A 50 g bullet travelling with an initial velocity of 400 m/s penetrates into a wall with an average force of 4*10^4 N.It goes out of the wall with a velocity of 50 m/s. What is the thickness of the wall
Initial ke = (1/2)(.05)(400)^2
Final Ke = (1/2)(.05) (50)^2
so work done = (1/2)(.05)( 400^2-50^2)
= force * distane = 40,000 T
solve for T
To find the thickness of the wall, we can use the principle of work-energy conservation. The work done by the average force on the bullet is equal to the change in kinetic energy of the bullet. Since the bullet slows down from 400 m/s to 50 m/s, its change in kinetic energy is given by:
ΔKE = (1/2) * m * (vf^2 - vi^2)
Where:
ΔKE is the change in kinetic energy
m is the mass of the bullet (given as 50 g, which is 0.05 kg)
vi is the initial velocity (given as 400 m/s)
vf is the final velocity (given as 50 m/s)
Now, let's calculate the change in kinetic energy:
ΔKE = (1/2) * 0.05 * (50^2 - 400^2)
= (1/2) * 0.05 * (2500 - 160000)
= (1/2) * 0.05 * (-157500)
= -3937.5 Joules
Since work is equal to force multiplied by distance, we can write:
Work = Force * Distance
Given that the work done by the average force is 4*10^4 N and the bullet exits the wall, we know that the work done by the force is equal to the negative change in kinetic energy of the bullet. Therefore:
-3937.5 Joules = (4*10^4 N) * Distance
Rearranging the equation to solve for distance, we get:
Distance = -3937.5 Joules / (4*10^4 N)
Distance = -0.0984375 meters
Since distance cannot be negative, we take the magnitude of the distance:
Distance = 0.0984375 meters
Therefore, the thickness of the wall is approximately 0.098 meters, or 9.8 cm.