How to Solve Recoil and Explosion Momentum Problems
Learn how recoil and explosion problems work when a system starts at rest, and how to calculate the velocities of objects moving apart. Includes worked examples and challenging HSC Physics questions on signed velocities, momentum conservation, and kinetic energy.
Two carts are held at rest with a compressed spring between them. Release the spring, and both carts shoot away from each other. Before the release, the total momentum was zero. Afterwards, both carts are moving. Has the system somehow created momentum?
Predict which cart moves faster if one cart has three times the mass of the other. The lighter cart moves three times as fast. Its momentum is equal in magnitude and opposite in direction to the heavier cart’s momentum, so the total momentum remains zero.
This is the central idea behind recoil and explosions: individual objects can gain momentum while the total momentum of the system remains unchanged.
Recoil: momentum remains conserved
| Before release | p_total = 0 Both carts are stationary. A compressed spring lies between the carts. |
|---|---|
| After release | m₁v₁ + m₂v₂ = 0 The 1.0 kg cart moves left at speed 3v; the 3.0 kg cart moves right at speed v. Their momenta are −3v and +3v kg·m/s, respectively, so their sum is zero. |
01Why objects recoil when a system explodes
Imagine standing on a skateboard and throwing a heavy ball forwards. You roll backwards. The ball gains forwards momentum, and you gain backwards momentum.
An explosion works on the same principle. A compressed spring, expanding gas, or chemical reaction pushes different parts of a system apart. Each part can move, but the system’s total momentum does not change if the net external impulse is negligible.
The word recoil describes the backwards motion of an object in response to something being launched or pushed forwards. A launcher recoils when it fires a projectile. A rocket recoils as exhaust gases are expelled backwards.
The important point is that the parts do not have to move at the same speed. A heavier object generally needs a smaller velocity to carry the same magnitude of momentum.
02Choose the system before choosing the equation
Suppose a launcher fires a projectile. Which object should you treat as the system?
You could choose the projectile alone, the launcher alone, or the launcher and projectile together. These choices lead to different conclusions about whether momentum is conserved.
For a recoil problem, the most useful choice is usually the complete system involved in the launch: the launcher and the projectile.
Choose the system boundary
| Launcher + projectile | p_total = constant Spring and projectile are inside the boundary. Negligible external impulse: total momentum is conserved. |
|---|---|
| Projectile only | The launcher force crosses the system boundary. Projectile momentum changes during firing. |
When the launcher and projectile are treated as one system, the forces they exert on each other are internal forces. They can change each object’s momentum, but they cannot change the total momentum of the combined system.
This follows from Newton’s third law: the launcher pushes the projectile forwards, and the projectile pushes the launcher backwards with an equal and opposite force. Over the launch, these forces produce equal and opposite impulses.
Momentum conservation still has a condition. The net external impulse on the chosen system must be zero or negligible over the time interval being considered. On a smooth horizontal track, gravity and the normal force balance vertically, while horizontal friction is negligible.
The momentum equation
Momentum is the product of an object’s mass and velocity:
\[
p=mv
\]
Here:
- \(p\) is momentum, measured in \(\mathrm{kg\,m\,s^{-1}}\).
- \(m\) is mass, measured in kilograms (\(\mathrm{kg}\)).
- \(v\) is velocity, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
Momentum is a vector, so direction matters. In a one-dimensional problem, we can account for direction using positive and negative signs.
Choose right as the positive direction. An object moving right has positive momentum; an object moving left has negative momentum.
The conservation of momentum equation is
\[
p_{\text{before}}=p_{\text{after}}
\]
If every part of the system starts at rest, its initial total momentum is zero. Therefore,
\[
p_{\text{after}}=0
\]
For two objects moving along the same straight line,
\[
m_1v_1+m_2v_2=0
\]
The velocities are signed, not just positive speed values. This is what allows the equation to represent objects moving in opposite directions.
Rearranging gives
\[
v_2=-\frac{m_1}{m_2}v_1
\]
The negative sign tells you that the objects move in opposite directions. The mass ratio tells you their relative speeds:
\[
\frac{|v_1|}{|v_2|}
=\frac{m_2}{m_1}
\]
If one object has three times the mass of the other, it moves at one-third of the other’s speed in a two-object recoil from rest.
Notice what this does not mean. It does not mean the heavier object has three times the momentum. Their momentum magnitudes are equal.
03Why kinetic energy can increase while momentum stays constant
A stationary system can have zero total momentum and zero kinetic energy. After an explosion, it can still have zero total momentum but have substantial kinetic energy.
There is no contradiction. Momentum and kinetic energy describe different things.
Momentum depends on velocity, including its sign:
\[
p=mv
\]
Kinetic energy depends on the square of velocity:
\[
K=\frac12 mv^2
\]
Kinetic energy \(K\) is measured in joules (\(\mathrm{J}\)). Because velocity is squared, kinetic energy is never negative.
Suppose two fragments fly apart with equal and opposite momenta. Their momenta cancel when added, but their kinetic energies do not. You add the positive kinetic energy of each fragment.
Momentum cancels; kinetic energy increases
| Before release | p₁ = p₂ = 0; K₁ + K₂ = 0 Both fragments are stationary; energy is stored in the compressed spring. |
|---|---|
| After release | p₁ + p₂ = 0; K₁ + K₂ > 0 Fragments move in opposite directions with equal and opposite momenta; both kinetic energies are positive. |
Where does the new kinetic energy come from?
It comes from another form of energy within the system. A spring can release stored elastic potential energy. An explosive reaction can convert chemical energy into kinetic energy, heat, sound, and other forms of energy.
Total energy is conserved, but kinetic energy is not necessarily conserved. During an explosion, kinetic energy usually increases. During an inelastic collision, kinetic energy usually decreases as some is converted into internal energy, sound, and deformation.
This distinction is essential: momentum conservation does not require kinetic energy to remain constant.
04Worked example: a launcher fires a projectile
A \(2.0\,\mathrm{kg}\) launcher fires a \(0.40\,\mathrm{kg}\) projectile horizontally to the right at \(12.0\,\mathrm{m\,s^{-1}}\). Both are initially at rest. The projectile’s speed is measured relative to the ground. Neglect horizontal external impulse.
Find the launcher’s recoil velocity and the total kinetic energy immediately after firing.
Step 1
Treat the launcher and projectile as one system. Take right as positive.
Initially, both are at rest, so
\[
p_{\text{before}}=0
\]
Step 2
Let \(v_L\) be the launcher’s final velocity. The projectile’s velocity is \(+12.0\,\mathrm{m\,s^{-1}}\).
\[
m_Pv_P+m_Lv_L=0
\]
Substitute the masses and velocity:
\[
(0.40)(12.0)+(2.0)v_L=0
\]
Rearranging,
\[
v_L=-\frac{0.40\times12.0}{2.0}
\]
\[
v_L=-2.4\,\mathrm{m\,s^{-1}}
\]
The launcher recoils to the left at \(2.4\,\mathrm{m\,s^{-1}}\). Its negative velocity gives the correct direction.
Step 3
For the projectile,
\[
K_P=\frac12(0.40)(12.0)^2
\]
\[
K_P=28.8\,\mathrm{J}
\]
For the launcher,
\[
K_L=\frac12(2.0)(2.4)^2
\]
\[
K_L=5.76\,\mathrm{J}
\]
Add the kinetic energies:
\[
\begin{aligned}
K_{\text{total}}&=K_P+K_L\\
&=28.8+5.76\\
&=34.56\,\mathrm{J}
\end{aligned}
\]
The total kinetic energy is approximately \(34.6\,\mathrm{J}\).
The system started with zero momentum and zero kinetic energy. It now has zero total momentum but \(34.6\,\mathrm{J}\) of kinetic energy. The launcher’s backwards momentum balances the projectile’s forwards momentum, while stored energy has been converted into motion.
In a real launcher, some energy may also become sound, heat, or deformation. The calculated kinetic energy is the amount transferred into the two objects’ motion, not necessarily the entire amount of stored energy released.
Recoil: equal momentum, different speeds
| Projectile | m = 0.40 kg v = +12.0 m/s; p = +4.8 kg·m/s |
|---|---|
| Launcher | m = 2.0 kg v = −2.4 m/s; p = −4.8 kg·m/s |
05Worked example: two fragments with a known separation speed
A stationary device breaks into two fragments of masses \(2.0\,\mathrm{kg}\) and \(3.0\,\mathrm{kg}\). Immediately after the explosion, their relative speed is \(10.0\,\mathrm{m\,s^{-1}}\).
Find the velocity of each fragment and the total kinetic energy after the explosion. Assume one-dimensional motion and negligible external impulse.
Step 1
Take right as positive. Choose the \(2.0\,\mathrm{kg}\) fragment to move left with velocity \(v_1\), and the \(3.0\,\mathrm{kg}\) fragment to move right with velocity \(v_2\).
Their relative speed is \(10.0\,\mathrm{m\,s^{-1}}\), so
\[
v_2-v_1=10.0
\]
The device initially had zero total momentum, giving
\[
2.0v_1+3.0v_2=0
\]
Step 2
Rearrange the momentum equation:
\[
v_1=-\frac{3.0}{2.0}v_2
\]
Substitute into the relative-speed equation:
\[
v_2-(-1.5v_2)=10.0
\]
\[
2.5v_2=10.0
\]
Therefore,
\[
v_2=+4.0\,\mathrm{m\,s^{-1}}
\]
Substitute back to find the other velocity:
\[
v_1=-1.5(4.0)
\]
\[
v_1=-6.0\,\mathrm{m\,s^{-1}}
\]
The lighter fragment moves left at \(6.0\,\mathrm{m\,s^{-1}}\), and the heavier fragment moves right at \(4.0\,\mathrm{m\,s^{-1}}\). Reversing both directions would describe the same physical situation.
Step 3
For the \(2.0\,\mathrm{kg}\) fragment,
\[
K_1=\frac12(2.0)(6.0)^2=36\,\mathrm{J}
\]
For the \(3.0\,\mathrm{kg}\) fragment,
\[
K_2=\frac12(3.0)(4.0)^2=24\,\mathrm{J}
\]
Hence,
\[
K_{\text{total}}=36+24
\]
\[
K_{\text{total}}=60\,\mathrm{J}
\]
The total kinetic energy after the explosion is \(60\,\mathrm{J}\).
The lighter fragment has more kinetic energy, even though both fragments have equal momentum magnitudes. For an object with momentum magnitude \(p\), its kinetic energy can also be written as
\[
K=\frac{p^2}{2m}
\]
At a fixed momentum magnitude, a smaller mass has greater kinetic energy. This is a useful consequence of combining momentum conservation with the kinetic-energy equation.
Fragment momentum and kinetic energy
| 2.0 kg fragment | v = −6.0 m/s p = −12 kg·m/s; K = 36 J |
|---|---|
| 3.0 kg fragment | v = +4.0 m/s p = +12 kg·m/s; K = 24 J |
06A reliable exam method
Use this sequence whenever you see a recoil or explosion problem.
| Step | What to do |
|---|---|
| 1. Choose the system | Include the objects whose total momentum you want to track. |
| 2. Check external impulse | Decide whether momentum is conserved over the interval being analysed. |
| 3. Set a positive direction | Assign a sign to every velocity and keep it consistent. |
| 4. Write momentum conservation | Use the initial total momentum, which is zero if the complete system starts at rest. |
| 5. Solve for the unknown velocities | Keep the signs until you have interpreted the directions. |
| 6. Handle energy separately | Use \(K=\frac12mv^2\). Do not assume kinetic energy is conserved through an explosion or an inelastic collision. |
One particularly tempting mistake is to use speeds as if they were signed velocities. Another is to assume that zero total momentum means zero kinetic energy. The first loses the direction information; the second confuses two different physical quantities.
There is one further trap: a velocity may be measured relative to a moving object rather than the ground. If that happens, you need a relative-velocity equation as well as the momentum equation. The final question will bring these ideas together.
07Questions and solutions
Three-fragment explosion
| Before explosion | p_total = 0 The device is stationary. |
|---|---|
| Fragment A | m = 1.0 kg; v = −8.0 m/s Moves left; p = −8.0 kg·m/s. |
| Fragment B | m = 1.5 kg; v = +2.0 m/s Moves right; p = +3.0 kg·m/s. |
| Fragment C | m = 1.5 kg; v = ? Direction and velocity are unknown. |
Question 1
Two carts of masses \(0.60\,\mathrm{kg}\) and \(1.40\,\mathrm{kg}\) are initially at rest on a low-friction track. A compressed spring pushes them apart. The \(0.60\,\mathrm{kg}\) cart moves left at \(3.5\,\mathrm{m\,s^{-1}}\).
Find the velocity of the \(1.40\,\mathrm{kg}\) cart.
Solution 1
The \(1.40\,\mathrm{kg}\) cart moves right at \(1.5\,\mathrm{m\,s^{-1}}\). Its momentum must balance the momentum of the lighter cart because the initial total momentum was zero.
Take right as positive. The lighter cart has velocity \(-3.5\,\mathrm{m\,s^{-1}}\).
Apply conservation of momentum:
\[
(0.60)(-3.5)+(1.40)v=0
\]
Rearrange:
\[
1.40v=2.10
\]
\[
v=\frac{2.10}{1.40}
\]
\[
v=+1.5\,\mathrm{m\,s^{-1}}
\]
The positive sign confirms that the heavier cart moves right. Its speed is lower, but its momentum magnitude is equal to that of the lighter cart.
Question 2
An initially stationary \(4.0\,\mathrm{kg}\) device breaks into three fragments on a straight, frictionless track.
- Fragment A has a mass of \(1.0\,\mathrm{kg}\) and moves left at \(8.0\,\mathrm{m\,s^{-1}}\).
- Fragment B has a mass of \(1.5\,\mathrm{kg}\) and moves right at \(2.0\,\mathrm{m\,s^{-1}}\).
- Fragment C has a mass of \(1.5\,\mathrm{kg}\).
Find the velocity of fragment C and the total kinetic energy immediately after the explosion. Assume negligible external impulse.
Solution 2
Fragment C moves right at approximately \(3.3\,\mathrm{m\,s^{-1}}\), and the total kinetic energy is approximately \(43\,\mathrm{J}\). Its direction follows from the sum of all three signed momenta, not from which fragment has the greatest speed.
Take right as positive. The known momenta are
\[
p_A=(1.0)(-8.0)=-8.0\,\mathrm{kg\,m\,s^{-1}}
\]
\[
p_B=(1.5)(+2.0)=+3.0\,\mathrm{kg\,m\,s^{-1}}
\]
The total momentum must be zero, so
\[
-8.0+3.0+1.5v_C=0
\]
\[
1.5v_C=5.0
\]
\[
v_C=+3.33\,\mathrm{m\,s^{-1}}
\]
Fragment C moves right. Fragments B and C both move right because their combined positive momentum balances fragment A’s negative momentum.
Now calculate the kinetic energies:
\[
K_A=\frac12(1.0)(8.0)^2=32\,\mathrm{J}
\]
\[
K_B=\frac12(1.5)(2.0)^2=3.0\,\mathrm{J}
\]
\[
K_C=\frac12(1.5)(3.33)^2
\approx8.3\,\mathrm{J}
\]
Therefore,
\[
K_{\text{total}}\approx32+3.0+8.3
\]
\[
K_{\text{total}}\approx43\,\mathrm{J}
\]
The key constraint is the total momentum of all three fragments. With three objects, there is no requirement for each fragment to have a matching partner travelling in the opposite direction.
Question 3
An initially stationary \(3.0\,\mathrm{kg}\) device explodes into a \(0.50\,\mathrm{kg}\) fragment and a \(2.50\,\mathrm{kg}\) fragment.
The \(0.50\,\mathrm{kg}\) fragment moves right and has \(36\,\mathrm{J}\) of kinetic energy. Find:
(a) the velocity of the \(0.50\,\mathrm{kg}\) fragment;
(b) the velocity of the \(2.50\,\mathrm{kg}\) fragment;
(c) the total kinetic energy immediately after the explosion.
Assume negligible external impulse.
Solution 3
The fragments have velocities of \(+12\,\mathrm{m\,s^{-1}}\) and \(-2.4\,\mathrm{m\,s^{-1}}\), respectively, and their total kinetic energy is \(43.2\,\mathrm{J}\). The given kinetic energy determines the lighter fragment’s speed; momentum conservation then determines the heavier fragment’s velocity.
Part (a): Find the lighter fragment’s velocity.
Start with the kinetic-energy equation:
\[
K=\frac12mv^2
\]
Substitute the known values:
\[
36=\frac12(0.50)v^2
\]
\[
v^2=\frac{2(36)}{0.50}=144
\]
\[
v=12\,\mathrm{m\,s^{-1}}
\]
The fragment is specified to move right, so its signed velocity is
\[
v_1=+12\,\mathrm{m\,s^{-1}}
\]
Kinetic energy gives the magnitude of the velocity, not its direction.
Part (b): Find the heavier fragment’s velocity.
The device initially had zero momentum:
\[
(0.50)(12)+(2.50)v_2=0
\]
\[
6.0+2.50v_2=0
\]
\[
v_2=-\frac{6.0}{2.50}
\]
\[
v_2=-2.4\,\mathrm{m\,s^{-1}}
\]
The heavier fragment moves left. Its momentum is \(-6.0\,\mathrm{kg\,m\,s^{-1}}\), exactly balancing the lighter fragment’s positive momentum.
Part (c): Find the total kinetic energy.
The heavier fragment’s kinetic energy is
\[
K_2=\frac12(2.50)(2.4)^2
\]
\[
K_2=7.2\,\mathrm{J}
\]
Add the two fragments’ kinetic energies:
\[
K_{\text{total}}=36+7.2
\]
\[
K_{\text{total}}=43.2\,\mathrm{J}
\]
The final kinetic energy is positive even though the total momentum remains zero. The increase in kinetic energy comes from energy released by the device, not from the creation of net momentum.
Wall impact changes the fragments’ momentum
| Before impact | p_total = 0 A: 1.0 kg, v = +6.0 m/s; B: 2.0 kg, v = −3.0 m/s. |
|---|---|
| After impact | p_total = −8.0 kg·m/s A rebounds left at 2.0 m/s; B continues left at 3.0 m/s. Wall at track’s right end. |
Question 4
A stationary \(3.0\,\mathrm{kg}\) device explodes into two fragments on a smooth track. Fragment A has a mass of \(1.0\,\mathrm{kg}\), and fragment B has a mass of \(2.0\,\mathrm{kg}\).
Immediately after the explosion, A moves right at \(6.0\,\mathrm{m\,s^{-1}}\), and B moves left at \(3.0\,\mathrm{m\,s^{-1}}\).
Fragment A then strikes a rigid wall and rebounds left at \(2.0\,\mathrm{m\,s^{-1}}\). Fragment B continues moving left at \(3.0\,\mathrm{m\,s^{-1}}\).
(a) Verify that total momentum is zero immediately after the explosion.
(b) Calculate the total momentum of the two fragments after A rebounds.
(c) Calculate the impulse delivered by the wall to fragment A.
(d) Explain why the fragments’ momentum after the collision does not contradict conservation of momentum during the explosion.
Solution 4
Immediately after the rebound, the fragments have a total momentum of \(-8.0\,\mathrm{kg\,m\,s^{-1}}\), and the wall has delivered an impulse of \(-8.0\,\mathrm{N\,s}\) to fragment A. Momentum was conserved during the explosion, but the later collision introduces an external impulse on the two-fragment system.
Part (a): Check momentum immediately after the explosion.
Take right as positive.
\[
p_{\text{total}}=(1.0)(+6.0)+(2.0)(-3.0)
\]
\[
p_{\text{total}}=+6.0-6.0=0
\]
The momenta balance, as expected for an initially stationary system with negligible external impulse.
Part (b): Calculate the momentum after the rebound.
After the collision, A’s velocity is \(-2.0\,\mathrm{m\,s^{-1}}\), while B’s remains \(-3.0\,\mathrm{m\,s^{-1}}\).
\[
p_{\text{total}}=(1.0)(-2.0)+(2.0)(-3.0)
\]
\[
p_{\text{total}}=-2.0-6.0
\]
\[
p_{\text{total}}=-8.0\,\mathrm{kg\,m\,s^{-1}}
\]
The fragments now have a net momentum to the left.
Part (c): Calculate the impulse from the wall.
Impulse equals the change in momentum:
\[
J=\Delta p=m(v_f-v_i)
\]
For fragment A,
\[
J=(1.0)(-2.0-6.0)
\]
\[
J=-8.0\,\mathrm{N\,s}
\]
The negative sign means the impulse acts to the left. The wall receives an equal and opposite impulse to the right.
Part (d): Explain the apparent contradiction.
Momentum conservation has not failed. The two-fragment system was isolated horizontally during the explosion, but it was not isolated during the collision with the wall.
The wall is outside the chosen system boundary, so its impulse changes the fragments’ total momentum. If the system included the wall and Earth, their momentum changes would also be counted. The wall and Earth acquire momentum to the right, although Earth’s resulting velocity is extremely small because its mass is enormous.
The crucial decision is to check whether an external impulse acts during the particular event being analysed. Do not assume that the fragments’ total momentum must remain zero after they interact with something outside the system.
Launcher recoil and projectile impact
| Before impact | mL = 4.0 kg; mP = 1.0 kg; mT = 3.0 kg Projectile moves right relative to launcher at 20.0 m/s; target is stationary to the right. Ground-frame velocities vL and vP are unknown. |
|---|---|
| After impact | mP+T = 4.0 kg Projectile embeds in target; combined mass moves right while launcher recoils left. Ground-frame velocities remain to be determined using momentum conservation. |
Question 5
A \(4.0\,\mathrm{kg}\) launcher contains a \(1.0\,\mathrm{kg}\) projectile. The launcher and projectile are initially at rest on a frictionless horizontal track. A stationary \(3.0\,\mathrm{kg}\) target sits to the right.
The launcher fires the projectile to the right. Immediately after firing, the projectile’s speed relative to the launcher is \(20.0\,\mathrm{m\,s^{-1}}\). The projectile then strikes the target and sticks to it.
Assume negligible external horizontal impulse on the complete system throughout the launch and collision. The launcher does not contact the target.
(a) Find the ground-frame velocities of the launcher and projectile immediately after firing.
(b) Find the velocity of the projectile and target together after the collision.
(c) Calculate the total kinetic energy of all objects immediately after firing and immediately after the collision.
(d) A student treats \(20.0\,\mathrm{m\,s^{-1}}\) as the projectile’s ground-frame velocity and uses momentum conservation to find the launcher’s recoil velocity. Explain precisely why this approach fails.
Solution 5
Immediately after firing, the launcher moves left at \(4.0\,\mathrm{m\,s^{-1}}\), and the projectile moves right at \(16.0\,\mathrm{m\,s^{-1}}\). After the projectile sticks to the target, the combined \(4.0\,\mathrm{kg}\) object moves right at \(4.0\,\mathrm{m\,s^{-1}}\). The total kinetic energy is \(160\,\mathrm{J}\) immediately after firing and \(64\,\mathrm{J}\) after the collision.
This problem requires two different constraints during the launch: conservation of momentum and the specified relative speed. Using only one of them is not enough.
Part (a): Find the ground-frame velocities after firing.
Take right as positive. Let \(v_P\) be the projectile’s ground-frame velocity and \(v_L\) the launcher’s ground-frame velocity.
The projectile’s velocity relative to the launcher is the difference between their ground-frame velocities:
\[
v_P-v_L=+20.0
\]
Because the launcher and projectile start at rest, their total initial momentum is zero. The target is also initially stationary, so it contributes zero momentum.
Conservation of momentum gives
\[
(1.0)v_P+(4.0)v_L=0
\]
Rearrange:
\[
v_P=-4.0v_L
\]
Substitute into the relative-velocity equation:
\[
-4.0v_L-v_L=20.0
\]
\[
-5.0v_L=20.0
\]
\[
v_L=-4.0\,\mathrm{m\,s^{-1}}
\]
Then
\[
v_P=-4.0(-4.0)
\]
\[
v_P=+16.0\,\mathrm{m\,s^{-1}}
\]
The launcher recoils left at \(4.0\,\mathrm{m\,s^{-1}}\), and the projectile travels right at \(16.0\,\mathrm{m\,s^{-1}}\).
Notice that the relative speed is indeed
\[
v_P-v_L=16.0-(-4.0)=20.0\,\mathrm{m\,s^{-1}}
\]
Both constraints are satisfied.
Part (b): Find the velocity after the collision.
The projectile sticks to the stationary target, so the collision is inelastic. Treat the projectile and target as one system during the collision.
Their initial momentum is
\[
p_{\text{before}}=(1.0)(16.0)+(3.0)(0)
\]
\[
p_{\text{before}}=16.0\,\mathrm{kg\,m\,s^{-1}}
\]
Their combined mass is
\[
m_{\text{combined}}=1.0+3.0=4.0\,\mathrm{kg}
\]
Let \(v_f\) be their common final velocity. Conservation of momentum gives
\[
16.0=(4.0)v_f
\]
\[
v_f=+4.0\,\mathrm{m\,s^{-1}}
\]
The projectile and target move right together at \(4.0\,\mathrm{m\,s^{-1}}\). The launcher continues moving left at \(4.0\,\mathrm{m\,s^{-1}}\).
The complete system’s final momentum is
\[
p_{\text{final}}=(4.0)(-4.0)+(4.0)(+4.0)
\]
\[
p_{\text{final}}=-16.0+16.0=0
\]
Total momentum remains zero throughout, as required.
Part (c): Compare the kinetic energies.
Immediately after firing, the launcher has kinetic energy
\[
K_L=\frac12(4.0)(4.0)^2=32\,\mathrm{J}
\]
The projectile has kinetic energy
\[
K_P=\frac12(1.0)(16.0)^2=128\,\mathrm{J}
\]
The target is initially stationary, so its kinetic energy is zero.
Therefore,
\[
K_{\text{after launch}}=32+128
\]
\[
K_{\text{after launch}}=160\,\mathrm{J}
\]
After the collision, the launcher still has \(32\,\mathrm{J}\) of kinetic energy. The combined projectile and target have
\[
K_{\text{combined}}=\frac12(4.0)(4.0)^2
\]
\[
K_{\text{combined}}=32\,\mathrm{J}
\]
The final total is
\[
K_{\text{after collision}}=32+32
\]
\[
K_{\text{after collision}}=64\,\mathrm{J}
\]
The total kinetic energy decreases by
\[
\Delta K=64-160=-96\,\mathrm{J}
\]
The \(96\,\mathrm{J}\) decrease is converted into other forms of energy during the collision, such as deformation, heating, and sound. Momentum remains conserved because the collision forces are internal to the complete system, and the net external horizontal impulse is negligible.
During the launch, the kinetic energy increases from zero to \(160\,\mathrm{J}\). That increase comes from the launcher’s stored energy.
Part (d): Explain why the tempting approach fails.
The student’s approach confuses a relative speed with a ground-frame speed. If the projectile’s ground-frame velocity were \(+20.0\,\mathrm{m\,s^{-1}}\), momentum conservation would give
\[
(1.0)(20.0)+(4.0)v_L=0
\]
\[
v_L=-5.0\,\mathrm{m\,s^{-1}}
\]
But those velocities would imply a relative speed of
\[
v_P-v_L=20.0-(-5.0)
\]
\[
v_P-v_L=25.0\,\mathrm{m\,s^{-1}}
\]
That contradicts the given relative speed of \(20.0\,\mathrm{m\,s^{-1}}\). The tempting answer satisfies momentum conservation but violates the other condition, so it cannot describe the stated launch.
The broader lesson is that a correct equation cannot rescue an incorrectly interpreted quantity. Check the reference frame of every velocity, and apply momentum conservation to the complete system over the event you are analysing.
The next useful connection is the impulse-momentum theorem. When a wall or another object lies outside your chosen system and exerts a force on it, calculate the external impulse to determine how that system’s momentum changes.