A .22 rifle bullet, traveling at 373m/s, strikes a block of soft wood, which it penetrates to a depth of 0.125m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.90. Assume a constant retarding force.How much time is required for the bullet to stop? and What is the magnitude of the force exerted on the bullet by the wood while the bullet is being stopped?
Magnitude of the force (F):bullet mass: 1.9g
1/2 x mv^2 = F x d
1/2 x 1.9 g x 373^2 = F x 0.125
F = 1057.4 N
Time (t):
1.9g x 373 =1057.4 x t
t = 6.7 x 10^-4 s
To determine the time required for the bullet to stop and the magnitude of the force exerted on the bullet by the wood, we can use the principles of physics. We'll use Newton's second law of motion, which states that the acceleration experienced by an object is directly proportional to the net force applied to it and inversely proportional to its mass.
1. First, let's find the acceleration of the bullet:
We can use the equation: v^2 = u^2 + 2as, where
v is the final velocity (which is 0 m/s as the bullet stops),
u is the initial velocity (373 m/s),
a is the acceleration, and
s is the distance the bullet has penetrated into the wood (0.125 m).
Rearranging the equation, we get: a = (v^2 - u^2) / (2s)
Substituting the values: a = (0 - (373)^2) / (2 * 0.125)
Therefore, a ≈ - 4229730.56 m/s²
Since the bullet is being stopped, the acceleration is negative.
2. Next, let's find the time required for the bullet to stop:
We can use the equation: v = u + at, where
v is the final velocity (0 m/s),
u is the initial velocity (373 m/s),
a is the acceleration (approximately - 4229730.56 m/s²), and
t is the time taken.
Rearranging the equation, we get: t = (v - u) / a
Substituting the values: t = (0 - 373) / (- 4229730.56)
Therefore, t ≈ 8.83 × 10^-5 seconds (rounded to 3 significant figures)
So, the time required for the bullet to stop is approximately 8.83 × 10^-5 seconds.
3. Finally, let's find the magnitude of the force exerted on the bullet:
We can use Newton's second law equation: F = ma, where
F is the force exerted,
m is the mass of the bullet (1.90 kg), and
a is the acceleration (approximately - 4229730.56 m/s²).
Substituting the values, we get: F = 1.90 kg × - 4229730.56 m/s²
Therefore, F ≈ - 8036498.05 N (rounded to 3 significant figures)
The negative sign indicates that the force exerted on the bullet is in the opposite direction of its motion. Thus, the magnitude of the force exerted on the bullet by the wood while it is being stopped is approximately 8036498.05 N.