Mechanics

1D Kinematics

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Learn position, velocity, and acceleration through graph slopes and a stop-in-zones challenge.

2D Kinematics

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Extend kinematics into vector components, projectile motion, and two-dimensional acceleration.

Collisions

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Compare elastic and inelastic collisions while keeping momentum and energy bookkeeping straight.

Collisions

A collision is a short interaction during which objects push strongly on one another and change their motion. The details of the contact can be messy, but the bookkeeping is not: in an isolated system, total momentum is conserved. The big question is what happens to the kinetic energy.

Momentum

Momentum quantifies how difficult it is to stop or redirect motion. For a single object,

p⃗=mv⃗.\vec{p} = m\vec{v}.

For a system of objects, we add the momenta of every part:

p⃗total=∑imiv⃗i.\vec{p}_{\text{total}} = \sum_i m_i \vec{v}_i.

If external forces are negligible during the collision, then the system is isolated and

p⃗before=p⃗after.\vec{p}_{\text{before}} = \vec{p}_{\text{after}}.

During collisions, objects may bounce apart, deform, or stick together. Although individual momenta can change dramatically during impact, the vector sum of the system’s total momentum stays fixed throughout.

Energy

Kinetic energy is the energy of motion:

K=12mv2.K = \frac{1}{2}mv^2.

Total energy is always conserved, but kinetic energy is not always conserved by itself. During a collision, some of the original kinetic energy can become

  • thermal energy from friction and internal vibration.
  • sound produced during impact.
  • permanent deformation of the objects.

So when we classify collisions, we are really asking whether the kinetic energy remains kinetic after the interaction.

Elastic Collisions

An elastic collision conserves both total momentum and total kinetic energy:

p⃗before=p⃗after,Kbefore=Kafter.\vec{p}_{\text{before}} = \vec{p}_{\text{after}}, \qquad K_{\text{before}} = K_{\text{after}}.

These collisions are good models for idealized billiard balls, gas particles, and many microscopic scattering processes. In an elastic collision, the objects separate after the collision, and no kinetic energy is “lost” to heat or deformation.

Inelastic Collisions

An inelastic collision still conserves momentum, but it does not conserve kinetic energy:

p⃗before=p⃗after,Kafter<Kbefore.\vec{p}_{\text{before}} = \vec{p}_{\text{after}}, \qquad K_{\text{after}} < K_{\text{before}}.

The “missing” kinetic energy has not vanished. It has been transformed into other forms of energy inside or around the colliding objects. A special case is the perfectly inelastic collision, where the objects stick together and move with one shared final velocity. In one dimension,

m1v1,i+m2v2,i=(m1+m2)vf.m_1 v_{1,i} + m_2 v_{2,i} = (m_1 + m_2) v_f.

1D Collisions

Explore the differences between elastic and inelastic collisions.

Collision Guide

Track what is conserved and what changes form.

In an isolated system, momentum is conserved in every collision. The key distinction is whether kinetic energy remains kinetic or is redistributed into other forms of energy.

What to Track

Elastic

Momentum and kinetic energy are both conserved.

Inelastic

Momentum is conserved, but kinetic energy is not.

Momentum

Total momentum stays constant because the interaction forces are internal to the system.
Total momentum still stays constant, even when the objects crumple, stick, or warm up.

Kinetic Energy

Kinetic energy before and after the collision is the same.
Some kinetic energy is transformed into sound, thermal energy, or deformation.

After the Impact

Objects separate with new speeds that still satisfy both conservation laws.
Objects may separate more slowly, or in the perfectly inelastic case move together.

Always True

Total momentum: pbefore = pafter

As long as outside forces are negligible, internal pushes and pulls cancel in pairs.

Perfectly Inelastic Limit

m1v1,i + m2v2,i = (m1 + m2)vf

The objects share a common final speed, which tells you the lost kinetic energy was converted into other forms.

Problem Solving

General 1D Elastic Collision

Problem

A 2 kg2\,\text{kg} cart moving at 5 m/s5\,\text{m/s} collides elastically with a 1 kg1\,\text{kg} cart moving at −1 m/s-1\,\text{m/s}. Find the two final velocities.

Given

  • m1=2 kgm_1 = 2\,\text{kg}
  • v1,i=5 m/sv_{1,i} = 5\,\text{m/s}
  • m2=1 kgm_2 = 1\,\text{kg}
  • v2,i=−1 m/sv_{2,i} = -1\,\text{m/s}
  • the collision is elastic

Use

Momentum is conserved:

m1v1,i+m2v2,i=m1v1,f+m2v2,f.m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f}.

For a one-dimensional elastic collision, the relative speed of approach equals the relative speed of separation:

v1,i−v2,i=v2,f−v1,f.v_{1,i} - v_{2,i} = v_{2,f} - v_{1,f}.

Solve

Substitute the known values into momentum conservation:

(2)(5)+(1)(−1)=(2)v1,f+(1)v2,f⇒9=2v1,f+v2,f.(2)(5) + (1)(-1) = (2)v_{1,f} + (1)v_{2,f} \qquad \Rightarrow \qquad 9 = 2v_{1,f} + v_{2,f}.

Now use the elastic-collision relation:

5−(−1)=v2,f−v1,f⇒6=v2,f−v1,f.5 - (-1) = v_{2,f} - v_{1,f} \qquad \Rightarrow \qquad 6 = v_{2,f} - v_{1,f}.

Substitute v2,f=v1,f+6v_{2,f} = v_{1,f} + 6 into the momentum equation:

9=2v1,f+(v1,f+6)⇒3=3v1,f⇒v1,f=1 m/s.9 = 2v_{1,f} + (v_{1,f} + 6) \qquad \Rightarrow \qquad 3 = 3v_{1,f} \qquad \Rightarrow \qquad v_{1,f} = 1\,\text{m/s}.

Then

v2,f=7 m/s.v_{2,f} = 7\,\text{m/s}.

Check

The initial kinetic energy is

Ki=12(2)(52)+12(1)(12)=25.5 J.K_i = \frac{1}{2}(2)(5^2) + \frac{1}{2}(1)(1^2) = 25.5\,\text{J}.

The final kinetic energy is

Kf=12(2)(12)+12(1)(72)=25.5 J.K_f = \frac{1}{2}(2)(1^2) + \frac{1}{2}(1)(7^2) = 25.5\,\text{J}.

Both momentum and kinetic energy are conserved, as expected for an elastic collision. Because the masses are different and both carts are already moving, the final speeds are not just swapped; they must satisfy both equations at the same time.

Perfectly Inelastic Collision

Problem

A 2 kg2\,\text{kg} cart moving at 3 m/s3\,\text{m/s} collides with a 1 kg1\,\text{kg} cart at rest. The carts stick together after the collision. Find the shared final speed vfv_f.

Given

  • m1=2 kgm_1 = 2\,\text{kg}
  • v1,i=3 m/sv_{1,i} = 3\,\text{m/s}
  • m2=1 kgm_2 = 1\,\text{kg}
  • v2,i=0v_{2,i} = 0
  • perfectly inelastic collision, so both carts share one final speed vfv_f

Use

For a perfectly inelastic collision in one dimension,

m1v1,i+m2v2,i=(m1+m2)vf.m_1 v_{1,i} + m_2 v_{2,i} = (m_1 + m_2) v_f.

Solve

Substitute the known values:

(2)(3)+(1)(0)=(2+1)vf.(2)(3) + (1)(0) = (2 + 1) v_f.

So

6=3vf⇒vf=2 m/s.6 = 3v_f \qquad \Rightarrow \qquad v_f = 2\,\text{m/s}.

Check

The initial kinetic energy is

Ki=12(2)(32)+12(1)(02)=9 J.K_i = \frac{1}{2}(2)(3^2) + \frac{1}{2}(1)(0^2) = 9\,\text{J}.

The final kinetic energy is

Kf=12(2+1)(22)=6 J.K_f = \frac{1}{2}(2 + 1)(2^2) = 6\,\text{J}.

The kinetic energy changes by

ΔK=Kf−Ki=6−9=−3 J.\Delta K = K_f - K_i = 6 - 9 = -3\,\text{J}.

That missing 3 J3\,\text{J} was transformed into other forms of energy such as sound, thermal energy, or deformation. The total momentum still remained conserved throughout the collision.

2D Perfectly Inelastic Collision

Problem

A 0.20 kg0.20\,\text{kg} puck moves east at 4.0 m/s4.0\,\text{m/s} and collides with a 0.30 kg0.30\,\text{kg} puck moving north at 2.0 m/s2.0\,\text{m/s}. The pucks stick together. Find the final velocity vector, including its speed and direction.

Given

  • m1=0.20 kgm_1 = 0.20\,\text{kg}, moving east at 4.0 m/s4.0\,\text{m/s}
  • m2=0.30 kgm_2 = 0.30\,\text{kg}, moving north at 2.0 m/s2.0\,\text{m/s}
  • perfectly inelastic collision, so the pucks move together afterward

Use

In two dimensions, momentum is conserved separately in each component:

px,before=px,after,py,before=py,after.p_{x,\text{before}} = p_{x,\text{after}}, \qquad p_{y,\text{before}} = p_{y,\text{after}}.

If the objects stick together, then

(m1+m2)vf,x=px,before,(m1+m2)vf,y=py,before.(m_1 + m_2)v_{f,x} = p_{x,\text{before}}, \qquad (m_1 + m_2)v_{f,y} = p_{y,\text{before}}.

Solve

First find the momentum components before the collision:

px,before=(0.20)(4.0)=0.80 kg⋅m/s,p_{x,\text{before}} = (0.20)(4.0) = 0.80\,\text{kg}\cdot\text{m/s},py,before=(0.30)(2.0)=0.60 kg⋅m/s.p_{y,\text{before}} = (0.30)(2.0) = 0.60\,\text{kg}\cdot\text{m/s}.

The combined mass is

m1+m2=0.50 kg.m_1 + m_2 = 0.50\,\text{kg}.

So the final velocity components are

vf,x=0.800.50=1.6 m/s,vf,y=0.600.50=1.2 m/s.v_{f,x} = \frac{0.80}{0.50} = 1.6\,\text{m/s}, \qquad v_{f,y} = \frac{0.60}{0.50} = 1.2\,\text{m/s}.

Now find the speed:

∣v⃗f∣=vf,x2+vf,y2=1.62+1.22=2.0 m/s.|\vec{v}_f| = \sqrt{v_{f,x}^2 + v_{f,y}^2} = \sqrt{1.6^2 + 1.2^2} = 2.0\,\text{m/s}.

The direction is

θ=tan⁡−1(1.21.6)≈36.9∘\theta = \tan^{-1}\left(\frac{1.2}{1.6}\right) \approx 36.9^\circ

north of east.

Check

The final velocity points northeast, which makes sense because one puck initially carries eastward momentum and the other carries northward momentum. The speed is smaller than either incoming component speed because the collision is perfectly inelastic and the two pucks share one combined motion after impact.

Collisions Checkpoint

Question 1 of 3

Two carts collide and stick together. Which statement must still be true if outside forces are negligible during the collision?

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