Question

the velocity of a bullet is reduced by 50% after entering 223 cm deep inside of a wooden block. what is the total distance the bullet penetrates the block before coming to stop? assume constant acceleration.

Answer 1

Calculate the value using the formula: Total distance= 2×((0.5V₀)² - V₀²)/2a

Perform the calculations to find the total distance the bullet penetrates the block before stopping.

To determine the total distance the bullet penetrates the block before stopping:

Use the concept of constant acceleration to solve this problem.

Let the initial velocity of the bullet be V₀, and it reduces by 50% upon penetrating the wooden block.

The final velocity of the bullet inside the block is Vf = 0.5 * V₀.

Utilize the equation of motion for uniformly accelerated motion: v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Rearrange the equation to solve for s:( v² - u² )/ 2a

As the final velocity inside the block (Vf) is 0.5 times the initial velocity (V₀), the acceleration remains constant.

Substitute the given values into the equation to solve for the distance (s):

s = ((0.5V₀)² - V₀²)/2a

The depth of penetration is given as 137 cm, so the total distance is twice the depth:

Total distance=2×137.

Calculate the value using the formula:

Total distance= 2×((0.5V₀)² - V₀²)/2a

Perform the calculations to find the total distance the bullet penetrates the block before stopping.

Complete Question:

The velocity of a bullet is reduced by 50% after entering 137 cm deep inside of a wooden block. What is the total distance the bullet penetrates the block before coming to stop? Assume constant acceleration. Give your answer in cm to three significant figures without units

Answer 2

The total distance the bullet penetrates the block before coming to a stop is approximately
`\(137.6 \, \text{cm}\)` to three significant figures.

Given that the velocity of the bullet is reduced by 50% after entering
`$137 \, \text{cm}$` inside the wooden block, let's denote the initial velocity of the bullet as
`$v_0$` and the final velocity as
`$0.5v_0$` (since it's reduced by 50%).

We know the equation for uniformly accelerated motion:

`\[v_f^2 = v_0^2 + 2a d\]`

Where:

-
`\(v_f\)` is the final velocity (which is 0.5v₀),

- v₀ is the initial velocity,

- a is the acceleration (which is constant),

- d is the distance traveled.

Given that the bullet comes to a stop
`($v_f = 0$),` we can solve for the initial velocity
`(\(v_0\))` and the distance traveled (d).

`\[0 = v_0^2 + 2ad\]`

`\[0 = (0.5v_0)^2 + 2ad\]`

`\[0 = 0.25v_0^2 + 2ad\]`

Also, we know that the bullet loses half its velocity after traveling
`\(137 \, \text{cm}\), so:`

`\[0.5v_0 = √(v_0^2 - 2ad)\]`

`\[0.5v_0 = √(0.25v_0^2 - 2ad)\]`

`\[0.25v_0^2 = 0.25v_0^2 - 2ad\]`

`\[0 = -2ad\]`

`\[d = (0.25v_0^2)/(2a)\]`

From the information provided, we have
`\(137 \, \text{cm}\) for \(d\)`. So,

`\[137 = (0.25v_0^2)/(2a)\]`

To find d, we need the values of
`\(v_0\) and \(a\)`. We know that the velocity reduces by 50% inside the block, so:

`\[0.5v_0 = v_0 - a * 137\]`

Given that the velocity decreases by 50%:

`\[v_0 - a * 137 = 0.5v_0\]`

`\[a * 137 = 0.5v_0\]`

`\[a = (0.5v_0)/(137)\]`

Substitute a into the equation for d:

`\[137 = (0.25v_0^2 * 137)/(1v_0)\]`

`\[137v_0 = 0.25v_0^2 * 137\]`

`\[v_0 = 0.25v_0^2\]`

`\[v_0 = 4 \, \text{cm/s}\]`

Now that we have
`\(v_0\), we can find \(a\)`:

`\[a = (0.5v_0)/(137)\]`

`\[a = (0.5 * 4)/(137)\]`

`\[a \approx 0.0146 \, \text{cm/s}^2\]`

Finally, substitute
`\(v_0\) and \(a\) into the equation for \(d\)`:

`\[d = (0.25v_0^2)/(2a)\]`

`\[d = (0.25 * 4^2)/(2 * 0.0146)\]`

`\[d \approx 137.6 \, \text{cm}\]`

Question: