Conservation of momentum and collisions | AP Physics 1 Linear momentum and collisions

This subchapter builds on the previous one and focuses on three core ideas:

Conservation of momentum
Elastic collisions
Inelastic collisions

Conservation of momentum

Momentum is conserved when you choose a system with no net external force (an isolated system). In that case, the system’s total linear momentum stays constant over time.

Consider $N$ particles with masses $m_{i}$ and velocities $v_{i}$ . The total momentum is

$P = i = 1 \sum N m_{i} v_{i} .$

Apply Newton’s second law to each particle:

$m_{i} \frac{d v _{i}}{d t} = F_{i, int} + F_{i, ext} .$

Now sum over all particles. The internal forces $F_{i, int}$ cancel in pairs because of Newton’s third law (action-reaction). That leaves only the external forces:

$\frac{d P}{d t} = i = 1 \sum N F_{i, ext} = F_{ext, net} .$

So if the net external force is zero, $F_{ext, net} = 0$ , then

$\frac{d P}{d t} = 0 ⟹ P_{initial} = P_{final} .$

Conservation of momentum before and after collision

Key Point:

$m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f} .$

Elastic collisions

An elastic collision conserves:

total momentum
total kinetic energy

In one dimension, those conservation laws are

$m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f},$

$\frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2} .$

Solving these simultaneously gives the standard final-velocity formulas:

$v_{1 f} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1 i} - \frac{2 m _{2}}{m _{1} + m _{2}} v_{2 i}, v_{2 f} = \frac{2 m _{1}}{m _{1} + m _{2}} v_{1 i} - \frac{m _{2} - m _{1}}{m _{1} + m _{2}} v_{2 i} .$

Example problem 1

A $2.0$ kg puck moving at $5.0$ m/s collides elastically with a $3.0$ kg puck at rest. Find their final speeds. Solution:

(spoiler)

$m_{1} = 2.0, v_{1 i} = 5.0, m_{2} = 3.0, v_{2 i} = 0.$

Apply the elastic-collision formulas:

$v_{1 f} = \frac{2 - 3}{2 + 3} (5.0) + \frac{2 \cdot 3}{5} (0) = - \frac{1}{5} 5.0 = - 1.0 m/s,$

$v_{2 f} = \frac{2 \cdot 2}{5} (5.0) + \frac{3 - 2}{5} (0) = \frac{4}{5} 5.0 = 4.0 m/s .$

Inelastic collisions

In an inelastic collision, momentum is conserved but kinetic energy decreases (some of it becomes thermal energy, sound, deformation, etc.).

In a perfectly inelastic collision, the objects stick together and move with a common final velocity $v_{f}$ :

$m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f} ⟹ v_{f} = \frac{m _{1} v _{1 i} + m _{2} v _{2 i}}{m _{1} + m _{2}} .$

The kinetic energy lost (difference between initial and final kinetic energy) is

$Δ K = \frac{1}{2} m_{1} v_{1 i}^{2} + \frac{1}{2} m_{2} v_{2 i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2} .$

Example problem 2

A $1.5$ kg cart moving at $3.0$ m/s collides and sticks to a $2.5$ kg cart at rest. Find their common speed and the kinetic energy lost. Solution.

(spoiler)

$m_{1} = 1.5, v_{1 i} = 3.0, m_{2} = 2.5, v_{2 i} = 0.$

Common speed:

$v_{f} = \frac{1.5 \cdot 3.0 + 2.5 \cdot 0}{1.5 + 2.5} = \frac{4.5}{4.0} = 1.125 m/s .$

Initial KE:

$K_{i} = \frac{1}{2} (1.5) (3.0)^{2} + 0 = 6.75 J .$

Final KE:

$K_{f} = \frac{1}{2} (4.0) (1.125)^{2} = 2.531 J .$

KE lost:

$Δ K = 6.75 - 2.531 = 4.219 J .$

Example problem 3

A bullet of mass $m_{b} = 0.010$ kg is fired horizontally at speed $v_{b} = 300$ m/s into a stationary block of mass $M = 1.50$ kg at the base of a rough incline of angle $θ = 3 0^{\circ}$ . The coefficient of kinetic friction between the block (with embedded bullet) and the incline is $μ_{k} = 0.15$ . After the inelastic collision, the combined mass slides up the incline and comes to rest. Determine the distance $d$ it travels up the incline.

Inelastic collision between the bullet and the block

Solution:

(spoiler)

Inelastic collision $\to$ speed just after impact. Conserve momentum of bullet + block:

$m_{b} v_{b} + M \cdot 0 = (M + m_{b}) v_{c} ⟹ v_{c} = \frac{m _{b} v _{b}}{M + m _{b}} = \frac{0.010 \times 300}{1.510} \approx 1.987 m/s .$
Energy loss in collision. Initial KE of bullet:

$K_{i} = \frac{1}{2} m_{b} v_{b}^{2} = \frac{1}{2} (0.010) (300)^{2} = 450 J .$

KE of combined mass just after collision:

$K_{c} = \frac{1}{2} (M + m_{b}) v_{c}^{2} = \frac{1}{2} (1.510) (1.987)^{2} \approx 2.985 J .$

KE dissipated:

$Δ K = K_{i} - K_{c} \approx 450 - 2.985 = 447.015 J .$
Work against gravity and friction up the incline. The kinetic energy just after the collision is used to do work against gravity and friction until the mass comes to rest:

$K_{c} = W_{g} + W_{f},$

where

$W_{g} = (M + m_{b}) g d sin θ, W_{f} = μ_{k} (M + m_{b}) g cos θ d .$

Thus

$\frac{1}{2} (M + m_{b}) v_{c}^{2} = (M + m_{b}) g d (sin θ + μ_{k} cos θ) .$

Cancel $(M + m_{b})$ and solve for $d$ :

$d = \frac{\frac{1}{2} v _{c}^{2}}{g ( sin θ + μ _{k} cos θ )} = \frac{( 1.987 ) ^{2} /2}{9.8 ( 0.5 + 0.15 \cdot 0.866 )} \approx 0.320 m .$

Answer: $d \approx 0.320$ m up the incline.

Example problem 4

A cart of mass $m_{1}$ moves on a frictionless horizontal track with speed $v_{1 i}$ and collides elastically with a stationary cart of mass $m_{2}$ . Cart $m_{2}$ is attached to a light, inextensible string that runs horizontally to a fixed pulley A (mounted at the same height as the string attachment point on $m_{2}$ ), then up to a higher fixed pulley B, and finally straight down to a hanging mass $m_{h}$ (which hangs above the track). Assume both pulleys are frictionless and the string is massless; the pulleys only redirect the string and provide no mechanical advantage. Assume the collision occurs so quickly that the string and $m_{h}$ do not affect the collision itself. After the collision, the string is taut; as $m_{2}$ moves to the right, $m_{h}$ moves upward.

Carts colliding and mass moving up as per given scenario

Given for numerical evaluation (use SI units): $m_{1} = 1.2, m_{2} = 1.8, m_{h} = 0.5, v_{1 i} = 2.5, g = 9.8$ . Answer the following:

(a) Using conservation of momentum and kinetic energy for an elastic head-on collision, determine $v_{1 f}$ and $v_{2 f}$ , the velocities of $m_{1}$ and $m_{2}$ immediately after the collision.

(b) Immediately after the collision, is the speed of the hanging mass equal to the speed of cart $m_{2}$ ? Justify briefly, and write an expression for the total kinetic energy of the moving pair ( $m_{2}$ and $m_{h}$ ) just after the collision.

(c) As the system evolves, $m_{h}$ rises and the pair comes momentarily to rest. Using energy, derive an expression for the maximum rise $h_{m a x}$ of $m_{h}$ in terms of $m_{2}, m_{h}, v_{2 f}$ , and $g$ .

(d) Calculate numerical values for $v_{1 f}, v_{2 f}$ , and $h_{m a x}$ using the given data.

(e) Describe the subsequent motion of $m_{1}$ after the collision and explain why. Solutions:

(spoiler)

(a) Post-collision speeds $v_{1 f}, v_{2 f}$ .

During the brief collision, the hanging mass and string have no effect, so the system is only $m_{1} + m_{2}$ . We apply conservation of linear momentum and conservation of kinetic energy.

Momentum conservation:

$m_{1} v_{1 i} + m_{2} (0) = m_{1} v_{1 f} + m_{2} v_{2 f} .$

Kinetic energy conservation (elastic collision):

$\frac{1}{2} m_{1} v_{1 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2} .$

From momentum:

$m_{1} v_{1 i} = m_{1} v_{1 f} + m_{2} v_{2 f} .$

Divide the energy equation by $\frac{1}{2}$ :

$m_{1} v_{1 i}^{2} = m_{1} v_{1 f}^{2} + m_{2} v_{2 f}^{2} .$

These two simultaneous equations can be solved algebraically. A useful result of solving them is the relative velocity relation for elastic collisions:

$v_{1 i} - 0 = - (v_{1 f} - v_{2 f}),$

or equivalently

$v_{1 i} = v_{2 f} - v_{1 f} .$

Now solve with $m_{1} = 1.2, m_{2} = 1.8, v_{1 i} = 2.5$ :

$v_{1 f} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1 i} = - 0.50 m/s,$

$v_{2 f} = \frac{2 m _{1}}{m _{1} + m _{2}} v_{1 i} = 2.0 m/s .$

$v_{1 f} = - 0.50 m/s, v_{2 f} = + 2.0 m/s$

(b) Speed matching and total KE just after the collision.

Justification. The string is massless and inextensible; the pulleys are ideal and only redirect the string. There is no mechanical advantage, so the instantaneous speed of the string is the same everywhere. Therefore, immediately after the collision the hanging mass speed equals the cart’s speed on the string:

$v_{h} = v_{2 f} .$

Total KE of the moving pair:

$K_{after} = \frac{1}{2} m_{2} v_{2 f}^{2} + \frac{1}{2} m_{h} v_{2 f}^{2} = \frac{1}{2} (m_{2} + m_{h}) v_{2 f}^{2} .$

(c) Maximum rise $h_{m a x}$ of $m_{h}$ (energy).

From just after the collision to the top of the rise, the pair’s kinetic energy converts into the gravitational potential of $m_{h}$ (both come momentarily to rest at the top):

$\frac{1}{2} (m_{2} + m_{h}) v_{2 f}^{2} = m_{h} g h_{m a x} ⟹ h_{m a x} = \frac{( m _{2} + m _{h} ) v _{2 f}^{2}}{2 m _{h} g} .$

Note: This result relies on $v_{h} = v_{2 f}$ because pulleys only change direction.

(d) Numerical results.

Using $m_{1} = 1.2, m_{2} = 1.8, m_{h} = 0.5, v_{1 i} = 2.5, g = 9.8$ :

$v_{1 f} = - 0.50 m/s, v_{2 f} = + 2.0 m/s,$

$h_{m a x} = \frac{( 1.8 + 0.5 ) ( 2.0 ) ^{2}}{2 ( 0.5 ) ( 9.8 )} = \frac{2.3 \cdot 4}{9.8} \approx 0.94 m .$

$h_{m a x} \approx 0.94 m .$

(e) Motion of $m_{1}$ after the collision.

After the instantaneous, elastic collision, $m_{1}$ has velocity $- 0.50 m/s$ (to the left). The track is frictionless and no horizontal forces act on $m_{1}$ thereafter. By Newton’s first law, $m_{1}$ continues with constant velocity to the left and plays no further role in the string-pulley-mass motion.

Example problem 5

A block $m_{1} = 1.0$ kg starts from rest at the top of a rough incline of angle $θ = 3 0^{\circ}$ and height $h = 2.0$ m. The coefficient of kinetic friction is $μ_{k} = 0.20$ . At the bottom, it collides inelastically and sticks to a second block $m_{2} = 1.5$ kg initially at rest on a horizontal surface attached to a spring of constant $k = 200$ N/m. Find the maximum spring compression $x_{m a x}$ .

Solution:

(spoiler)

(a) Speed at bottom of incline: Work-energy (including friction):

$m_{1} g h - f_{k} d = \frac{1}{2} m_{1} v_{1}^{2}, f_{k} = μ_{k} m_{1} g cos θ, d = \frac{h}{sin θ} .$

Compute:

$f_{k} = 0.20 \times 1.0 \times 9.8 \times cos 3 0^{\circ} \approx 1.695 N, d = \frac{2.0}{0.5} = 4.0 m .$

$m_{1} g h - f_{k} d = (1.0) (9.8) (2.0) - (1.695) (4.0) = 19.6 - 6.78 = 12.82 J,$

$v_{1} = \frac{2 \times 12.82}{1.0} \approx 5.065 m/s .$

(b) Inelastic collision: $m_{1}$ sticks to $m_{2}$ , so

$m_{1} v_{1} + m_{2} \cdot 0 = (m_{1} + m_{2}) v_{c} ⟹ v_{c} = \frac{1.0 \cdot 5.065}{2.5} \approx 2.026 m/s .$

$\frac{1}{2} (m_{1} + m_{2}) v_{c}^{2} = \frac{1}{2} k x_{m a x}^{2} ⟹ x_{m a x} = v_{c} \frac{m _{1} + m _{2}}{k} = 2.026 \frac{2.5}{200} \approx 2.026 \times 0.1118 \approx 0.227 m .$

This subchapter builds on the previous one and focuses on three core ideas:

Conservation of momentum
Elastic collisions
Inelastic collisions