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## SS2: PHYSICS - 1ST TERM

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Lesson 9, Topic 3
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# Conservation of Linear Momentum

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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, M1, and M2, moving in the same direction with initial velocities U1 and U2, 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,

M1U1 + M2U2 = M1V1 + M2V2

When two bodies moving in opposite directions collide,

M1U1 + M2U2 = M1V1 + M2V2

U2 is negative because it is moving in the opposite direction.

Therefore:

M1U1 + M2(-U2) = M1V1 + M2V2

M1U1 – M2U2 = M1V1 + M2V2

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

M1U1 + M2U2 = M1V1 + M2V2

Remember that U2 = 0 m/s (the body is at rest / stationary)
Then we will get

M1U1 + M2(0) = M1V1 + M2V2

M1U1 + 0 = M1V1 + M2V2

M1U1 = M1V1 + M2V2

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

M1U1 + M2U2 = (M1 + M2)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

M1 = M2 = 0.5 kg
U1 = 10 ms-1, U2 = 0 ms-1
V1 = V2 = V

using the formula;

M1U1 + M2U2 = M1V1 + M2V2

M1U1 + M2(0) = M1V1 + M2V2

but V1 = V2 = V

M1U1 = (M1 + M2)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

M1 = 10 kg, M2 = 15 kg
U1 = 3 m/s, U2 = -4 m/s (opposite direction)
Common velocity, V = ?

using the formula;

M1U1 – M2U2 = M1V1 + M2V2

M1U1 – M2U2 = (M1 + M2)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

M1 = 8 kg, M2 = ?
U1 = 30 m/s, U2 = -15 m/s (opposite direction)
Common Velocity V = 7.5 m/s

using the formula;

M1U1 – M2U2 = M1V1 + M2V2

M1U1 – M2U2 = (M1 + M2)V

(8 × 30) – (M2 × 15) = (8 + M2) × 7.5

240 – 15M2 = 60M2

240 = 75M2

M2 = $$\frac{240}{75}$$

M2 = 3.2kg

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