Question

A rifle with a barrel length of 60 cm fires a 10 g bullet with a horizontal speed of 400 m/s. The bullet strikes a block of wood and penetrates to a depth of 12 cm.
a. what resistive force (assumed to be constant) does the wood exert on the bullet?
b. How long does it take the bullet to come to rest?
c. Draw a velocity verses time graph for the bullet in the wood.

Answer 1

Please help!!

Answer 2

a) The work done by the resistive force "F" equals the bullet's change in kinetic energy;

-Fx = 0 - (1/2)mv^2

F(.12) = (1/2)(.01)(160000)

F = 6666.7 N

b) The impulse delivered to the bullet equals the bullets change in momentum;

- (6666.7)t = 0 - (.01)(400)

t = .0006 s

c) the bullets acceleration is

a = - F/m = - 666667 m/s2


please explain me this?

Answer 3

@douglaswinslowcooper

Answer 4

@agent0smith

Answer 5

@douglaswinslowcooper

Answer 6

??

Answer 7

(F)(x) = work done in stopping bullet = loss of kinetic energy

F x = (1/2) m v^2

(F) (0.12 m) = (1/2) (0.01 kg)(400 m/s)^2

F = 6667. newtons as you have.

F t = impulse = change in momentum

6667 t = (0.01 kg)(400 m/s)

t = 0.00060 s as you have.


Bullet's acceleration is (-400 m/s)/(0.0006s) = -666700 m/s^2
which you have except for rounding differences, probably.

Which part is not clear? You used a = -F/m, and I used
a = (0-v)/t , but we both assume constant force thus constant acceleration.