Question

what is the mass of a bullet that takes 15 cm to decelerate from a muzzle velocity of 320 m/s to resting and the average force decelerating the bullet is 2013N?

Answer 1

given that bullet decelerate from muzzle velocity 320 m/s to rest in 15 cm distance

so here we know that

$d = 15 cm = 0.15 m$

$v_f = 0$

$v_i = 320 m/s$

now we can use kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 320^2 = 2*a*0.15$

$a = - 3.41 * 10^5 m/s^2$

now we know that force acting on the bullet is

$F = 2013 N$

now by formula of force

$F = ma$

$2013 = m* 3.41* 10^5$

$m = 5.90 * 10^{-3} kg$

$m = 5.90 gram$

so mass of bullet is 5.90 g

Answer 2

m = mass of the bullet = ?

F = force on the bullet = 2013

a = acceleration of the bullet = - F/m = - 2013/m

v₀ = initial velocity of the bullet = 320 m/s

v = final velocity of the bullet = 0 m/s

d = stopping distance = 15 cm = 0.15 m

Using the kinematics equation

v² = v²₀ + 2 a d

0² = 320² + 2 ( - 2013/m) (0.15)

m = 5.89 x 10⁻³ kg