Momentum in Explosions (College Board AP® Physics 1: Algebra-Based)

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Written by: Katie M

Reviewed by: Caroline Carroll

Momentum in explosions

The conservation of momentum principle can be used to analyze explosions
An explosion is defined as

An interaction in which forces internal to the system move objects within that system apart

Some examples of explosions include:
- the recoil of a gun after shooting a bullet
- the radioactive emission of a particle from an unstable nucleus

Explosions in one dimension

Explosions in one dimension (1D) usually involve one stationary object becoming two objects that move away from each other along the same line
- Before the explosion, the velocity, and therefore momentum, is zero
- After the collision, the velocity vectors are directed along the same axis and can be either positive or negative
When solving problems involving explosions in one dimension:
- identify the objects included in the system, making sure there are no net external forces
- write an expression for the momentum of each object after the explosion
- write an expression for the total momentum of the system before and after the explosion
- apply the conservation of momentum principle
- solve for the unknown quantity

An explosion in one dimension where a single object AB becomes two objects A and B moving along the same line in opposite directions. Before the explosion, object AB is at rest. After the explosion, A and B move away from each other with velocities vA, and vB respectively. — **During an explosion in one dimension, an internal force causes a single object, AB, to become two objects, A and B, which move along the same line in opposite directions**

Applying the conservation of momentum principle (defining the positive direction to the right):

$0 = m_{B} {\vec{v}}_{B} - m_{A} {\vec{v}}_{A}$

Since the velocity of each object is directed along the same axis, the vector notation can be removed

$\frac{m_{A}}{m_{B}} = \frac{v_{B}}{v_{A}}$

Explosions in two dimensions

Explosions in two dimensions (2D) usually involve one stationary object becoming two or more objects that move away from each other in different directions
- Before the explosion, the velocity, and therefore momentum, is zero
- After the explosion, the velocity vectors may have both horizontal and vertical components
When solving problems involving explosions in two dimensions:
- identify the objects included in the system, making sure there are no net external forces
- write expressions for the x- and y- components of the momentum of each object after the explosion
- write expressions for the total momentum of the system in each direction before and after the explosion
- apply the conservation of momentum principle to each direction separately
- solve for the unknown quantity using Pythagoras' theorem

An explosion in two dimensions where single object ABC becomes three objects A, B and C which move apart in different directions. Before the explosion, object ABC is at rest. After the explosion, object A moves to the left at an angle θA above the horizontal with velocity vA, object B moves to the right at an angle θB above the horizontal with velocity vB, and object C moves vertically down with velocity vC. — **After an explosion of object ABC in two dimensions, three objects, A, B, and C, move off in different directions**

If no external forces act on a system, horizontal and vertical components of momentum are conserved
Applying the conservation of momentum principle:

$0 = m_{A} {\vec{v}}_{A} + m_{B} {\vec{v}}_{B} + m_{C} {\vec{v}}_{C}$

The components of the velocities in the x-direction are:

$0 = m_{B} v_{B} \cos θ_{B} - m_{A} v_{A} \cos θ_{A}$

The components of the velocities in the y-direction are:

$0 = m_{A} v_{A} \sin θ_{A} + m_{B} v_{B} \sin θ_{B} - m_{C} v_{C}$

For an explosion in two dimensions where a single object ABC becomes three objects A, B and C, the velocities must be resolved into x- and y- components.
Object A moves to the left at an angle θA above the horizontal with velocity vA, so it has a velocity of vAcosθA in the x-direction and vAsinθA in the y-direction.
Object B moves to the right at an angle θB above the horizontal with velocity vB, so it has a velocity of vBcosθB in the x-direction and vBsinθB in the y-direction.
Object C moves vertically down with velocity vC, so it has a vertical velocity component only. — **To solve problems involving explosions in two dimensions, the velocity vectors must be resolved into their horizontal and vertical components**

Worked Example

A stationary uranium nucleus undergoes fission and splits into two fragments of unequal mass and two neutrons, where $m_{1} > m_{2} ≫ m_{n}$ , as shown in the diagram.

Fission of a uranium nucleus. Before fission, the uranium nucleus is stationary. After fission, a heavier nucleus of mass m1 moves with velocity v1 in the opposite direction to a lighter nucleus with mass m2 and velocity v2. At right angles to the nuclei, two neutrons of mass mn move in opposite directions with velocities vn1 and vn2 respectively

Which of the following is true for the velocities of the fragments?

A $\frac{v_{1}}{v_{2}} > 1$

B $\frac{v_{1}}{v_{2}} < 1$

C $v_{2} < v_{n 1} < v_{n 2}$

D $v_{1} > v_{n 1} = v_{n 2}$

The correct answer is B

Answer:

Step 1: Analyze the scenario

The uranium nucleus is initially at rest, so the total momentum before the fission is zero
The total momentum of the system must be conserved
So, the final total momentum of the four fragments must also be zero
The fragments move along the same line, and the neutrons move along the same line, so, the individual components of their momenta are equal in magnitude and opposite in direction

Step 2: Eliminate incorrect options

For momentum to be conserved
- the heavier the fragment, the lower the velocity
- the lighter the fragment, the greater the velocity

Applying conservation of momentum in the direction of the neutrons:

$m_{n} v_{n 1} = - m_{n} v_{n 2}$

$v_{n 1} = - v_{n 2}$

Deducing the component of the momenta in the direction of the neutrons. The positive direction is defined to the right. Neutrons have equal mass so they must have equal and opposite velocities.

The neutrons have equal mass, hence, they must have equal velocities
- This eliminates option C
The neutrons are much lighter than the two fragments, so they must have higher velocities
- This eliminates option D

Step 3: Deduce the correct option

Applying conservation of momentum in the direction of the fragments:

$m_{1} v_{1} = - m_{2} v_{2}$

$\frac{v_{1}}{v_{2}} = - \frac{m_{2}}{m_{1}}$

Deducing the component of the momenta in the direction of the fragments. The positive direction is defined upwards. The fragments have unequal masses so the heavier mass must have a smaller velocity.

Since $m_{1} > m_{2}$ , the ratio of the masses will be $\frac{m_{2}}{m_{1}} > 1$ , therefore:

$\frac{v_{1}}{v_{2}} > - 1$

Which is equivalent to:

$\frac{v_{1}}{v_{2}} < 1$

Therefore, option B is correct

Last updated: 17 September 2024

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