A bullet is shot into a block of plastic. The bullet penetrates the block 0.75 m. The mass of the bullet is 8 g. It is traveling with a speed of 410 m/s before it hits the block.
A Use kinematic equations to find the magnitude of the acceleration on the bullet as it is penetrating the block (ignore gravity, and assume that the force on the bullet as it penetrates the block is constant).
B Use Newton's 2nd Law to find the magnitude of the force exerted on the bullet by the plastic block.
a 0.03 kg alumimum bullet traveling at 494 m/s strikes an armor plate and comes to a stop. if all its energy is converted to heat that is absorbed by the bullet, what is the bullets's temperature change in degrees
A To find the magnitude of the acceleration on the bullet as it is penetrating the block, we can use the kinematic equation:
vf^2 = vi^2 + 2ax
where
vf = final velocity of the bullet (0 m/s, since it comes to rest)
vi = initial velocity of the bullet (410 m/s)
a = acceleration of the bullet
x = displacement of the bullet (0.75 m)
Rearranging the equation, we get:
a = (vf^2 - vi^2) / (2x)
Substituting the given values, we have:
a = (0 - (410 m/s)^2) / (2 * 0.75 m)
Simplifying the equation:
a = (-168,100 m^2/s^2) / 1.5 m
a = -112,066.67 m/s^2
Therefore, the magnitude of the acceleration on the bullet as it penetrates the block is 112,066.67 m/s^2 (rounding to two decimal places).
B To find the magnitude of the force exerted on the bullet by the plastic block, we can use Newton's second law:
F = ma
where
F = force exerted on the bullet (to be determined)
m = mass of the bullet (8 g = 0.008 kg)
a = acceleration of the bullet (-112,066.67 m/s^2, as calculated in part A)
Substituting the given values, we have:
F = (0.008 kg) * (-112,066.67 m/s^2)
Simplifying the equation:
F = -896.53 N
Therefore, the magnitude of the force exerted on the bullet by the plastic block is approximately 896.53 N (rounded to two decimal places).
A) To find the magnitude of acceleration on the bullet as it is penetrating the block, we can use the kinematic equation:
v^2 = u^2 + 2as
Where v is the final velocity (which is 0 as the bullet stops inside the block), u is the initial velocity (410 m/s), a is the acceleration, and s is the distance penetrated by the bullet (0.75 m).
Rearranging the equation, we can solve for the acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - (410 m/s)^2) / (2 * 0.75 m)
Simplifying:
a = - (410^2) / (2 * 0.75)
Calculating this expression, we find:
a ≈ - 112,128.88 m/s^2
Since we are only interested in the magnitude of the acceleration, we discard the negative sign:
Magnitude of acceleration = 112,128.88 m/s^2
B) According to Newton's 2nd Law, the force exerted on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
First, we need to convert the mass of the bullet from grams to kilograms:
Mass of the bullet = 8 g = 0.008 kg
Substituting the given values, we can calculate the force:
F = 0.008 kg * 112,128.88 m/s^2
Calculating this expression, we find:
F ≈ 897 kg*m/s^2
This is equivalent to 897 Newtons.
Therefore, the magnitude of the force exerted on the bullet by the plastic block is approximately 897 Newtons.
Just switch my numbers with yours. Same problem different values
V2x= V2ox + 2axX
0=3752 + 2a(0.55)
a=127840.9091 m/s^2
F=ma
F=.010kg x 127840.9091=1278.409091 N