momentum and impulse

Momentum

Momentum is a measure of the motion of an object and is the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

p⃗=mv⃗[kg⋅ms] \vec {p} = m \vec {v} \quad [kg \cdot \frac {m} {s}]

Where p⃗ \vec {p} is momentum, m m is mass, and v⃗ \vec {v} is velocity. The units of momentum are kg⋅ms kg \cdot \frac {m} {s} .

Impulse

Impulse is the change in momentum of an object when a force is applied over a time interval. It is also a vector quantity and is equal to the product of the net force and the time during which it acts.

Δp⃗=F⃗netΔt[kg⋅ms] \Delta \vec {p} = \vec {F}_{ \text {net} } \Delta t \quad [kg \cdot \frac {m} {s}]

Where Δp⃗ \Delta \vec {p} is the change in momentum, F⃗net \vec {F}_{ \text {net} } is the net force, and Δt \Delta t is the time interval during which the force acts.

Example

The impulse of a car crash is 24.0N⋅s 24.0 N \cdot s . If the maximum force a person can safely handle is 8.40N 8.40N , what is the minimum amount of time for the collision to be safe?

Answer

Δp=FΔt ⇒ Δt=ΔpF=24.0N⋅s8.40N=2.86s \Delta p = F \Delta t \ \Rightarrow \ \Delta t = \frac { \Delta p } { F } = \frac { 24.0 N \cdot s } { 8.40N } = 2.86s


∴ \therefore The collision must take a minimum of 2.86s 2.86s in order for it to be safe.

Law of Conservation of Momentum

The law of conservation of momentum states that the total momentum of a closed system is conserved if no external forces are acting on it. This means the total momentum before any interaction is equal to the total momentum after the interaction.

∑p⃗i=∑p⃗f \sum \vec {p}_{ \text {i} } = \sum \vec {p}_{ \text {f} }

This principle is crucial in analyzing collisions and interactions in isolated systems.

Collisions

There are three types of collisions that we will look at:

  1. Inelastic collisions

  2. Elastic collisions

  3. Perfectly elastics collisions

Inelastic Collisions

In an inelastic collision, objects collide and stick together. The total momentum of the system is conserved, and the velocities of both objects become the same.

m1v⃗1+m2v⃗2=(m1+m2)v⃗f m_1 \vec {v}_1 + m_2 \vec {v}_2 = (m_1 + m_2) \vec {v}_{ \text {f} }

Where m1 m_1 and m2 m_2 are the masses of the colliding objects, v⃗1 \vec{v}_1 and v⃗2 \vec{v}_2 are their velocities before the collision, and v⃗final \vec{v}_{\text{final}} is their common velocity after the collision.

Elastic Collisions

Elastic collisions involve both objects bouncing off of each other.

m1v⃗1+m2v⃗2=m1v⃗1′+m2v⃗2′ m_1 \vec {v}_1 + m_2 \vec {v}_2 = m_1 \vec{v}_1' + m_2 \vec{v}_2'

Where v⃗1′ \vec{v}_1' and v⃗2′ \vec{v}_2' are the velocities of the objects after the collision.

Perfectly Elastic Collisions

In a perfectly elastic collision, both momentum and kinetic energy are conserved. The colliding objects rebound without any loss of kinetic energy.

You would still use the equation we just learned for elastic collisions:

m1v⃗1+m2v⃗2=m1v⃗1′+m2v⃗2′ m_1 \vec {v}_1 + m_2 \vec {v}_2 = m_1 \vec{v}_1' + m_2 \vec{v}_2'

However, since the kinetic energy is conserved, you can also use:

12m1v12+12m2v22=12m1v1′2+12m2v2′2 \frac {1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2

Where the kinetic energy before and after the collision remains the same.

Example

A 1.00×103kg 1.00 \times 10^3 kg car travelling east at a velocity of 25ms 25 \frac {m} {s} [east] collides with another 1.5×103kg 1.5 \times 10^3 kg travelling 20ms 20 \frac {m} {s} [west]. When they collide, they stick together. What is the velocity of both of the cars after the collision?

Answer

Since they stick together after the collision, this is an inelastic collision:

   m1v⃗1+m2v⃗2=(m1+m2)v⃗f \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, m_1 \vec {v}_1 + m_2 \vec {v}_2 = (m_1 + m_2) \vec {v}_f

(1.00×103kg)(25ms)+(1.5×103kg)(−20ms)=(1.00×103kg+1.50×103kg)v⃗f (1.00 \times 10^3 kg)(25 \frac {m} {s}) + (1.5 \times 10^3 kg)(-20 \frac {m} {s}) = (1.00 \times 10^3 kg + 1.50 \times 10^3 kg) \vec {v}_f

  v⃗f=25.0×103kg⋅ms − 30×103kg⋅ms2.50×103kg \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \vec {v}_f = \frac { 25.0 \times 10^3 kg \cdot \frac {m} {s} \ - \ 30 \times 10^3 kg \cdot \frac {m} {s} } { 2.50 \times 10^3 kg }

  v⃗f=−2.00ms \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \vec {v}_f = -2.00 \frac {m} {s}


∴ \therefore Both cars will travel 2.00ms 2.00 \frac {m} {s} [west] after the collision.

Example

A 0.40kg 0.40kg ball is travelling at a velocity of 5.0ms 5.0 \frac {m} {s} east and collides with a 0.45kg 0.45kg ball that is moving at a velocity of 6.0ms 6.0 \frac {m} {s} west. If the 0.40kg 0.40kg ball travels at a velocity of 3.0ms 3.0 \frac {m} {s} west after the collision, what is the velocity of the 0.45kg 0.45kg ball?

Answer

This looks like an elastic collision, so let's determine the final velocity of the 0.45kg 0.45kg ball:

   m1v⃗1+m2v⃗2=m1v⃗1′+m2v⃗2′ \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \, m_1 \vec {v}_1 + m_2 \vec {v}_2 = m_1 \vec {v}_1' + m_2 \vec {v}_2'

(0.40kg)(5.0ms)+(0.45kg)(−6.0ms)=(0.40kg)(−3.0ms)+(0.45kg)v⃗2′ (0.40kg)(5.0 \frac {m} {s}) + (0.45kg)(-6.0 \frac {m} {s}) = (0.40kg)(-3.0 \frac {m} {s}) + (0.45kg) \vec {v}_2'

 v⃗2′=2.0kg⋅ms−2.7kg⋅ms+1.2kg⋅ms0.45kg \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \vec {v}_2' = \frac { 2.0 kg \cdot \frac {m} {s} - 2.7kg \cdot \frac {m} {s} + 1.2 kg \cdot \frac {m} {s} } { 0.45kg }

 v⃗2′=1.1ms \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \vec {v}_2' = 1.1 \frac {m} {s}


∴ \therefore The 0.45kg 0.45kg ball travels 1.1ms 1.1 \frac {m} {s} east after the collision.

Example

On a pool table, one small 0.500kg 0.500kg ball is going 2.50ms 2.50 \frac {m} {s} east and collides with another bigger 1.00kg 1.00kg ball going 4.00ms 4.00 \frac {m} {s} west. If the collision is super elastic, what are the velocities of the balls after the collision?

Answer

Since this is a perfectly elastic collision, we know that both momentum and kinetic energy is conserved, so we can use both equations:

m1v⃗1+m2v⃗2=m1v⃗1′+m2v⃗2′ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad m_1 \vec {v}_1 + m_2 \vec {v}_2 = m_1 \vec {v}_1' + m_2 \vec {v}_2'

(0.500kg)(2.50ms)+(1.00kg)(−4.00ms)=(0.500kg)v⃗1′+(1.00kg)v⃗2′ (0.500kg)(2.50 \frac {m} {s}) + (1.00kg)(-4.00 \frac {m} {s}) = (0.500kg) \vec {v}_1' + (1.00kg) \vec {v}_2'

   v⃗2′=2.75ms−0.500v⃗1′ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \vec {v}_2' = 2.75 \frac {m} {s} - 0.500 \vec {v}_1'


Now we can use the kinetic energy of the system to form another equation:

   12m1v⃗12+12m2v⃗22=12m1v⃗1′2+12m2v⃗2′2 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, \frac {1} {2} m_1 \vec {v}_1^2 + \frac {1} {2} m_2 \vec {v}_2^2 = \frac {1} {2} m_1 \vec {v}_1'^2 + \frac {1} {2} m_2 \vec {v}_2'^2

    m1v⃗12+m2v⃗22=m1v⃗1′2+m2v⃗2′2 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, \, m_1 \vec {v}_1^2 + m_2 \vec {v}_2^2 = m_1 \vec {v}_1'^2 + m_2 \vec {v}_2'^2

(0.500kg)(2.50ms)2+(1.00kg)(−4.00ms)2=(0.500kg)v⃗1′2+(1.00kg)v⃗2′2 (0.500kg)(2.50 \frac {m} {s})^2 + (1.00kg)(-4.00 \frac {m} {s})^2 = (0.500kg) \vec {v}_1'^2 + (1.00kg) \vec {v}_2'^2

  19.125kgm2s2=(0.500kg)v⃗1′2+(1.00kg)v⃗2′2 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, 19.125 kg \frac {m^2} {s^2} = (0.500kg) \vec {v}_1'^2 + (1.00kg) \vec {v}_2'^2


Plug in the first equation into the second:

19.125kgm2s2=(0.500kg)v⃗1′2+(1.00kg)(2.75ms−0.500v⃗1′)2 19.125 kg \frac {m^2} {s^2} = (0.500kg) \vec {v}_1'^2 + (1.00kg)(2.75 \frac {m} {s} - 0.500 \vec {v}_1')^2

  0=(0.500kg)v⃗1′2+7.563kgm2s2−(2.75kgms)v⃗1′−(0.250kg)v⃗1′2−19.125kgm2s2 \quad \quad \quad \quad \ \ 0 = (0.500kg) \vec {v}_1'^2 + 7.563 kg \frac {m^2} {s^2} - (2.75 kg \frac {m} {s}) \vec {v}_1' - (0.250 kg) \vec {v}_1'^2 - 19.125 kg \frac {m^2} {s^2}

  0=(0.250kg)v⃗1′2−(2.75kgms)v⃗1′−11.562kgm2s2 \quad \quad \quad \quad \ \ 0 = (0.250kg) \vec {v}_1'^2 - (2.75 kg \frac {m} {s}) \vec {v}_1' - 11.562 kg \frac {m^2} {s^2}

v⃗1′=14.25ms \vec {v}_1' = 14.25 \frac {m} {s} or v⃗1′=−3.25ms \vec {v}_1' = -3.25 \frac {m} {s}

Since the balls bounce off of each other and go in opposite directions after the collision, v⃗1′=−3.25ms \vec {v}_1' = -3.25 \frac {m} {s} .


Now, let's find v⃗2′ \vec {v}_2' :

v⃗2′=2.75ms−0.500v⃗1′=2.75ms−0.500(−3.25)=4.38ms \vec {v}_2' = 2.75 \frac {m} {s} - 0.500 \vec {v}_1' = 2.75 \frac {m} {s} - 0.500(-3.25) = 4.38 \frac {m} {s}


∴ \therefore The final velocities of the balls are −3.25ms -3.25 \frac {m} {s} for the 0.500kg 0.500kg ball and 4.38ms 4.38 \frac {m} {s} for the 1.00kg 1.00kg ball.

Two Dimensional Collisions

In two-dimensional collisions, the principles of conservation of momentum apply in both the x and y directions. This means the total momentum in each direction is conserved.

∑p⃗x,initial=∑p⃗x,final \sum \vec{p}_{x, \text{initial}} = \sum \vec{p}_{x, \text{final}}

And,

∑p⃗y,initial=∑p⃗y,final \sum \vec{p}_{y, \text{initial}} = \sum \vec{p}_{y, \text{final}}

The analysis of such collisions involves breaking down the velocities into their components and applying conservation of momentum in each direction separately just like we have done before.

Example

A 1.50kg 1.50kg ball hits a stationary 2.75kg 2.75kg ball. The collision results in the 1.50kg 1.50kg moving with a velocity of 3.1ms 3.1 \frac {m} {s} [E 35∘ 35^{\circ} N] and the 2.75kg 2.75kg ball moving with a velocity of 4.5ms 4.5 \frac {m} {s} [e 41∘ 41^{\circ} S]. What was the initial velocity of the 1.50kg 1.50kg ball?

Answer

First, let's write our given:

  • m1=1.50kg m_1 = 1.50kg

  • m2=2.75kg m_2 = 2.75kg

  • v1=? v_1 = ?

  • v2=0ms v_2 = 0 \frac {m} {s}

  • v1′=3.1ms v_1' = 3.1 \frac {m} {s} [E 35∘ 35^{\circ} N]

  • v1′=4.5ms v_1' = 4.5 \frac {m} {s} [E 40∘ 40^{\circ} S]


Now, let's calculate v1,x v_{1,x} using the conservation of momentum with the x-components:

  m1v⃗1,x+m2v⃗2,x=m1v⃗1,x′+m2v⃗2,x′ \quad \quad \quad \quad \quad \, \, m_1 \vec {v}_{1,x} + m_2 \vec {v}_{2,x} = m_1 \vec {v}_{1,x}' + m_2 \vec {v}_{2,x}'

(1.50kg)v⃗1,x+(2.75kg)(0ms)=(1.50kg)(3.1cos⁡35∘)+(2.75kg)(4.5cos⁡40∘) (1.50kg) \vec {v}_{1,x} + (2.75kg)(0 \frac {m} {s}) = (1.50kg)(3.1 \cos 35^{\circ}) + (2.75kg)(4.5 \cos 40^{\circ})

   v⃗1,x=(1.50kg)(3.1cos⁡35∘)+(2.75kg)(4.5cos⁡40∘)1.50kg \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, \vec {v}_{1,x} = \frac { (1.50kg)(3.1 \cos 35^{\circ}) + (2.75kg)(4.5 \cos 40^{\circ}) } { 1.50kg }

   v⃗1,x=8.86ms \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, \vec {v}_{1,x} = 8.86 \frac {m} {s}


Now for v1,y v_{1,y} :

  m1v⃗1,y+m2v⃗2,y=m1v⃗1,y′+m2v⃗2,y′ \quad \quad \quad \quad \quad \, \, m_1 \vec {v}_{1,y} + m_2 \vec {v}_{2,y} = m_1 \vec {v}_{1,y}' + m_2 \vec {v}_{2,y}'

(1.50kg)v⃗1,y+(2.75kg)(0ms)=(1.50kg)(3.1sin⁡35∘)+(2.75kg)(4.5sin⁡40∘) (1.50kg) \vec {v}_{1,y} + (2.75kg)(0 \frac {m} {s}) = (1.50kg)(3.1 \sin 35^{\circ}) + (2.75kg)(4.5 \sin 40^{\circ})

   v⃗1,y=(1.50kg)(3.1sin⁡35∘)+(2.75kg)(−4.5sin⁡40∘)1.50kg \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \vec {v}_{1,y} = \frac { (1.50kg)(3.1 \sin 35^{\circ}) + (2.75kg)(-4.5 \sin 40^{\circ}) } { 1.50kg }

   v⃗1,y=−3.52ms \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \vec {v}_{1,y} = -3.52 \frac {m} {s}


We have the components of v1 v_1 , let's find its magnitude:

∣v1⃗∣=(8.86ms)2+(−3.52ms)2=9.53ms | \vec {v_1} | = \sqrt { (8.86 \frac {m} {s})^2 + (-3.52 \frac {m} {s})^2 } = 9.53 \frac {m} {s}


Finally, its direction:

θ=tan⁡−1(−3.52ms8.86ms)=−21.7∘ \theta = \tan^{-1} ( \frac { -3.52 \frac {m} {s} } { 8.86 \frac {m} {s} } ) = -21.7^{\circ}


∴ \therefore The initial velocity of the 1.50kg 1.50kg ball was 9.53ms 9.53 \frac {m} {s} [E 21.7∘ 21.7^{\circ} S].