cm more before coming to rest.
$$\therefore$$ $$Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)$$
Dividing eq. $$(1)$$ and $$(2)$$ we get,
$${x \over 3} = {1 \over 3}$$ or x = 1 cm
i.e., $$3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2}...\left( 1 \right)$$
Let the bullet will penetrate x cm more before coming to rest.
$$\therefore$$ $$Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)$$
Dividing eq. $$(1)$$ and $$(2)$$ we get,
$${x \over 3} = {1 \over 3}$$ or x = 1 cm