Question

In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spring constant of the bow is 330 N/m and she pulls the arrow back a distance of 0.55 m. The apple on the stage is 5.00 m higher than the launching point of the arrow. At what speed does the arrow (a) leave the bow? (b) strike the apple?

Answer 1

Answer:

a) $v=99.8584\ m.s^{-1}$

b) $v'=99.366\ m.s^{-1}$

Explanation:

Given:

mass of the arrow, $m=0.02\ kg$

stiffness constant of the bow, $k=330\ N,m^{-1}$

distance of pulling back the arrow on the bow from its mean position, $\Delta x=0.55\ m$

height of the apple targeted, $h=5\ m$

Force on the arrow due to the stiffness of the bow:

$F=k.\Delta x$

$F=330\times 0.55$

$F=181.5\ N$

Now the acceleration of the arrow upwards:

$a=\frac{F}{m}$

$a=\frac{181.5}{0.02}$

$a=9075\ m.s^{-2}$

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

$v^2=u^2+2(a-g).\Delta x$

where:

$v=$ velocity with which the arrow leaves the bow

$u=$ initial velocity of the arrow after it left

$v^2=0^2+2\times (9075-9.81)\times 0.55$

$v=99.8584\ m.s^{-1}$

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

Using eq. of motion:

$v'^2=v^2-2\times g.h$

where:

$v'=$ final velocity when the arrow hits the target

$v=$ initial velocity after the arrow has been launched

$v'^2=99.8584^2-2\times 9.81\times 5$

$v'=99.366\ m.s^{-1}$