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# M1 V1 M2 V2 Calculator + Online Solver With Free Steps

The **M1 V1 M2 V2 Calculator **uses the law of conservation of momentum to solve for an unknown quantity in the equation of momentum conservation. In the case of multiple unknown quantities (variables), the calculator finds expressions for each unknown in terms of the other unknowns.

## What Is the M1 V1 M2 V2 Calculator?

**The M1 V1 M2 V2 Calculator is an online tool that solves for an unknown quantity in the momentum conservation equation using the values provided for the other variables. If the user provides multiple unknowns, it finds an expression for each unknown in terms of the others.**

The **calculator interface** consists of 6 text boxes. From top to bottom, they take:

- $m_1$: Mass of the first body in
**kg**. - $m_2$: Mass of the second body in
**kg**. - $\boldsymbol{u_1}$: Initial velocity of the first body in
**m/s**. - $\boldsymbol{u_2}$: Initial velocity of the second body in
**m/s**. - $\boldsymbol{v_1}$: Final velocity of the first body in
**m/s**. - $\boldsymbol{v_2}$: Final velocity of the second body in
**m/s**.

The unit of each quantity is right next to the text box. Currently, only metric SI units are supported.

## How To Use the M1 V1 M2 V2 Calculator?

You can use the **M1 V1 M2 V2 Calculator **to find the value of an unknown variable such as the mass or velocity of an object in a collision between two objects by entering the values of the other parameters (mass and initial and final velocities). See the step-by-step guidelines below for help.

### Step 1

Check which quantity is unknown. In the corresponding quantity’s text box, enter a character commonly used for unknowns like x, y, z, etc. Otherwise, enter the value for that quantity.

### Step 2

Enter the mass of the two bodies in the first two text boxes. These must be in **kg**.

### Step 3

Enter the initial velocities (pre-collision) in the third ($\boldsymbol u_1$) and fourth ($\boldsymbol u_2$) text boxes. These must be in **m/s**.

### Step 4

Enter the final velocities (post-collision) in the fifth ($\boldsymbol v_1$) and sixth ($\boldsymbol v_2$) text boxes. These must also be in **m/s**.

### Step 5

Press the **Submit **button to get the results.

### Results

The results show as an extension of the calculator interface. They include two sections: the first contains the input in LaTeX format for manual verification while the second shows the solution (value of the unknown quantity).

## How Does the M1 V1 M2 V2 Calculator Work?

The **M1 V1 M2 V2 Calculator** works by solving the following equation for the unknowns:

\[ m_1 \boldsymbol{u_1} + m_2 \boldsymbol{u_2} = m_1 \boldsymbol{v_1} + m_2 \boldsymbol{v_2} \tag*{(1)} \]

### Momentum

Momentum is defined as the product of mass m and velocity **v**:

momentum = **p** = m**v**

Generally speaking, the greater the value of momentum, the larger the time it takes to bring the body to rest. You may observe that a car moving at a fast speed will always stop faster than a truck moving at the same or even a lesser speed.

### Law of Conservation of Momentum

The law of conservation of momentum is a fundamental principle of physics and states that in an isolated system, the total momentum of two bodies before and after a collision remains the same. It builds on the law of conservation of energy, which states that energy can neither be created nor destroyed. It implies that energy only transfers between different forms.

#### Isolated Systems

The law of conservation of momentum applies to isolated systems, in which objects do not interact with their surroundings and ONLY with each other. An example of such a system is two balls on a boundless frictionless plane. Momentum in such systems, like energy, is conserved as there are no energy losses due to friction, etc.

That is not to say that momentum conservation does not occur in practice – only that in systems with external forces and factors, momentum is not wholly conserved depending on the strength of the factors in play.

In an isolated system, an object moving with a constant velocity keeps moving at that velocity infinitely. Therefore, the only possibility of change is upon a collision with another object.

#### Physical Scenario of Momentum Conservation

Consider two balls rolling along a line in the same direction such that the one in the lead is slower than the one behind it. Eventually, the ball at the back will crash into the back of the one at the front. The velocity and momentum of the balls change after this collision.

Let the mass of the balls be $m_1$ and $m_2$. Suppose the initial velocities of the balls were $\boldsymbol{u_1}$ and $\boldsymbol{u_2}$, and the final velocities after the collision are $\boldsymbol{v_1}$ and $\boldsymbol{v_2}$ respectively.

Let $\boldsymbol{p_1}$ and $\boldsymbol{p_2}$ be the momentum of the first and second ball before the collision, and $\boldsymbol{p_1’}$ and $\boldsymbol{p_2’}$ be the momentum of the two after the collision. Then, the law of momentum conservation states that:

total momentum before the collision = total momentum after the collision

\[ \boldsymbol{p_1} + \boldsymbol{p_2} = \boldsymbol{p_1’} + \boldsymbol{p_2’} \]

\[ m_1 \boldsymbol{u_1} + m_2 \boldsymbol{u_2} = m_1 \boldsymbol{v_1} + m_2 \boldsymbol{v_2} \]

Which is the equation (1). Clearly, if any one of $m_1$, $m_2$, $\boldsymbol{u_1}$, $\boldsymbol{u_2}$, $\boldsymbol{v_1}$, and $\boldsymbol{v_2}$ is unknown, we can find it out using equation (1).

**Solved Examples**

### Example 1

Imagine a car with a mass of 1000 kg moving at a velocity of 20.8333 m/s on the highway. It crashes into the back of a jeep with a mass of 1500 kg moving at a velocity of 15 m/s. After the collision, the jeep is now moving at a velocity of 18 m/s. Assuming an isolated system, what is the velocity of the car post-collision?

### Solution

Let $m_1$ = 1000 kg, $m_2$ = 1500 kg, $\boldsymbol{u_1}$ = 20.8333 m/s, $\boldsymbol{u_2}$ = 15.0 m/s, $\boldsymbol{v_1}$ = y, and $\boldsymbol{v_2}$ = 18 m/s. Using equation (1), we get:

1000(20.8333) + 1500(15.0) = 1000(y) + 1500(18)

20833 + 22500 = 1000y + 27000

43333 = 1000y + 27000

Rearranging to isolate y:

y = 16333 / 1000 = **16.333 m/s**