**Formula for Momentum:**

**Momentum = Mass x Velocity**

Here we will be using the formula above as well as the rule “Momentum before collision is equal to the momentum after collision” to solve problems involving momentum.

**Momentum Before Collision = Momentum After Collision**

**M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2**

Where:

- M1 & M2 represents mass of the two bodies colliding.
- U1 & U2 represents the velocities of the bodies before collision.
- V1 & V2 represents the velocities of the bodies after collision.

**Problem # 1: - Finding Mass (Elastic Collision)**

A car of mass 500 kg is moving at 8m/s when it hits another car. The first car stops moving after which the second car starts moving at a speed of 15m/s. What is the mass of the second car?

**Answer:**

Using the formula above we can easily solve for the mass of the second car:

Momentum Before Collision = Momentum After Collision

M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2

Where:

- M1 & M2 is the mass of the first and second car respectively.
- U1 & U2 is the velocity before collision for the first and second car respectively.
- V1 & V2 is the velocity after collision for the first and second car respectively.

Now that we have our formula all that is left is to plug in the values and solve for the mass M2.

M1 x U1 + M2 x U2 = M1 x V1 + M2 x V2

500 x 8 + M2 x 0 = 500 x 0 + M2 x 15

4000 + 0 = 0 + M2 x 15

4000 / 15 = M2

266.7kg = M2

**Problem #2 –Finding Common Velocity (Inelastic Collision)**

A bullet of mass 0.25kg travelling at 150km/s strikes a stationary block of mass 3kg. What is the common velocity of the bullet and block after collision?

**Answer:**

The first step to solving the problem is to try and find out exactly what they are asking for. The question stated that you need to find the common velocity of the bullet and the block after collision. This simply means that the objects remain together after collision and so have a common velocity. This is known as an inelastic collision, when given problems involving inelastic collisions use the formula below to solve instead of the general formula given above:

M1 x U1 + M2 x U2 = (M1 + M2) V

Where V is the common velocity.

- In this case M1 would be the mass of the bullet.
- U1 is velocity of the bullet
- M2 is mass of block
- U2 is velocity of Block (would be zero since it is stationary)

The next step would be to plug in the values for both objects into the formula and then solve:

M1 x U1 + M2 x U2 = (M1 + M2) V

0.25 x 150 + 3 x 0 = (0.25 + 3) V

37.5 + 0 = 3.25 x V

*Transpose for V*

37.5 + 0 = 3.25 x V

37.5/3.25=(

V = 11.54m/s

**Problem #3 -Finding Recoil Velocity**

A gun of mass 4 kg fires a bullet of mass 0.15kg at 250 m/s. What is the recoil velocity of the gun after?

**Answer**

Using the principle of conservation of Momentum:

Momentum Before = Momentum After

0 = M1 x V1 – M2 x V2

M2 x V2 = M1 x V1

- Where M1 is the mass of the bullet
- V1 is the velocity of the bullet
- M2 is the mass of the gun
- V2 is the recoil velocity of the gun

M2 x V2 = M1 x V1

4 x V2 = 0.15 x 250

V2 = 37.5/4

V2 = 9.38m/s

## 11 Response to Momentum Problems with Solutions

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A motor truck requires a larger force to set in motion when it is heavily loaded than when it is empty. Similarly, far more powerful brakes are needed to stop a heavy goods vehicle than a light car moving with the same speed. The heavier vehicle is said to posses a greater quantity of motion or momentum than the lighter one.

And by Newton's second law of motion

The rate of change in momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.

. A machine gun of mass 10 kg fires 20g bullets at the rate of 10 bullets per second, with the speed of 500 m/s. What is the force necessary to hold the gun in position?

. A machine gun of mass 10 kg fires 20g bullets at the rate of 10 bullets per second, with the speed of 500 m/s. What is the force necessary to hold the gun in position?

These are one of the most common mistakes that we rarely notice.

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