Question

A 7.80-g bullet moving at 575 m/s strikes the hand of a superhero, causing the hand to move 5.50 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. (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.

Answer 1

a) 23444.3 N

b) 0.191 ms

Step-by-step explanation:

Given:

Mass of the bullet m = 7.8 g = 7.8 * 10⁻³ Kg

Initial speed u = 575 m/s

Final speed v = 0

Distance covered s = 5.50 cm = 0.055 m

(a)

According to work energy theorem

Work done W = Change in Kinetic energy

hence, force*distance = 1/2*mass*velocity²

that is, F*S = (1/2)mu²

F*0.055 = 0.5 * 7.8 * 10⁻³ * (575)^2

F = 23444.3 N

(2)

From Impulse definition

force*time = change in momentum

where momentum = mass * velocity

F*t = Change in momentum = m* u

23444.3t = 7.8*10⁻³ * 575

Time, t = 0.191 ms

Answer 2

a)40.77N

b)0.11seg

Step-by-step explanation:

The bullet has a kinetic energy:

`K=1/2*m*v^(2) = 1/2*0.0078kg*575m/s=2.2425J`

When the bullet stops this energy goes to zero, as the energy must be conserved the work done by the head of the superhero must be equal to the original energy of the bullet:

`W=K`

Considering an average constant force, the work can be calculated as:

`W=F*d=K`

Solving for F:

`F=K/d=2.2425J/0.055m=40.77N`

The acceleration(on the opposite direction of motion) on the bullet will be:

`a=F/m=40.77N/0.0078kg=5226.9 m/s^(2)`

The velocity of the bullet considering a constant acceleration can be calculate as:

`V=V_(0) - a*t`

It stops moving when V=0 so:

`V_(0) - a*t=0=>t=V_(0)/a=(575m/s)/(5226.9m/s^(2) ) =0.11seg`