Kinetic Energy

Step 1: Notice that it is a kinematics problem

It may be hard to see it, but to solve this problem, you’re going to have to realize that we need the kinematic equations of Chapter 3. So which one are we going to use?

To figure out the right equation to use, we’ll look at our givens.

$\overrightarrow{a}=3\text{m}/{\text{s}}^2$​
${\overrightarrow{v}}_i=20\text{m}/\text{s}$​
$\overrightarrow{d}=8\text{m}$

Since we’re concerned about the car’s final kinetic energy, we will need the car’s final velocity. Therefore, we’ll need the kinematic equation:

${\left({\overrightarrow{v}}_f\right)}^2={\left({\overrightarrow{v}}_i\right)}^2+2\overrightarrow{a}\left(\mathrm{\Delta }x\right)$

We have all our values, so let’s solve for our final velocity,

v_f²=(20 m/s)²+2(3 m/s²)(8 m)=448m²/s², so v_f=21.17 m/s.

Step 2: Solve for the kinetic energy of the car

Now that we have the car’s final velocity, we can solve for the car’s final kinetic energy using the kinetic energy equation:

$KE=\frac{1}{2}m{\left(\overrightarrow{v}\right)}^2$

where we know that the mass of the car is $300\text{kg}$ and the velocity of the car after traveling $8\text{m}$ is $21.17\text{m}/\text{s}$. Therefore,

$KE=\frac{1}{2}300\text{kg}\cdot {\left(21.17\text{m}/\text{s}\right)}^2$

$KE=67.2\text{kJ}$

Step 1: Realize that this is an inelastic collision

If you don’t recognize this, then we cannot solve this problem. This is a conservation of linear momentum problem since it’s only traveling in one dimension.

The key phrase to look for in the problem is that the apple and bullet accelerate together after collision, meaning that they travel as one. This means that a completely inelastic collision has occurred.

Step 2: Solve for the velocity of the golden apple + bullet

Now that we know that this is a completely inelastic collision, we can solve for the velocity of the whole apple + bullet.

If you remember from previous chapter, the equation we will use is:

${\overrightarrow{V}}_f=\frac{m_1}{m_1+m_2}{\overrightarrow{v}}_{\mathrm{1,}i}$

We know that:

$m_1=10\text{g}$​
$m_2=50\text{g}$​
${\overrightarrow{v}}_{\mathrm{1,}i}=500\text{m}/\text{s}$ ​
${\overrightarrow{V}}_f=\frac{0.01\text{kg}}{0.01\text{kg}+0.05\text{kg}}500\text{m}/\text{s}$​
${\overrightarrow{V}}_f=83.33\text{m}/\text{s}$

Step 3: Solve for the kinetic energy of the system

Simple enough, just use the kinetic energy equation except that this time the mass is the combined mass of the apple and bullet.

$KE=\frac{1}{2}\left(0.01\text{kg}+0.05\text{kg}\right){\left(83.33\text{m}/\text{s}\right)}^2$

$KE=208.3\text{J}$

Step 4: Solve for the distance the golden apple + bullet Travel

In order to solve for the distance, we must realize that this is a kinematics approach. At this point, let’s use our kinematic equation: $\mathrm{\Delta }x={\overrightarrow{v}}_{i,x}\mathrm{\Delta }t+\frac{1}{2}{\overrightarrow{a}}_x{\left(\mathrm{\Delta }t\right)}^2$​
Since this is all in one direction, we don’t need to worry about y-direction movement, so we can assume this is all in the x-direction.

Let’s see what our givens are:

${\overrightarrow{a}}_x=-8\text{m}/{\text{s}}^2$

$\mathrm{\Delta }t=5\text{s}$

And we also have to use the velocity that we solved for in step 2 because we’re concerned with how far the apple moves after the bullet has collided with it. In that case:

${\overrightarrow{v}}_{i,x}=83.33\text{m}/\text{s}$

Therefore, we can plug in our values in:

Δx=(83.333 m/s)(5 s)+½(−8 m/s²)(5 s)².

Δx=316.667 m. The speed after 5 s is 43.333 m/s, so the object has not yet reversed.

We would expect it to travel far since the bullet is traveling so fast. Can you imagine if the object hit wasn’t an apple? What if it was something much heavier?

Section breakdown

1. What is Kinetic Energy

2. Example 5.2.1 - The (slowly accelerating) Race Car

3. Example 5.2.2 - The Golden Apple gets Shot