A bullet fired into a fixed target loses half of its initial velocity in penetrating 15cm. How much further it will penetrate before coming to rest?
To find out how much further the bullet will penetrate before coming to rest, we can use the concept of deceleration and equations of motion.
Given that the bullet loses half of its initial velocity in penetrating 15cm, we can infer that the deceleration is constant (since it is not mentioned otherwise).
Let's denote:
- Initial velocity of the bullet as u
- Final velocity of the bullet as v (which is zero, as it comes to rest)
- Distance traveled before losing half its velocity as d1 (15cm)
- Distance traveled before coming to rest as d2
Using the equation v^2 = u^2 + 2ad (where v = 0, a is the constant deceleration, and d is the distance traveled), we can solve for d:
0^2 = u^2 + 2ad2
Simplifying the equation, we have:
u^2 = -2ad2
Since the initial velocity is reduced by half in d1, we can rewrite it as:
u/2 = u^2 + 2ad1
Using this relationship, we can solve for u in terms of d1:
u = 2ad1
Now we can substitute the value of u into the equation:
u^2 = -2ad2 becomes (2ad1)^2 = -2ad2
Simplifying further:
4a^2d1^2 = -2ad2
Dividing both sides by -2ad1^2, we have:
2a = -d2
Finally, substituting the value of a back into the equation, we can solve for d2:
2(-d2)d1^2 = -2ad2
-d2 * d1^2 = -d2
d2 = d1^2
Therefore, the bullet will penetrate an additional distance equal to the square of the distance it took to lose half its initial velocity (d1).
To find out how much further the bullet will penetrate before coming to rest, we can use the concept of linear deceleration. We know that the bullet loses half of its initial velocity in penetrating 15cm. Let's break down the problem step by step using physics equations.
First, let's assign some variables:
- Initial velocity (Vi) = V
- Final velocity (Vf) = 0 (since the bullet comes to rest)
- Distance covered in the first part (d1) = 15cm = 0.15m
- Distance covered in the second part (d2) = unknown (what we want to find)
Now, let's use the equation of motion for constant acceleration (deceleration, in this case) to relate the variables:
Vf^2 = Vi^2 + 2ad
Since the bullet comes to rest, Vf = 0, and we know that Vi loses half of its initial velocity, so Vi = V/2. Plugging these values into the equation, we get:
0 = (V/2)^2 + 2a(0.15)
Simplifying the equation, we have:
0 = V^2/4 + 0.3a
Next, let's solve for acceleration (a):
0.3a = -V^2/4
a = -V^2/(1.2)
Now, we have the equation to solve for the distance in the second part, d2:
0 = (V/2)^2 + 2(-V^2/(1.2))(d2)
Simplifying this equation, we get:
d2 = V^2/(4V^2/(1.2))
d2 = 1.2*(V^2/4V^2)
d2 = 1.2/4
d2 = 0.3m
Therefore, the bullet will penetrate an additional 0.3 meters before coming to rest.
Just a note: This solution assumes that the deceleration is constant throughout the bullet's penetration. In practice, deceleration might not be constant, especially considering factors like air resistance and the bullet's interaction with the target. But for the sake of the problem, we consider a simplified scenario.
the velocity is half, so the remaining energy (kinetic) is a quarter of the original
3/4 of the energy to penetrate 15 cm
it will penetrate another 5 cm