Question

A 7.80-g bullet moving at 600 m/s strikes the hand of a superhero, causing the hand to move 5.90 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet. N (b) Assuming the force is constant, determine how much time elapses between the moment the bullet strikes the hand and the moment it stops moving. s

Answer 1

a) F = 2.3797 10⁴ N, b) t = 1.97 10⁻⁴ s

Step-by-step explanation:

a) For this exercise let's use the relationship between work and scientific energy

W = ΔK

W = F .d

ΔK = ½ m v² - ½ m v₀²

the bold are vectors. in this case the force of the bullet on the hand and the displacement of the bullet has the opposite direction, therefore the angle between the two is 180. The velocity when the bullet stops is zero.

We substitute

-F x = - ½ m vo2

F =
`(m v_o^2)/(2 x)`

let's calculate

F =
`(7.80 \ 10^(-3) 600^(2) )/(2 \ 5.90 \ 10^(-2) )`

F = 2.3797 10⁴ N

b) let's find the acceleration

F = ma

a = F / m

a = 2.3797 10⁴ / 7.80 10⁻³

a = 3.05 10⁶ m / s²

now we can use the kinematics relations

v = v₀ - a t

the final velocity is zero v = 0

t = v₀ / a

t = 600 / 3.05 10⁶

t = 1.97 10⁻⁴ s