The law of conservation of linear momentum states that for a closed or isolated system of colliding bodies the total momentum remains constant, provided no external forces act.

OR

If two or more bodies in a closed or isolated system collide, their total momentum before collision is equal to their total momentum after collision provided no external forces act.

**A Closed / Isolated System:**

A closed or isolated system means a system in which no external force acts. Or it is a system in which there is no interference by any external factor.

### Different Cases Of Collision:

If masses, M_{1}, and M_{2}, moving in the same direction with initial velocities U_{1} and U_{2}, respectively, collide with each other and move in different directions

From conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision; therefore,

M_{1}U_{1 }+ M_{2}U_{2} = M_{1}V_{1 }+ M_{2}V_{2}

When two bodies moving in opposite directions collide,

M_{1}U_{1 }+ M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

U_{2} is negative because it is moving in the opposite direction.

Therefore:

M_{1}U_{1 }+ M_{2}(-U_{2}) = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 }– M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

When a stationary body collides with a moving body, and move apart.

M_{1}U_{1 }+ M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

Remember that U_{2} = 0 m/s (the body is at rest / stationary)

Then we will get

M_{1}U_{1 }+ M_{2}(0) = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 }+ 0 = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 } = M_{1}V_{1} + M_{2}V_{2}

If two bodies collide with each other, and stick together, and move with common velocity.

M_{1}U_{1 }+ M_{2}U_{2} = (M_{1} + M_{2})V

V is the common velocity of the body.

### Example 1

A ball of mass 0.5 kg moving at 10 ms^{-1} collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

**Solution**

M_{1} = M_{2} = 0.5 kg

U_{1} = 10 ms^{-1}, U_{2} = 0 ms^{-1}

V_{1} = V_{2} = V

**using the formula;**

M_{1}U_{1 }+ M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 }+ M_{2}(0) = M_{1}V_{1} + M_{2}V_{2}

but V_{1} = V_{2} = V

M_{1}U_{1} = (M_{1} + M_{2})V

**Substitute the values given in question**

0.5(10) = (0.5 + 0.5)V

0.5 × 10 = 1 × V

∴ V = 5 m/s

### Example 2

An object A of mass 10 kg moving with a velocity of 3 m/s collides with an object B of mass 15 kg moving with a velocity of 4 m/s in opposite direction. If A and B stick together after collision. Calculate their common velocity?

**Solution**

M_{1} = 10 kg, M_{2} = 15 kg

U_{1} = 3 m/s, U_{2} = -4 m/s (opposite direction)

Common velocity, V = ?

**using the formula;**

M_{1}U_{1 }– M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 }– M_{2}U_{2} = (M_{1} + M_{2})V

(10 × 3) – (15 × 4) = (10 + 15)V

30 – 60 = 25V

-30 = 25V

V = \( \frac{-30}{25} \)

V = -1.2 m/s

The negative sign implies that the two objects collide and move in the same direction as B.

### Example 3

A body P of mass 8 kg moving with a velocity of 30 m/s collides with another body Q moving with velocity of 15m/s in the opposite direction. If both bodies now move in the direction of P at a velocity of 7.5 m/s, calculate the mass of Q

**Solution**

M_{1} = 8 kg, M_{2} = ?

U_{1} = 30 m/s, U_{2} = -15 m/s (opposite direction)

Common Velocity V = 7.5 m/s

**using the formula;**

M_{1}U_{1 }– M_{2}U_{2} = M_{1}V_{1} + M_{2}V_{2}

M_{1}U_{1 }– M_{2}U_{2} = (M_{1} + M_{2})V

(8 × 30) – (M_{2} × 15) = (8 + M_{2}) × 7.5

240 – 15M_{2} = 60M_{2}

240 = 75M_{2}

M_{2} = \( \frac{240}{75} \)

M_{2} = 3.2kg

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