Question

A 0.02kg bullet is traveling with a speed of 300m/s when it hits a wall. Assuming it comes to rest after traveling 0.1m with constant deceleration into the wall.

1. Calculate the work done by the wall on the bullet as the bullet comes to a stop.
2. Calculate the deceleration of the bullet as it comes to a stop.

Answer 1

1. W = 900 J

2. a = -450000 m/s²

Step-by-step explanation:

2.

First, we calculate the deceleration of the bullet by using 3rd equation of motion:

`2as = V_(f)^(2) - V_(i)^(2)\\`

where,

a = deceleration = ?

s = distance traveled = 0.1 m

Vf = Final Speed = 0 m/s (it eventually stops)

Vi = Initial Speed = 300 m/s

Therefore,

`2a(0.1\ m) = (0\ m/s)^(2) - (300\ m/s)^(2)\\a = (-90000\ m^(2)/s^(2))/(0.2\ m)\\\\`

a = -450000 m/s² (negative sign shows deceleration)

1.

Now, we calculate the force using Newton's Second Law:

`F = ma\\F = (0.02\ kg)(-450000\ m/s^(2))\\F = - 9000\ N`

Negative sign shows decelerating force.

For work done:

`Work\ Done = W = Fs\\W = (9000 N)(0.1\ m)`

W = 900 J