How to Calculate Projectile Recoil Speed | HSC Physics

A 20.0 g ball leaves a 4.00 kg spring launcher at 12.0 m/s to the right. If that speed is measured relative to the ground, the launcher recoils at just 0.0600 m/s to the left.

That is the basic method for calculating recoil speed: treat the projectile and launcher as one system, then conserve momentum in the direction of firing.

However, there’s an important detail that can change your answer. Is the projectile’s speed measured relative to the ground, or relative to the moving launcher? These are different speeds, and using the wrong one is a common source of errors in HSC Physics questions.

Before working through the calculation, consider two predictions:

• If the projectile’s mass doubles while everything else stays the same, what happens to recoil speed?
• If the launcher mass doubles instead, what happens?

The first change doubles the recoil speed. The second halves it. We’ll derive both results and then look at situations where the calculation becomes less straightforward.

01The experiment behind the recoil calculation

Imagine a spring launcher mounted on a wheeled trolley. The trolley sits on a level, low-friction track, with its barrel pointing horizontally to the right.

A small ball is placed inside the launcher. When the spring releases, the ball travels forwards and the trolley rolls backwards.

How could we test whether momentum is conserved?

A simple classroom investigation would use a stationary video camera positioned beside the track. A measuring scale runs parallel to the track, allowing the positions of both the ball and trolley to be tracked immediately after separation.

The important prediction is that the total horizontal momentum before and after firing should be the same.

Initially, both objects are stationary, so their total momentum is zero. After firing, the ball has positive momentum. The trolley must therefore have negative momentum of equal magnitude, provided the external horizontal impulse is negligible.

Suppose the video analysis produces the following illustrative data.

Illustrative data, not recorded experimental measurements.

The two momenta cancel exactly. The ball moves much faster, but the trolley has a much greater mass.

This is the central idea behind recoil: the two objects acquire equal and opposite momenta, not equal and opposite velocities.

Why the measuring equipment matters

There is a subtle experimental problem here.

Suppose the first design uses a photogate attached directly to the moving trolley. The gate measures the ball’s speed relative to the trolley, not its speed relative to the ground.

Those measurements are not interchangeable.

A stationary camera or a sensor referenced to the track measures ground-frame motion. A photogate moving with the launcher measures relative motion.

The first measurement can be used directly in the basic recoil equation. The second requires a slightly different equation, which we’ll derive shortly.

This is a useful example of how an experiment can produce an accurate measurement but still lead to an incorrect calculation if the reference frame is misunderstood.

01Choose the system and conserve momentum

The first step in a recoil problem is deciding which objects belong to your system.

For a projectile fired from a launcher, choose:

• The projectile.
• The launcher and anything attached to it that recoils.

Let:

• \(m\) = projectile mass, in kilograms.
• \(M\) = mass of the recoiling launcher and its trolley, in kilograms.
• \(v_p\) = projectile velocity relative to the ground, in metres per second.
• \(v_L\) = launcher velocity relative to the ground, in metres per second.

The projectile’s mass is separate from \(M\). Don’t include it twice.

Assume the combined system is initially stationary and experiences negligible external horizontal impulse during firing.

The initial momentum is therefore zero.

Conservation of momentum gives:

\[ p_{\text{initial}}=p_{\text{final}} \]

\[ 0=mv_p+Mv_L \]

Rearranging:

\[ Mv_L=-mv_p \]

\[ v_L=-\frac{mv_p}{M} \]

The negative sign tells us that the launcher moves in the opposite direction to the projectile.

The recoil speed is the magnitude of this velocity:

\[ \boxed{V_{\text{recoil}}=\frac{mv_p}{M}} \]

This is the main equation to remember when the projectile’s velocity is measured relative to the ground.

Why does the launcher move more slowly?

Consider the momentum equation:

\[ mv_p=MV_{\text{recoil}} \]

The projectile and launcher have equal momentum magnitudes.

However, the launcher has a much greater mass. It therefore needs a much smaller velocity to have the same momentum.

Think of pushing two shopping trolleys: a lightly loaded trolley accelerates much more easily than a heavily loaded one when both experience the same force.

The analogy helps explain the effect of mass, although recoil itself involves two objects exchanging momentum through an internal interaction.

What if the projectile is fired at an angle?

Momentum is a vector, so direction matters.

Suppose the projectile travels at an angle \(\theta\) above the horizontal, with speed \(v\) relative to the ground.

Its horizontal and vertical velocity components are:

\[ v_x=v\cos\theta \]

\[ v_y=v\sin\theta \]

If the launcher is mounted on a horizontal track, it can recoil horizontally but cannot move vertically.

We therefore conserve horizontal momentum:

\[ 0=mv\cos\theta+Mv_L \]

The recoil speed becomes:

\[ \boxed{V_{\text{recoil}}= \frac{mv\cos\theta}{M}} \]

Notice that we use the horizontal component, not the full projectile speed.

Why don’t we conserve vertical momentum as well? Because the track and Earth exert external vertical forces on the projectile-launcher system. The launcher is constrained to remain on the track, so the vertical momentum of the projectile is not balanced by vertical launcher motion.

If the launcher were freely floating in space, with no external impulse in either direction, both momentum components would need to be conserved.

02Momentum is conserved, but kinetic energy is different

A common mistake is assuming that kinetic energy must also be conserved during firing.

It doesn’t.

Momentum conservation and energy conservation are different principles.

When a spring launcher fires, its stored elastic potential energy is transferred into other forms of energy. These include:

• Kinetic energy of the projectile.
• Kinetic energy of the recoiling launcher.
• Sound.
• Thermal energy.
• Vibrations and deformation.

The total energy is conserved, but the total kinetic energy does not have to remain constant.

We can use the first worked example to see this.

The projectile’s kinetic energy is:

\[ E_{k,p}=\frac12 mv_p^2 \]

\[ E_{k,p}=\frac12(0.030)(12.0)^2 \]

\[ E_{k,p}=2.16\ \mathrm{J} \]

The launcher’s kinetic energy is:

\[ E_{k,L}=\frac12 MV_{\text{recoil}}^2 \]

\[ E_{k,L}=\frac12(6.0)(0.0600)^2 \]

\[ E_{k,L}=0.0108\ \mathrm{J} \]

The total translational kinetic energy is:

\[ E_{k,\text{total}}=2.16+0.0108 \]

\[ \boxed{E_{k,\text{total}}=2.1708\ \mathrm{J}} \]

The spring must supply at least this much energy under the stated assumptions.

If it releases 2.50 J, the remaining 0.329 J can be transferred into sound, heat, vibrations, and other non-translational forms.

This is how energy is used as a separate consistency check.

Remember: Use momentum to calculate recoil velocity. Use energy to check whether the resulting motion is consistent with the available energy.

03The important distinction: ground speed versus relative speed

So far, we’ve assumed that the projectile’s velocity is measured relative to the ground.

But what if a question gives the projectile’s speed relative to the launcher?

This changes the equation.

Imagine the launcher moving backwards as the projectile moves forwards. A photogate attached to the launcher measures how quickly the projectile moves past it.

The projectile’s speed relative to the launcher is greater than its speed relative to the ground because the launcher itself is moving backwards.

Let’s derive the correct relationship.

Let:

• \(u\) = projectile speed relative to the launcher.
• \(v_p\) = projectile velocity relative to the ground.
• \(v_L\) = launcher velocity relative to the ground.
• \(V\) = positive recoil speed.

Taking right as positive:

\[ u=v_p-v_L \]

Since the launcher moves left:

\[ v_L=-V \]

Therefore:

\[ u=v_p+V \]

Rearranging:

\[ v_p=u-V \]

Now substitute this into conservation of momentum:

\[ 0=m(u-V)-MV \]

Expand:

\[ 0=mu-mV-MV \]

Rearrange:

\[ mu=(M+m)V \]

Finally:

\[ \boxed{V_{\text{recoil}}=\frac{mu}{M+m}} \]

This is the correct equation when the projectile’s speed is measured relative to the launcher.

Notice the denominator. It is now \(M+m\), rather than just \(M\).

The difference exists because the projectile’s ground-frame speed is reduced by the launcher’s backwards motion.

Which equation should you use?

The key is to identify the reference frame before substituting numbers.

Don’t assume that every speed described as a launch speed is automatically measured relative to the ground. Read the question carefully.

07Common mistakes in recoil questions

Before attempting exam-style questions, make sure you can recognise these traps.

For angled projectiles, it is worth revising Projectile Initial Velocity Components for HSC Physics.

Once the projectile leaves the launcher, its subsequent motion is a separate projectile-motion problem. The ground-frame velocity at separation becomes its initial velocity for that calculation.

08Questions and solutions

The following questions build from a standard recoil calculation towards situations involving momentum components, moving launchers, external impulses, and collisions.

Question 1

A 0.0180 kg ball is fired horizontally from a 3.00 kg spring launcher and trolley. Both are initially stationary.

Immediately after firing, the ball has a velocity of 12.0 m/s to the right, measured relative to the ground.

Calculate the recoil velocity of the launcher.

Solution 1

The launcher recoils at 0.0720 m/s to the left.

The projectile and launcher initially have zero total momentum.

Take right as positive.

Using conservation of momentum:

\[ 0=mv_p+Mv_L \]

Substitute:

\[ 0=(0.0180)(12.0)+(3.00)v_L \]

\[ 0=0.216+3.00v_L \]

Rearranging:

\[ v_L=-\frac{0.216}{3.00} \]

\[ v_L=-0.0720\ \mathrm{m\,s^{-1}} \]

The negative sign indicates that the launcher moves left.

The projectile has momentum of \(+0.216\ \mathrm{kg\,m\,s^{-1}}\), and the launcher has momentum of \(-0.216\ \mathrm{kg\,m\,s^{-1}}\).

Their total momentum remains zero.

Question 2

A 0.0600 kg ball is launched from a 4.50 kg launcher mounted on a frictionless horizontal track.

The ball leaves the launcher at 14.0 m/s relative to the ground, at an angle of 35.0° above the horizontal.

The launcher can move horizontally but is constrained vertically.

Calculate:

(a) The recoil speed of the launcher.

(b) The vertical momentum of the ball immediately after launch.

(c) Explain why the launcher does not acquire equal and opposite vertical momentum.

Solution 2

The launcher recoils at 0.153 m/s to the left. The ball has upward vertical momentum of approximately 0.482 kg m/s.

(a) Calculate the recoil speed.

Only horizontal momentum is conserved because the launcher is constrained to the horizontal track.

The projectile’s horizontal velocity is:

\[ v_x=14.0\cos35.0^\circ \]

\[ v_x=11.47\ \mathrm{m\,s^{-1}} \]

The initial horizontal momentum is zero.

Using conservation of horizontal momentum:

\[ 0=mv_x+Mv_L \]

Substitute:

\[ 0=(0.0600)(11.47)+(4.50)v_L \]

\[ v_L=-\frac{0.688}{4.50} \]

\[ v_L=-0.153\ \mathrm{m\,s^{-1}} \]

The recoil speed is therefore 0.153 m/s to the left.

(b) Calculate the ball’s vertical momentum.

The vertical velocity component is:

\[ v_y=14.0\sin35.0^\circ \]

\[ v_y=8.03\ \mathrm{m\,s^{-1}} \]

Vertical momentum is:

\[ p_y=mv_y \]

\[ p_y=(0.0600)(8.03) \]

\[ \boxed{p_y=0.482\ \mathrm{kg\,m\,s^{-1}}} \]

It acts upwards.

(c) Explain the vertical momentum difference.

The launcher cannot move vertically because the track constrains its motion.

The track and Earth exert external vertical forces on the projectile-launcher system. Consequently, vertical momentum is not conserved for this system.

A tempting approach is to apply momentum conservation in both directions and give the launcher an equal downward momentum. That would only be appropriate if the launcher were free to move vertically and the system experienced negligible external vertical impulse.

Question 3

A 5.00 kg launcher and a 0.100 kg projectile are initially moving together to the right at 0.150 m/s.

The launcher fires the projectile horizontally forwards. Immediately after firing, the projectile’s speed relative to the launcher is 9.00 m/s.

Calculate:

(a) The final velocity of the launcher relative to the ground.

(b) The final velocity of the projectile relative to the ground.

(c) Determine whether the launcher continues moving forwards or reverses direction.

Assume negligible external horizontal impulse.

Solution 3

The launcher reverses direction and finishes moving left at 0.0265 m/s. The projectile moves right at 8.97 m/s relative to the ground.

This question is different because the system is not initially stationary.

(a) Find the launcher’s final velocity.

The initial momentum is:

\[ p_i=(M+m)v_0 \]

\[ p_i=(5.00+0.100)(0.150) \]

\[ p_i=0.765\ \mathrm{kg\,m\,s^{-1}} \]

Take right as positive.

Let the launcher’s final velocity be \(v_L\).

The projectile’s velocity relative to the launcher is:

\[ v_p-v_L=9.00 \]

Therefore:

\[ v_p=v_L+9.00 \]

Apply conservation of momentum:

\[ 0.765=5.00v_L+0.100v_p \]

Substitute:

\[ 0.765=5.00v_L+0.100(v_L+9.00) \]

Expand:

\[ 0.765=5.10v_L+0.900 \]

Rearrange:

\[ 5.10v_L=-0.135 \]

\[ \boxed{v_L=-0.0265\ \mathrm{m\,s^{-1}}} \]

(b) Find the projectile’s final velocity.

Using the relative velocity relationship:

\[ v_p=v_L+9.00 \]

\[ v_p=-0.0265+9.00 \]

\[ \boxed{v_p=8.97\ \mathrm{m\,s^{-1}}} \]

(c) Determine the launcher’s direction.

The launcher’s final velocity is negative, so it moves left.

It initially moved right at 0.150 m/s, but the recoil changes its velocity by approximately 0.176 m/s to the left.

That change is large enough to reverse its direction.

The tempting approach is to calculate recoil as though the launcher initially had zero velocity. That would give the wrong final ground-frame velocity because the system already had positive momentum before firing.

Question 4

A 0.0400 kg ball is fired horizontally from a 3.00 kg spring launcher and trolley. Initially, both are stationary.

A stationary camera records the following velocities immediately after firing:

• Ball: \(+10.0\ \mathrm{m\,s^{-1}}\).
• Launcher: \(-0.120\ \mathrm{m\,s^{-1}}\).

The firing interval lasts 0.0500 s. During this interval, the track exerts an average horizontal force of \(+0.800\ \mathrm{N}\) on the combined system.

The spring releases 2.50 J of stored energy.

(a) Calculate the total final horizontal momentum of the ball-launcher system.

(b) Determine whether the measured momentum is consistent with the external impulse.

(c) Calculate the total translational kinetic energy after firing and determine how much of the released spring energy does not appear as translational kinetic energy.

Solution 4

The total final momentum is +0.0400 kg m/s, which agrees exactly with the external impulse of +0.0400 N s. The translational kinetic energy is 2.0216 J, leaving 0.4784 J outside translational kinetic energy.

(a) Calculate the total final momentum.

The total momentum is:

\[ p_f=mv_p+Mv_L \]

Substitute:

\[ p_f=(0.0400)(10.0)+(3.00)(-0.120) \]

\[ p_f=0.400-0.360 \]

\[ \boxed{p_f=+0.0400\ \mathrm{kg\,m\,s^{-1}}} \]

The total final momentum is not zero.

That does not automatically mean the measurements are incorrect.

(b) Compare the momentum change with the external impulse.

The initial momentum is zero.

The external impulse is:

\[ J=F_{\text{average}}\Delta t \]

\[ J=(0.800)(0.0500) \]

\[ J=+0.0400\ \mathrm{N\,s} \]

The impulse-momentum relationship is:

\[ J=\Delta p \]

Since:

\[ \Delta p=0.0400-0 \]

\[ \boxed{J=\Delta p} \]

The measured data are consistent with the external impulse.

The tempting approach would be to insist that the final momentum must be zero because the system initially had zero momentum. That ignores the horizontal external force exerted by the track.

(c) Calculate the kinetic energy.

The ball’s kinetic energy is:

\[ E_{k,b}=\frac12mv_p^2 \]

\[ E_{k,b}=\frac12(0.0400)(10.0)^2 \]

\[ E_{k,b}=2.00\ \mathrm{J} \]

The launcher’s kinetic energy is:

\[ E_{k,L}=\frac12MV_L^2 \]

\[ E_{k,L}=\frac12(3.00)(0.120)^2 \]

\[ E_{k,L}=0.0216\ \mathrm{J} \]

Total translational kinetic energy:

\[ E_{k,\text{total}}=2.00+0.0216 \]

\[ \boxed{E_{k,\text{total}}=2.0216\ \mathrm{J}} \]

The spring releases 2.50 J.

The difference is:

\[ E_{\text{difference}}=2.50-2.0216 \]

\[ \boxed{E_{\text{difference}}=0.4784\ \mathrm{J}} \]

This energy is not necessarily all converted into heat. It can include sound, vibrations, deformation, and energy transferred to the surroundings.

Momentum and energy provide different checks on the same event.

Question 5

A 1.00 kg spring launcher and a 0.200 kg ball are initially stationary on a frictionless horizontal track.

The ball is launched to the right at 5.00 m/s relative to the launcher.

A stationary 0.800 kg target cart is positioned ahead of the launcher. The ball travels forwards and embeds itself in the target cart.

After the collision, the ball and target cart move together.

Determine:

(a) The launcher’s velocity immediately after firing.

(b) The ball’s velocity relative to the ground immediately before impact.

(c) The velocity of the target cart and embedded ball immediately after impact.

(d) The total horizontal momentum of the launcher, ball, and target cart after the collision.

(e) The kinetic energy converted into other forms during the collision.

Explain why treating the given 5.00 m/s as the ball’s ground-frame speed produces a plausible but incorrect result.

Assume negligible horizontal external impulse throughout, and that the launcher does not collide with the target cart.

Solution 5

The launcher finishes moving left at 0.833 m/s. The ball travels right at 4.17 m/s relative to the ground, and the target cart with the embedded ball moves right at 0.833 m/s after impact. Total momentum remains zero, and 1.39 J of kinetic energy is converted into other forms during the collision.

This problem requires two separate applications of conservation of momentum.

The first occurs when the ball leaves the launcher. The second occurs when the ball embeds in the target cart.

(a) Calculate the launcher’s velocity after firing.

Initially, the complete system is stationary.

Take right as positive.

Let:

• \(v_L\) = launcher’s velocity after firing.
• \(v_p\) = projectile’s ground-frame velocity.

The relative velocity condition is:

\[ v_p-v_L=5.00 \]

Therefore:

\[ v_p=v_L+5.00 \]

Apply conservation of momentum to the launcher and projectile:

\[ 0=Mv_L+mv_p \]

Substitute:

\[ 0=1.00v_L+0.200(v_L+5.00) \]

Expand:

\[ 0=1.20v_L+1.00 \]

Rearrange:

\[ v_L=-\frac{1.00}{1.20} \]

\[ \boxed{v_L=-0.833\ \mathrm{m\,s^{-1}}} \]

The launcher moves left at 0.833 m/s.

(b) Calculate the ball’s ground-frame velocity.

Using the relative velocity equation:

\[ v_p=v_L+5.00 \]

\[ v_p=-0.833+5.00 \]

\[ \boxed{v_p=+4.17\ \mathrm{m\,s^{-1}}} \]

The ball travels towards the target at 4.17 m/s relative to the ground.

This is slower than its stated relative launch speed because the launcher is moving backwards.

(c) Calculate the target cart’s velocity after impact.

Now consider a different system: the ball and target cart.

Immediately before impact:

• The ball has mass 0.200 kg and velocity +4.17 m/s.
• The target cart has mass 0.800 kg and is stationary.

The ball embeds in the cart, so the collision is perfectly inelastic.

The combined mass after impact is:

\[ m_{\text{combined}}=0.200+0.800 \]

\[ m_{\text{combined}}=1.00\ \mathrm{kg} \]

Let \(v_T\) be the final velocity of the combined target and ball.

Conservation of momentum gives:

\[ mv_p+Mv_T^{\text{initial}} =(m+M_T)v_T \]

Since the target is initially stationary:

\[ (0.200)(4.17)+(0.800)(0) =(1.00)v_T \]

Therefore:

\[ \boxed{v_T=+0.833\ \mathrm{m\,s^{-1}}} \]

The target cart and embedded ball move right at 0.833 m/s.

(d) Verify total momentum after impact.

The launcher continues moving left at its previous velocity. The collision occurs between the ball and target cart, not the launcher.

The launcher’s momentum is:

\[ p_L=(1.00)(-0.833) \]

\[ p_L=-0.833\ \mathrm{kg\,m\,s^{-1}} \]

The combined target and ball momentum is:

\[ p_T=(1.00)(+0.833) \]

\[ p_T=+0.833\ \mathrm{kg\,m\,s^{-1}} \]

Total momentum:

\[ p_{\text{total}}=-0.833+0.833 \]

\[ \boxed{p_{\text{total}}=0} \]

The complete system still has zero total momentum.

Notice something interesting: the launcher and the combined target group have equal masses of 1.00 kg. Since their momenta are equal and opposite, they also have equal and opposite velocities.

This equality is a consequence of the particular masses in this question, not a general rule for recoil problems.

(e) Calculate the kinetic energy converted during impact.

The launcher is unaffected by the collision, so its kinetic energy does not change.

Immediately before impact, the projectile’s kinetic energy is:

\[ E_{k,p}=\frac12mv_p^2 \]

\[ E_{k,p}=\frac12(0.200)(4.17)^2 \]

\[ E_{k,p}\approx1.736\ \mathrm{J} \]

Immediately after impact, the combined target and ball have kinetic energy:

\[ E_{k,T}=\frac12(m+M_T)v_T^2 \]

\[ E_{k,T}=\frac12(1.00)(0.833)^2 \]

\[ E_{k,T}\approx0.347\ \mathrm{J} \]

The kinetic energy converted into other forms during impact is:

\[ \Delta E_k=1.736-0.347 \]

\[ \boxed{\Delta E_k\approx1.39\ \mathrm{J}} \]

The collision conserves momentum but does not conserve kinetic energy.

Why the tempting calculation fails

A student might treat the given 5.00 m/s as the ball’s ground-frame speed.

Using the basic recoil equation would give:

\[ V=\frac{(0.200)(5.00)}{1.00} \]

\[ V=1.00\ \mathrm{m\,s^{-1}} \]

That result looks reasonable. It even produces equal and opposite momentum for the launcher and projectile.

But it does not satisfy the actual information in the question.

If the projectile moved right at 5.00 m/s relative to the ground while the launcher moved left at 1.00 m/s, their relative speed would be:

\[ u=5.00-(-1.00) \]

\[ u=6.00\ \mathrm{m\,s^{-1}} \]

The question specifies 5.00 m/s, not 6.00 m/s.

The incorrect calculation satisfies momentum conservation but violates the relative velocity condition.

That’s the crucial distinction: a valid solution must satisfy every physical condition in the question, not just one equation.

Once you’ve calculated recoil, the next useful step is to follow the projectile in the ground frame.

Its velocity immediately after separation provides the initial velocity for projectile motion. From there, you can resolve its components, account for gravitational acceleration, and calculate its later velocity.

For those next steps, see Projectile Initial Velocity Components for HSC Physics and Final Velocity of a Projectile: HSC Physics Guide.