Direct impact & coalescence | CAIE 9709 Mechanics Notes

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Mechanics

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Translate the collision story into two snapshots

Direct means the motion before and after is along the same straight line.
At rest means velocity and momentum are zero.
Rebounds or reverses means the velocity sign changes.
Moves away describes a direction; it does not tell you to use a positive sign until you have chosen your axis.

Key idea

Draw the before and after pictures before writing the equation. The diagram is a sign-control device, not decoration.

If the bodies coalesce, add masses and share one velocity

Definition

What is meant by coalesce?

Model answer: To stick together during the impact and then move as one combined body.

Swipe left or right to see the whole diagram →

Coalesce means stick together. Add the masses and use one common final velocity.

m_1u_1+m_2u_2=(m_1+m_2)v

Here $u_1$ and $u_2$ are the two signed velocities before impact, and $v$ is the one common velocity after impact.

Coalescing collision lab

right is positive

Mass of A2 kgVelocity of A+4.0 m s⁻¹Mass of B3 kgVelocity of B-1.0 m s⁻¹

total p = +5.0 kg m s⁻¹

joined v = +1.00 m s⁻¹

KE lost = 15.00 J

The joined body moves to the right. Its momentum is unchanged, but its kinetic energy is lower unless both particles already had the same velocity.

Worked example

Two blocks moving towards one another

A 0.2 kg block moves right at 4 m s⁻¹. A 0.5 kg block moves left at 1 m s⁻¹. They coalesce.

p_1=0.2(4)=0.8

p_2=0.5(-1)=-0.5

p_{\text{before}}=0.8-0.5=0.3

p_{\text{after}}=0.7v

0.3=0.7v

v=\dfrac37=0.429\text{ m s}^{-1}

The joined block moves right.

Common mistake

After coalescence there are not two final velocities. There is one combined mass and one common velocity.

Interpret the answer and check the energy claim

Swipe left or right to see the whole diagram →

A negative velocity is information, not a failed answer. Translate its sign into a direction.

If you chose right as positive and obtain $v=-1.4\text{ m s}^{-1}$ , report either that velocity or a speed of 1.4 m s⁻¹ to the left. Do not delete the sign during working.

Swipe left or right to see the whole diagram →

Momentum conservation does not imply kinetic-energy conservation. In a coalescing impact, kinetic energy is lost.

E_k=\frac12mv^2

Symbols

$Eₖ$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = velocity; squaring makes the energy non-negative (m s⁻¹)

Momentum is conserved in the collision model. Kinetic energy is not automatically conserved. When bodies with different velocities coalesce, kinetic energy decreases. This is a check using the energy formula, not a second momentum law.

\text{KE lost}=\text{KE before}-\text{KE after}

Your turnIndependent check [5 marks]

A 3 kg particle moves right at 2 m s⁻¹ and a 5 kg particle moves left at 2 m s⁻¹. They collide and coalesce. Find the velocity of the combined particle, state its speed and direction, and find the kinetic energy lost in the impact.

Show worked answer

3(2)+5(-2)=8v

v=-0.5\text{ m s}^{-1}

The combined particle moves left at 0.5 m s⁻¹.

\text{first body: }\tfrac12(3)(2^2)=6\text{ J}

\text{second body: }\tfrac12(5)(2^2)=10\text{ J}

\text{KE before}=6+10=16\text{ J}

\text{KE after}=\tfrac12(8)(0.5^2)=1\text{ J}

\text{KE lost}=15\text{ J}

Answer

v=-0.5\text{ m s}^{-1}

0.5\text{ m s}^{-1}\text{ to the left}

15\text{ J lost}

Translate the collision story into two snapshots

Direct means the motion before and after is along the same straight line.
At rest means velocity and momentum are zero.
Rebounds or reverses means the velocity sign changes.
Moves away describes a direction; it does not tell you to use a positive sign until you have chosen your axis.

Key idea

Draw the before and after pictures before writing the equation. The diagram is a sign-control device, not decoration.

If the bodies coalesce, add masses and share one velocity

Definition

What is meant by coalesce?

Model answer: To stick together during the impact and then move as one combined body.

Swipe left or right to see the whole diagram →

Coalesce means stick together. Add the masses and use one common final velocity.

m_1u_1+m_2u_2=(m_1+m_2)v

Here $u_1$ and $u_2$ are the two signed velocities before impact, and $v$ is the one common velocity after impact.

Coalescing collision lab

right is positive

Mass of A2 kgVelocity of A+4.0 m s⁻¹Mass of B3 kgVelocity of B-1.0 m s⁻¹

total p = +5.0 kg m s⁻¹

joined v = +1.00 m s⁻¹

KE lost = 15.00 J

The joined body moves to the right. Its momentum is unchanged, but its kinetic energy is lower unless both particles already had the same velocity.

Worked example

Two blocks moving towards one another

A 0.2 kg block moves right at 4 m s⁻¹. A 0.5 kg block moves left at 1 m s⁻¹. They coalesce.

p_1=0.2(4)=0.8

p_2=0.5(-1)=-0.5

p_{\text{before}}=0.8-0.5=0.3

p_{\text{after}}=0.7v

0.3=0.7v

v=\dfrac37=0.429\text{ m s}^{-1}

The joined block moves right.

Common mistake

After coalescence there are not two final velocities. There is one combined mass and one common velocity.

Interpret the answer and check the energy claim

Swipe left or right to see the whole diagram →

A negative velocity is information, not a failed answer. Translate its sign into a direction.

If you chose right as positive and obtain $v=-1.4\text{ m s}^{-1}$ , report either that velocity or a speed of 1.4 m s⁻¹ to the left. Do not delete the sign during working.

Swipe left or right to see the whole diagram →

Momentum conservation does not imply kinetic-energy conservation. In a coalescing impact, kinetic energy is lost.

E_k=\frac12mv^2

Symbols

$Eₖ$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = velocity; squaring makes the energy non-negative (m s⁻¹)

\text{KE lost}=\text{KE before}-\text{KE after}

Your turnIndependent check [5 marks]

Show worked answer

3(2)+5(-2)=8v

v=-0.5\text{ m s}^{-1}

The combined particle moves left at 0.5 m s⁻¹.

\text{first body: }\tfrac12(3)(2^2)=6\text{ J}

\text{second body: }\tfrac12(5)(2^2)=10\text{ J}

\text{KE before}=6+10=16\text{ J}

\text{KE after}=\tfrac12(8)(0.5^2)=1\text{ J}

\text{KE lost}=15\text{ J}

Answer

v=-0.5\text{ m s}^{-1}

0.5\text{ m s}^{-1}\text{ to the left}

15\text{ J lost}