Why Explosions Increase Kinetic Energy Without Changing Momentum

A firework sitting still bursts into several fragments. A fraction of a second later, the fragments are flying in different directions, each with kinetic energy. Yet the total momentum of the fragments can still be exactly zero.

How can an explosion increase kinetic energy without changing total momentum?

The answer is that an explosion converts stored internal energy into kinetic energy, while the forces between the fragments produce equal and opposite changes in momentum. Momentum is a vector, so the fragments’ momenta can cancel. Kinetic energy is a scalar, so their kinetic energies add.

That difference is the key to understanding explosion problems in HSC Physics.

01The puzzle: moving fragments, zero total momentum

Imagine two skaters standing still on smooth ice. One pushes against the other, and they glide apart.

Before the push, neither skater is moving. The total momentum is zero.

After the push, both are moving. One travels left, the other travels right. Their momenta are equal in magnitude and opposite in direction, so the total momentum is still zero.

But both skaters now have kinetic energy.

An explosion works in a similar way. The important difference is the source of the energy: an explosion releases energy stored internally, often as chemical potential energy or elastic potential energy.

Let’s predict something before introducing the equations.

Suppose a stationary device breaks into two fragments. One fragment has twice the mass of the other. Will they move away at equal speeds?

No. The lighter fragment must move twice as fast as the heavier fragment if total momentum is to remain zero.

The fragments do not need equal speeds. They need equal and opposite momenta.

02Why momentum stays constant while kinetic energy increases

The explanation becomes much clearer when we treat momentum and kinetic energy separately.

Momentum is a vector

Momentum is defined as:

\[ \vec p=m\vec v \]

where:

• \(\vec p\) is momentum, measured in \(\mathrm{kg\,m\,s^{-1}}\).
• \(m\) is mass, measured in kilograms.
• \(\vec v\) is velocity, measured in \(\mathrm{m\,s^{-1}}\).

Momentum has direction. If an object moves right, its momentum is positive when right is chosen as the positive direction. If it moves left, its momentum is negative.

For an isolated system, total momentum is conserved:

\[ \vec p_{\text{before}}=\vec p_{\text{after}} \]

An isolated system here means that the net external impulse during the explosion is zero or negligible.

The forces between the fragments are internal forces. By Newton’s third law, these forces are equal in magnitude and opposite in direction.

Consequently, their impulses cancel when we consider the entire system.

For an explosion involving two fragments:

\[ m_A\vec v_A+m_B\vec v_B=\vec p_i \]

If the original object was stationary, its initial momentum was zero:

\[ m_A\vec v_A+m_B\vec v_B=0 \]

This means the two fragments must have equal and opposite momentum vectors.

In one dimension, taking right as positive:

\[ m_Av_A=-m_Bv_B \]

The negative sign indicates opposite directions.

Kinetic energy is a scalar

Kinetic energy is calculated using:

\[ E_k=\frac12mv^2 \]

where:

• \(E_k\) is kinetic energy, measured in joules.
• \(m\) is mass in kilograms.
• \(v\) is speed in metres per second.

Notice that velocity is squared.

This means kinetic energy does not have a positive or negative direction. An object travelling left has positive kinetic energy, just like an object travelling right.

For two fragments:

\[ E_{k,\text{total}}= \frac12m_Av_A^2+ \frac12m_Bv_B^2 \]

Both contributions are positive whenever the fragments are moving.

This is why opposite momenta can cancel while kinetic energies add.

What if the original object was stationary?

This is the simplest explosion scenario.

Before the explosion:

\[ p_i=0 \]

\[ E_{k,i}=0 \]

After the explosion, the momentum vectors still sum to zero:

\[ \vec p_f=0 \]

However, the fragments are moving, so:

\[ E_{k,f}>0 \]

Therefore:

\[ \Delta E_k=E_{k,f}-E_{k,i}>0 \]

The total kinetic energy has increased even though total momentum has not changed.

There is no violation of energy conservation. The increase in kinetic energy comes from a decrease in stored internal energy.

03Test the idea: an exploding two-cart system

A useful way to investigate this is with two dynamics carts and a compressed spring plunger.

Consider the following apparatus:

• Two carts sit on a level, low-friction horizontal track.
• Cart A is on the left, and cart B is on the right.
• A compressed spring plunger is mounted between the carts.
• A latch holds the spring compressed until it is released.
• A light gate sits near each end of the track to measure the speed of the departing cart.

The system consists of both carts and the spring plunger. The plunger’s mass is included in the mass of cart A.

When the latch is released, the spring expands. It pushes cart A to the left and cart B to the right.

What should the experiment show?

Before releasing the spring, both carts are stationary.

Therefore:

\[ p_i=0 \]

If the horizontal external impulse is negligible, the final momentum must also be zero:

\[ m_Av_A+m_Bv_B=0 \]

The speeds should be inversely proportional to the masses.

For example, if cart B has twice the mass of cart A, cart A should leave at twice the speed of cart B.

But what happens to their kinetic energy?

Both carts should gain kinetic energy. The spring’s stored elastic potential energy is converted into their motion, as well as any sound, vibration, or thermal energy produced by the release.

Measuring the results

A video of the carts moving apart would establish that they travel in opposite directions. However, it would not necessarily provide sufficiently accurate speeds to test momentum conservation.

A better design uses two photogates and flags of known length.

Suppose each cart carries a flag of length \(0.060\ \mathrm m\). The photogates measure how long each flag blocks the light beam.

The speed is calculated using:

\[ v=\frac{\text{flag length}}{\text{blocking time}} \]

For cart A, suppose the blocking time is \(0.050\ \mathrm s\):

\[ v_A=\frac{0.060}{0.050} =1.20\ \mathrm{m\,s^{-1}} \]

For cart B, suppose the blocking time is \(0.100\ \mathrm s\):

\[ v_B=\frac{0.060}{0.100} =0.60\ \mathrm{m\,s^{-1}} \]

The following measurements are illustrative data, not results from an actual experiment.

The total final momentum is:

\[ p_f=-0.60+0.60=0 \]

Momentum is conserved.

Now add the kinetic energies:

\[ E_{k,f}=0.36+0.18 \]

\[ E_{k,f}=0.54\ \mathrm J \]

Initially, both carts were stationary, so their total kinetic energy was zero.

The final kinetic energy is \(0.54\ \mathrm J\).

The experiment’s predicted behaviour is clear: total momentum remains zero, but total kinetic energy increases.

What could make the measurements inaccurate?

A real investigation has limitations.

A slightly tilted track can give the carts an external impulse. Rolling friction can also change their momentum. In addition, a photogate measures speed, so the direction must be identified from which way the cart travels.

A revised setup should therefore:

• Level the track carefully.
• Use two photogates so both speeds are measured independently.
• Measure the moving masses accurately, including attached components.
• Check that rolling friction is sufficiently small.
• Release the spring without giving either cart an additional horizontal push.

The measured total momentum might not be exactly zero because of experimental uncertainty. The important question is whether the discrepancy is small enough to be explained by the apparatus.

The experiment supports the model when the measured momenta balance within the uncertainty of the measurements.

04Why equal and opposite forces do not mean zero work

There is another subtle point worth understanding.

Newton’s third law tells us that the spring exerts equal and opposite forces on the two carts.

You might predict that these forces do equal and opposite amounts of work, meaning the total work is zero.

That prediction is incorrect.

The reason is that work depends on both force and displacement:

\[ W=Fd\cos\theta \]

where \(W\) is work, \(F\) is force, \(d\) is displacement, and \(\theta\) is the angle between them.

Consider the two carts again.

• The spring pushes cart A to the left, and cart A moves left.
• The spring pushes cart B to the right, and cart B moves right.

For each cart, the force and displacement point in the same direction.

Therefore, \(\theta=0^\circ\), and:

\[ W=Fd \]

Both carts receive positive work.

The important distinction is:

• Impulses from the internal force pair are equal and opposite, so their contributions to total momentum cancel.
• Work done by the internal forces is positive on both carts, so their kinetic energies increase.

The spring loses stored elastic potential energy as it expands.

This is the physical mechanism behind the increase in kinetic energy.

It also explains why conservation of momentum does not imply conservation of kinetic energy.

05Explosions compared with collisions

An explosion and an inelastic collision can produce opposite changes in kinetic energy while both conserve momentum.

Assuming negligible net external impulse:

In an explosion, stored internal energy is converted into kinetic energy.

In an inelastic collision, some of the original kinetic energy is converted into other forms, such as thermal energy, sound, and deformation.

In an elastic collision, total kinetic energy remains unchanged.

In all three cases, total momentum can remain constant because the internal forces do not change the system’s total momentum.

For a closer look at the reverse energy transfer, see Inelastic Collisions: Momentum and Kinetic Energy for HSC Physics.

06A reliable method for explosion problems

When an HSC question involves an explosion, use this sequence.

Step 1: Define the system.

Include all fragments. If a spring, plunger, or other internal mechanism is relevant, consider whether it belongs in your system.

Step 2: Choose a positive direction.

For a one-dimensional problem, choose right or east as positive. Give objects travelling in the opposite direction negative velocities.

For two-dimensional problems, use separate horizontal and vertical components.

Step 3: Check external impulse.

Momentum is conserved if the net external impulse is zero or negligible.

If an external impulse is significant, use:

\[ p_f=p_i+J_{\text{ext}} \]

where \(J_{\text{ext}}\) is the net external impulse.

Step 4: Apply momentum conservation.

For two fragments in one dimension:

\[ m_Au_A+m_Bu_B=m_Av_A+m_Bv_B \]

For an object initially stationary:

\[ m_Av_A+m_Bv_B=0 \]

For two-dimensional explosions, conserve momentum separately in each direction:

\[ p_{ix}=p_{fx} \]

\[ p_{iy}=p_{fy} \]

Step 5: Calculate kinetic energy separately.

Use:

\[ E_k=\frac12mv^2 \]

Add the kinetic energies of all fragments. Do not give velocities negative signs when calculating kinetic energy, because speed is squared.

Step 6: Interpret the energy change.

Calculate:

\[ \Delta E_k=E_{k,f}-E_{k,i} \]

A positive result means total translational kinetic energy has increased.

That increase must be accounted for by energy transferred from an internal store or another energy source. Do not assume that all the released internal energy becomes kinetic energy; some may become sound, heat, or deformation.

For the broader decision about when to use momentum, kinetic energy, or both, see Momentum or Kinetic Energy? A Guide to Collision Problems.

07Worked examples

08Questions and solutions

Question 1

A stationary device explodes into two fragments.

Fragment A has a mass of \(0.40\ \mathrm{kg}\) and moves left at \(6.0\ \mathrm{m\,s^{-1}}\). Fragment B has a mass of \(0.80\ \mathrm{kg}\).

Assuming negligible external impulse, calculate:

a) The velocity of fragment B.

b) The total kinetic energy of the fragments after the explosion.

Solution 1

Fragment B moves right at \(3.0\ \mathrm{m\,s^{-1}}\), and the total final kinetic energy is \(10.8\ \mathrm J\).

Part a: Find the velocity of fragment B.

Take right as positive.

The initial momentum is zero because the device was stationary.

Using momentum conservation:

\[ m_Av_A+m_Bv_B=0 \]

Substitute:

\[ (0.40)(-6.0)+(0.80)v_B=0 \]

\[ -2.4+0.80v_B=0 \]

\[ v_B=\frac{2.4}{0.80} \]

\[ v_B=+3.0\ \mathrm{m\,s^{-1}} \]

The positive sign means fragment B moves right.

Part b: Calculate the total kinetic energy.

Add the kinetic energies of both fragments:

\[ E_{k,f}= \frac12m_Av_A^2+ \frac12m_Bv_B^2 \]

Substitute:

\[ E_{k,f}= \frac12(0.40)(6.0)^2+ \frac12(0.80)(3.0)^2 \]

\[ E_{k,f}=7.2+3.6 \]

\[ \boxed{E_{k,f}=10.8\ \mathrm J} \]

The initial kinetic energy was zero, so the explosion increased the total kinetic energy by \(10.8\ \mathrm J\).

Notice that the fragments have different speeds, but their momenta are equal in magnitude and opposite in direction.

Question 2

A stationary \(1.00\ \mathrm{kg}\) device explodes into three fragments.

The fragments have the following properties:

• Fragment A has a mass of \(0.20\ \mathrm{kg}\) and travels east at \(12\ \mathrm{m\,s^{-1}}\).
• Fragment B has a mass of \(0.30\ \mathrm{kg}\) and travels north at \(8.0\ \mathrm{m\,s^{-1}}\).
• Fragment C has a mass of \(0.50\ \mathrm{kg}\).

Assuming negligible external impulse, calculate:

a) The horizontal and vertical components of fragment C’s velocity.

b) Its speed and direction.

c) The total kinetic energy after the explosion.

Take east as positive \(x\), and north as positive \(y\).

Solution 2

Fragment C travels southwest with velocity components of \(-4.8\ \mathrm{m\,s^{-1}}\) horizontally and \(-4.8\ \mathrm{m\,s^{-1}}\) vertically. Its speed is approximately \(6.8\ \mathrm{m\,s^{-1}}\), and the total kinetic energy is \(35.52\ \mathrm J\).

Part a: Find the velocity components of fragment C.

The device was initially stationary, so:

\[ p_{ix}=0 \]

\[ p_{iy}=0 \]

Momentum must be conserved independently in each direction.

For the horizontal direction:

\[ m_Av_{Ax}+m_Cv_{Cx}=0 \]

Substitute:

\[ (0.20)(12)+(0.50)v_{Cx}=0 \]

\[ 2.4+0.50v_{Cx}=0 \]

\[ v_{Cx}=-4.8\ \mathrm{m\,s^{-1}} \]

For the vertical direction:

\[ m_Bv_{By}+m_Cv_{Cy}=0 \]

Substitute:

\[ (0.30)(8.0)+(0.50)v_{Cy}=0 \]

\[ 2.4+0.50v_{Cy}=0 \]

\[ v_{Cy}=-4.8\ \mathrm{m\,s^{-1}} \]

Both components are negative, so fragment C travels southwest.

Part b: Find the speed and direction.

The speed is the magnitude of the velocity vector:

\[ v_C=\sqrt{v_{Cx}^2+v_{Cy}^2} \]

Substitute:

\[ v_C=\sqrt{(-4.8)^2+(-4.8)^2} \]

\[ v_C=\sqrt{46.08} \]

\[ v_C\approx6.79\ \mathrm{m\,s^{-1}} \]

The two components have equal magnitudes, so the direction is:

\[ 45^\circ\text{ south of west} \]

Part c: Calculate the total kinetic energy.

Calculate each fragment’s kinetic energy separately.

Fragment A:

\[ E_{k,A}=\frac12(0.20)(12)^2 \]

\[ E_{k,A}=14.4\ \mathrm J \]

Fragment B:

\[ E_{k,B}=\frac12(0.30)(8.0)^2 \]

\[ E_{k,B}=9.6\ \mathrm J \]

Fragment C:

\[ E_{k,C}=\frac12(0.50)(6.79)^2 \]

\[ E_{k,C}\approx11.52\ \mathrm J \]

Add the three contributions:

\[ E_{k,f}=14.4+9.6+11.52 \]

\[ \boxed{E_{k,f}=35.52\ \mathrm J} \]

The initial kinetic energy was zero, so the explosion increased the total kinetic energy by \(35.52\ \mathrm J\).

A common mistake in this question is to combine the momentum magnitudes as though they were ordinary positive numbers. Momentum must be resolved into components because direction matters.

Question 3

A \(3.0\ \mathrm{kg}\) device travels east at \(4.0\ \mathrm{m\,s^{-1}}\) before exploding into two fragments.

Fragment A has a mass of \(1.0\ \mathrm{kg}\) and travels west at \(2.0\ \mathrm{m\,s^{-1}}\) after the explosion.

Fragment B has a mass of \(2.0\ \mathrm{kg}\).

Assuming negligible external impulse, calculate:

a) The final velocity of fragment B.

b) The initial kinetic energy of the device.

c) The final total kinetic energy of the fragments.

d) The increase in total kinetic energy.

Solution 3

Fragment B moves east at \(7.0\ \mathrm{m\,s^{-1}}\), and the explosion increases total kinetic energy by \(27\ \mathrm J\).

Part a: Find the final velocity of fragment B.

Take east as positive.

The initial momentum is:

\[ p_i=mv \]

\[ p_i=(3.0)(4.0) \]

\[ p_i=12.0\ \mathrm{kg\,m\,s^{-1}} \]

Momentum conservation gives:

\[ m_Av_A+m_Bv_B=p_i \]

Substitute:

\[ (1.0)(-2.0)+(2.0)v_B=12.0 \]

\[ -2.0+2.0v_B=12.0 \]

\[ 2.0v_B=14.0 \]

\[ \boxed{v_B=+7.0\ \mathrm{m\,s^{-1}}} \]

The positive sign indicates eastward motion.

Part b: Calculate the initial kinetic energy.

Before the explosion, the entire device was moving at \(4.0\ \mathrm{m\,s^{-1}}\).

\[ E_{k,i}=\frac12mv^2 \]

\[ E_{k,i}=\frac12(3.0)(4.0)^2 \]

\[ \boxed{E_{k,i}=24\ \mathrm J} \]

Part c: Calculate the final kinetic energy.

Both fragments contribute positively:

\[ E_{k,f}= \frac12m_Av_A^2+ \frac12m_Bv_B^2 \]

Substitute:

\[ E_{k,f}= \frac12(1.0)(2.0)^2+ \frac12(2.0)(7.0)^2 \]

\[ E_{k,f}=2.0+49.0 \]

\[ \boxed{E_{k,f}=51\ \mathrm J} \]

Part d: Calculate the increase in kinetic energy.

\[ \Delta E_k=E_{k,f}-E_{k,i} \]

\[ \Delta E_k=51-24 \]

\[ \boxed{\Delta E_k=27\ \mathrm J} \]

The system’s total momentum remains \(12.0\ \mathrm{kg\,m\,s^{-1}}\), but its kinetic energy increases by \(27\ \mathrm J\).

The important detail is that the original device was already moving. You must calculate the initial kinetic energy rather than automatically assuming it is zero.

Question 4

A \(4.0\ \mathrm{kg}\) device is initially stationary on a horizontal track. It explodes into two fragments with masses of \(1.0\ \mathrm{kg}\) and \(3.0\ \mathrm{kg}\).

During the explosion, an external actuator also applies a horizontal impulse of \(1.2\ \mathrm{N\,s}\) to the right.

The \(1.0\ \mathrm{kg}\) fragment moves left at \(3.0\ \mathrm{m\,s^{-1}}\).

Calculate:

a) The final velocity of the \(3.0\ \mathrm{kg}\) fragment.

b) The final total momentum.

c) The final total kinetic energy.

d) Explain why conserving momentum by setting the initial and final momenta equal would give an incorrect result.

Take right as positive.

Solution 4

The \(3.0\ \mathrm{kg}\) fragment moves right at \(1.4\ \mathrm{m\,s^{-1}}\). The final total momentum is \(+1.2\ \mathrm{kg\,m\,s^{-1}}\), and the final kinetic energy is \(7.44\ \mathrm J\).

Part a: Calculate the velocity of the second fragment.

This time, we cannot simply conserve momentum because an external impulse acts on the system.

Instead, use the impulse-momentum relationship:

\[ p_f=p_i+J_{\text{ext}} \]

The initial momentum is zero:

\[ p_f=0+1.2 \]

Therefore:

\[ p_f=+1.2\ \mathrm{kg\,m\,s^{-1}} \]

Now apply the momentum equation to the fragments:

\[ m_Av_A+m_Bv_B=1.2 \]

Substitute:

\[ (1.0)(-3.0)+(3.0)v_B=1.2 \]

\[ -3.0+3.0v_B=1.2 \]

\[ 3.0v_B=4.2 \]

\[ \boxed{v_B=+1.4\ \mathrm{m\,s^{-1}}} \]

Part b: Confirm the final total momentum.

\[ p_f=(1.0)(-3.0)+(3.0)(1.4) \]

\[ p_f=-3.0+4.2 \]

\[ \boxed{p_f=+1.2\ \mathrm{kg\,m\,s^{-1}}} \]

This agrees with the external impulse.

Part c: Calculate the final kinetic energy.

\[ E_{k,f}= \frac12m_Av_A^2+ \frac12m_Bv_B^2 \]

Substitute:

\[ E_{k,f}= \frac12(1.0)(3.0)^2+ \frac12(3.0)(1.4)^2 \]

\[ E_{k,f}=4.5+2.94 \]

\[ \boxed{E_{k,f}=7.44\ \mathrm J} \]

The initial kinetic energy was zero, so the fragments have \(7.44\ \mathrm J\) of translational kinetic energy after the event.

However, we cannot determine exactly how much internal energy was released from these measurements alone. The external actuator may also have transferred energy to the system. Impulse tells us about momentum change, not directly about work.

Part d: Explain why ordinary momentum conservation does not apply.

The internal explosion forces still occur in equal and opposite pairs. Their contributions to total momentum cancel.

However, the external actuator supplies an additional impulse of \(+1.2\ \mathrm{N\,s}\).

Therefore, the system’s total momentum changes by \(+1.2\ \mathrm{kg\,m\,s^{-1}}\).

The mistake would be to ignore the external impulse and assume the system is isolated.

This is why checking external impulse must come before applying momentum conservation.

Question 5

A \(3.0\ \mathrm{kg}\) device travels east at \(2.0\ \mathrm{m\,s^{-1}}\) along a frictionless track. It explodes into two fragments:

• Fragment A has a mass of \(1.0\ \mathrm{kg}\).
• Fragment B has a mass of \(2.0\ \mathrm{kg}\).

After the explosion, an observer standing beside the track measures fragment A travelling west at \(4.0\ \mathrm{m\,s^{-1}}\).

Assume negligible external impulse and that the total mass remains \(3.0\ \mathrm{kg}\).

Answer the following:

a) Calculate the final velocity of fragment B in the ground observer’s frame.

b) Calculate the change in total kinetic energy in this frame.

c) A second observer moves east at a constant \(2.0\ \mathrm{m\,s^{-1}}\) relative to the ground. Calculate the initial and final velocities of both fragments as measured by this observer.

d) Calculate the change in total kinetic energy in the second observer’s frame.

e) A student argues:

Kinetic energy depends on the observer’s reference frame. Therefore, the amount of kinetic energy gained in an explosion must also depend on the observer.

Assess this statement. Use the two sets of measurements to explain whether the student is correct.

Assume both observers use Newtonian mechanics and travel at constant velocities.

Solution 5

The second fragment moves east at \(5.0\ \mathrm{m\,s^{-1}}\). Both observers measure an increase in total kinetic energy of \(27\ \mathrm J\), even though their individual initial and final kinetic energy values are different.

This is the most interesting distinction in the question: kinetic energy itself depends on the reference frame, but the change in kinetic energy is the same for both observers when total momentum is conserved and the total mass is unchanged.

Part a: Find fragment B’s velocity in the ground frame.

Take east as positive.

The initial total momentum is:

\[ p_i=mv \]

\[ p_i=(3.0)(2.0) \]

\[ p_i=6.0\ \mathrm{kg\,m\,s^{-1}} \]

Apply momentum conservation:

\[ m_Av_A+m_Bv_B=p_i \]

Substitute:

\[ (1.0)(-4.0)+(2.0)v_B=6.0 \]

\[ -4.0+2.0v_B=6.0 \]

\[ 2.0v_B=10.0 \]

\[ \boxed{v_B=+5.0\ \mathrm{m\,s^{-1}}} \]

Fragment B travels east at \(5.0\ \mathrm{m\,s^{-1}}\).

Part b: Calculate the kinetic energy change in the ground frame.

Initially, the entire device moves at \(2.0\ \mathrm{m\,s^{-1}}\):

\[ E_{k,i}=\frac12(3.0)(2.0)^2 \]

\[ E_{k,i}=6.0\ \mathrm J \]

After the explosion:

\[ E_{k,f}= \frac12(1.0)(4.0)^2+ \frac12(2.0)(5.0)^2 \]

\[ E_{k,f}=8.0+25.0 \]

\[ E_{k,f}=33.0\ \mathrm J \]

Therefore:

\[ \Delta E_k=33.0-6.0 \]

\[ \boxed{\Delta E_k=+27.0\ \mathrm J} \]

Part c: Find the velocities in the second observer’s frame.

The second observer moves east at \(2.0\ \mathrm{m\,s^{-1}}\).

To convert a velocity measured from the ground into this moving observer’s frame, subtract the observer’s velocity:

\[ v'=v-U \]

where \(U=+2.0\ \mathrm{m\,s^{-1}}\).

Initially, the device travels east at \(2.0\ \mathrm{m\,s^{-1}}\) relative to the ground.

Therefore, its initial velocity in the second observer’s frame is:

\[ u'=2.0-2.0=0 \]

The device is initially stationary relative to this observer.

For fragment A:

\[ v_A'=-4.0-2.0 \]

\[ \boxed{v_A'=-6.0\ \mathrm{m\,s^{-1}}} \]

For fragment B:

\[ v_B'=+5.0-2.0 \]

\[ \boxed{v_B'=+3.0\ \mathrm{m\,s^{-1}}} \]

The second observer sees the fragments separating at \(6.0\ \mathrm{m\,s^{-1}}\) west and \(3.0\ \mathrm{m\,s^{-1}}\) east.

Part d: Calculate the kinetic energy change in the second frame.

The initial device is stationary in this frame:

\[ E_{k,i}'=0 \]

After the explosion:

\[ E_{k,f}'= \frac12(1.0)(6.0)^2+ \frac12(2.0)(3.0)^2 \]

\[ E_{k,f}'=18.0+9.0 \]

\[ E_{k,f}'=27.0\ \mathrm J \]

Therefore:

\[ \Delta E_k'=27.0-0 \]

\[ \boxed{\Delta E_k'=+27.0\ \mathrm J} \]

Both observers measure the same increase in kinetic energy.

Part e: Assess the student’s statement.

The student’s statement is incorrect.

The total kinetic energy is frame-dependent, but its change is not necessarily frame-dependent.

We can show why algebraically.

For a system of objects moving in one dimension, the total kinetic energy measured by an observer moving at speed \(U\) is:

\[ E_k'=\frac12\sum m_i(v_i-U)^2 \]

Expanding the square:

\[ E_k'= \frac12\sum m_iv_i^2 -U\sum m_iv_i +\frac12U^2\sum m_i \]

Recognising the total kinetic energy, momentum, and mass:

\[ E_k'=E_k-UP+\frac12MU^2 \]

Here:

• \(E_k'\) is kinetic energy in the moving observer’s frame.
• \(E_k\) is kinetic energy in the original frame.
• \(P\) is total momentum.
• \(M\) is total mass.
• \(U\) is the observer’s velocity relative to the original frame.

Now compare the initial and final states.

Because the system’s total mass is unchanged:

\[ \Delta E_k'=\Delta E_k-U\Delta P \]

But momentum is conserved, so:

\[ \Delta P=0 \]

Therefore:

\[ \boxed{\Delta E_k'=\Delta E_k} \]

The change in kinetic energy is identical for both observers.

This is the subtle result.

The ground observer measures:

• Initial kinetic energy: \(6.0\ \mathrm J\).
• Final kinetic energy: \(33.0\ \mathrm J\).

The moving observer measures:

• Initial kinetic energy: \(0\ \mathrm J\).
• Final kinetic energy: \(27.0\ \mathrm J\).

The actual kinetic energy values differ, but the increase is \(27.0\ \mathrm J\) in both frames.

If the explosion converts all its released internal energy into translational kinetic energy, that internal energy decrease is \(27.0\ \mathrm J\).

The physical energy conversion does not depend on which of these two observers describes it.

This result relies on the system having negligible external impulse and unchanged total mass within the Newtonian approximation. If an external impulse changes the system’s total momentum, the kinetic energy change need not be the same in different reference frames.

That brings us back to the central idea: momentum conservation and kinetic energy change are related, but they are not the same condition.

08What to take into your next problem

An explosion is not an exception to conservation laws. It is a particularly clear demonstration that different physical quantities behave differently.

Remember these three relationships:

• Total momentum stays constant when the net external impulse is negligible.
• Kinetic energy can increase because stored internal energy is converted into motion.
• Equal and opposite momenta do not mean zero kinetic energy, because kinetic energy is a scalar and momentum is a vector.

Once you are comfortable with this, the next useful step is analysing recoil and projectile-launch problems. A projectile and its launcher experience equal and opposite momentum changes, but the energy released by the launch is distributed differently between them.

The same momentum reasoning applies, with the additional challenge of determining how the masses affect their respective recoil speeds.

How to Calculate Projectile Recoil Speed | HSC Physics