Momentum. Conservation of momentum

Momentum of a body, a closed system, the law of conservation of momentum for motion along a straight line, an inelastic collision, jet propulsion.

Terms

Studied 0 of 4

Formulas

Momentum of a body

p = mv

Quantities
Symbol	Name	Unit
$p$	momentum of the body	kg·m/s
$m$	mass of the body	kg
$v$	velocity of the body	m/s

Solved for another quantity:

m = \dfrac{p}{v}

v = \dfrac{p}{m}

Note. The velocity $v$ is taken with a sign: plus if the body moves along the axis, minus if against it. For example, a $2\ \text{kg}$ body at $3\ \text{m/s}$ : $p = 6\ \text{kg}\cdot\text{m/s}$ .

Conservation of momentum (two bodies)

m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

Quantities
Symbol	Name	Unit
$m_1,\ m_2$	masses of the bodies	kg
$v_1,\ v_2$	velocities of the bodies before the interaction	m/s
$v_1',\ v_2'$	velocities of the bodies after the interaction	m/s

Solved for another quantity:

v_1' = \dfrac{m_1 v_1 + m_2 v_2 - m_2 v_2'}{m_1}

Note. Written for a closed system of two bodies moving along one straight line; velocities are taken with signs. For example, a $3\ \text{kg}$ cart at $4\ \text{m/s}$ catches up with a resting $1\ \text{kg}$ cart and afterwards moves at $2\ \text{m/s}$ : $3\cdot 4 + 0 = 3\cdot 2 + 1\cdot v_2'$ , so $v_2' = 6\ \text{m/s}$ .

Inelastic collision (bodies stick together)

m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v'

Quantities
Symbol	Name	Unit
$m_1,\ m_2$	masses of the bodies	kg
$v_1,\ v_2$	velocities of the bodies before the collision	m/s
$v'$	common velocity of the bodies after the collision	m/s

Solved for another quantity:

v' = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}

m_2 = \dfrac{m_1 (v_1 - v')}{v' - v_2}

Note. After an inelastic collision the bodies move together with one velocity $v'$ . Velocities are taken with signs. For example, a $3\ \text{kg}$ cart at $4\ \text{m/s}$ hits a resting $1\ \text{kg}$ cart: $v' = \dfrac{3\cdot 4 + 1\cdot 0}{3 + 1} = 3\ \text{m/s}$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A collision of two carts (inelastic)

Friction is ignored, the system is closed. The carts collide at t = 2 s and then move together. The positive direction of the axis is to the right. The edges of the track correspond to ±10 m.

Time0 s

Graphs

p(t)

p₁ is blue, p₂ is yellow, p = p₁ + p₂ is purple

v'

3 m/s

p_1

12 kg·m/s

p_2

0 kg·m/s

p

12 kg·m/s

Formulas with your numbers

p = m_1 v_1 + m_2 v_2 = 3\cdot 4 + 1\cdot 0 = 12\ \text{kg·m/s}

v' = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \dfrac{12}{4} = 3\ \text{m/s}

How to read the graphs

On the $p(t)$ graph each cart's momentum is a step: it is constant before the collision and jumps at the moment of impact (the carts act on each other).
The total momentum $p = p_1 + p_2$ is a horizontal line: the system is closed, so the sum does not change. What one cart loses, the other gains.
A negative momentum means the body moves against the chosen axis direction (to the left). The sign of the momentum is the sign of the velocity.
After the collision both carts move with the same velocity, so their momenta are in the ratio of their masses: $p_1 / p_2 = m_1 / m_2$ .

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Terms

Studied 0 of 4

Formulas

Momentum of a body

p = mv

Quantities
Symbol	Name	Unit
$p$	momentum of the body	kg·m/s
$m$	mass of the body	kg
$v$	velocity of the body	m/s

Solved for another quantity:

m = \dfrac{p}{v}

v = \dfrac{p}{m}

Conservation of momentum (two bodies)

m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

Quantities
Symbol	Name	Unit
$m_1,\ m_2$	masses of the bodies	kg
$v_1,\ v_2$	velocities of the bodies before the interaction	m/s
$v_1',\ v_2'$	velocities of the bodies after the interaction	m/s

Solved for another quantity:

v_1' = \dfrac{m_1 v_1 + m_2 v_2 - m_2 v_2'}{m_1}

Inelastic collision (bodies stick together)

m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v'

Quantities
Symbol	Name	Unit
$m_1,\ m_2$	masses of the bodies	kg
$v_1,\ v_2$	velocities of the bodies before the collision	m/s
$v'$	common velocity of the bodies after the collision	m/s

Solved for another quantity:

v' = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}

m_2 = \dfrac{m_1 (v_1 - v')}{v' - v_2}

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A collision of two carts (inelastic)

Friction is ignored, the system is closed. The carts collide at t = 2 s and then move together. The positive direction of the axis is to the right. The edges of the track correspond to ±10 m.

Time0 s

Graphs

p(t)

p₁ is blue, p₂ is yellow, p = p₁ + p₂ is purple

v'

3 m/s

p_1

12 kg·m/s

p_2

0 kg·m/s

p

12 kg·m/s

Formulas with your numbers

p = m_1 v_1 + m_2 v_2 = 3\cdot 4 + 1\cdot 0 = 12\ \text{kg·m/s}

v' = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \dfrac{12}{4} = 3\ \text{m/s}

How to read the graphs

On the

p(t)

graph each cart's momentum is a step: it is constant before the collision and jumps at the moment of impact (the carts act on each other).

The total momentum

p = p_1 + p_2

is a horizontal line: the system is closed, so the sum does not change. What one cart loses, the other gains.

A negative momentum means the body moves against the chosen axis direction (to the left). The sign of the momentum is the sign of the velocity.

After the collision both carts move with the same velocity, so their momenta are in the ratio of their masses:

p_1 / p_2 = m_1 / m_2