Why Moving Your Hands Back Reduces Force When Catching a Ball

A 0.15 kg ball reaches your hands at 20 m s\(^{-1}\). You need to stop it. If you keep your hands rigid, the ball stops quickly. If you let your hands move backwards with the ball, the stopping takes longer. The second catch produces a smaller average force because the same change in momentum is spread over a longer time.

But here’s something worth predicting before we go further: does moving your hands backwards reduce the impulse delivered to the ball?

No. If the ball has the same initial velocity and is brought to rest in both catches, its change in momentum is identical. What changes is how long the force acts, and how that force is distributed throughout the collision.

01Why moving your hands backwards reduces average force

Imagine catching a ball travelling horizontally towards you.

In both catches, the ball starts with the same velocity and finishes at rest. The difference is what your hands do after contact.

• Rigid catch: Your hands barely move. The ball is stopped over a short time.
• Drawn-back catch: Your hands travel backwards in the direction of the incoming ball. The ball takes longer to stop.

Your hands act like a braking system. A sudden, hard stop requires a large force over a short time. A gentler stop can deliver the same change in momentum over a longer time.

The momentum change does not disappear

Let’s establish a sign convention.

Take the positive horizontal direction as the direction in which the ball initially travels.

For a ball of mass \(m\), the initial velocity is \(v_i\), and its final velocity is \(v_f\).

The impulse delivered to the ball is equal to its change in momentum:

\[ J=\Delta p=m(v_f-v_i) \]

Impulse is measured in newton-seconds (N s).

For the 0.15 kg ball travelling at \(+20\text{ m s}^{-1}\), which is brought to rest:

\[ \begin{aligned} J&=m(v_f-v_i)\\ &=0.15(0-20)\\ &=-3.0\text{ N s} \end{aligned} \]

The negative sign tells us that the impulse acts opposite to the ball’s initial motion.

This result is exactly the same whether you catch the ball with rigid hands or move your hands backwards.

The ball still needs an impulse of \(-3.0\text{ N s}\) to go from its initial momentum to zero momentum.

Where the stopping time enters

Impulse is also the signed area under a net force-time graph.

For a varying force, the area represents the total impulse accumulated throughout the interaction.

The average net force is therefore:

\[ F_{\mathrm{net,avg}}=\frac{J}{\Delta t} \]

where:

• \(F_{\mathrm{net,avg}}\) is the average net force, in newtons (N).
• \(J\) is the impulse, in newton-seconds (N s).
• \(\Delta t\) is the duration of the collision, in seconds (s).

For the same momentum change, increasing the stopping time reduces the magnitude of the average force.

This is the entire reason moving your hands backwards helps.

Comparing two catches

Suppose the same 0.15 kg ball arrives at \(20\text{ m s}^{-1}\) and is brought to rest in two different ways.

Consider these illustrative force-time measurements. Both force profiles are assumed to be triangular.

The drawn-back catch takes five times as long.

Consequently, its average force magnitude is five times smaller.

Notice that the impulse remains unchanged. The ball has undergone the same momentum change in both situations.

02Average force and peak force are not the same thing

There is an important distinction here.

The average force tells us the constant force that would produce the same impulse over the entire collision duration.

The peak force is the largest instantaneous force experienced during the collision.

These are different measurements.

For a triangular force-time pulse that begins and ends at zero:

\[ J=\frac{1}{2}F_{\mathrm{peak}}\Delta t \]

Dividing by the collision duration gives:

\[ F_{\mathrm{avg}}=\frac{1}{2}F_{\mathrm{peak}} \]

This explains why the peak force is twice the average force in the triangular example.

However, you must not assume this relationship for every collision.

A rectangular pulse has the same average and peak force. A trapezium-shaped pulse has a different relationship.

Does a longer catch always produce a smaller peak force?

Not necessarily.

A longer stopping time guarantees a smaller average force magnitude if the impulse remains unchanged. It does not, by itself, guarantee a smaller peak force.

Consider two idealised force profiles that both deliver an impulse of 3.0 N s.

• A short catch has a triangular force pulse lasting 0.020 s, with a peak of 300 N.
• A longer catch has a force of 10 N for 0.100 s, followed by a brief 500 N spike lasting 0.004 s.

For the longer catch:

\[ J=(10)(0.100)+(500)(0.004) \]

\[ J=1.0+2.0=3.0\text{ N s} \]

Its total duration is 0.104 s, so:

\[ F_{\mathrm{avg}}=\frac{3.0}{0.104} =28.8\text{ N} \]

Its average force is much smaller than that of the short catch, but its peak force is 500 N, which is actually larger than 300 N.

This is an important limitation of the usual explanation.

Moving your hands backwards generally helps reduce both average and peak force because it allows the collision to happen more smoothly. However, the reduced peak depends on the shape of the force-time curve, not simply the duration.

A sudden hard contact with a finger, a rigid object, or another part of the catching surface could still produce a large peak.

03How could you test this experimentally?

Suppose you wanted to investigate whether moving a catching surface backwards actually reduces the force experienced by a ball.

You need to measure three things:

1. The ball’s change in momentum.
2. The duration of the collision.
3. The force throughout the collision.

Simply asking someone to catch a ball twice and describe which catch felt harder is not enough. The incoming speeds might differ, and human reaction times are not accurate enough to measure a collision lasting a few hundredths of a second.

A controlled apparatus gives much better evidence.

Setting up the experiment

A suitable investigation could use:

• A ball launcher to send identical balls horizontally at a padded catching surface.
• Two light gates near the end of the track to measure the ball’s incoming velocity.
• A force sensor positioned directly behind the padded catching surface.
• A sliding carriage that allows the catching surface to move backwards.
• An adjustable brake that controls the carriage’s motion.
• A data logger connected to the force sensor to record force against time.
• A position sensor to measure the movement of the catching surface.

The ball travels from the launcher, passes through the light gates, and strikes the padded surface.

When the ball makes contact, the surface compresses the force sensor. The sensor converts the applied force into an electrical signal, which travels to the data logger.

The logger records the force throughout the collision.

The apparatus can be tested in two configurations.

Trial A: Locked carriage

The catching surface cannot move backwards. The ball is stopped over a relatively short time.

Trial B: Moving carriage

The catching surface is allowed to move backwards against the adjustable brake. The brake is set so that the ball and catching surface eventually come to rest together.

The same ball, launch speed, and final ball velocity should be used in both trials.

What should the experiment show?

The prediction follows directly from the impulse-momentum relationship.

If the initial and final velocities are identical, the two trials must produce the same impulse.

However, the moving carriage should increase the collision duration.

The force-time graph should therefore have a greater width and a smaller average force magnitude.

For example, an illustrative set of results might be:

These are illustrative values, not measurements from an actual experiment.

The matching impulse values support the prediction that the ball experiences the same overall momentum change in both trials.

The lower average force in the moving-carriage trial is consistent with the longer collision duration.

What could make the experiment unreliable?

There are a few things to watch.

• Different launch speeds: Even a small change in incoming velocity changes the ball’s momentum and therefore the required impulse.
• Different final velocities: If the ball rebounds in one trial but not the other, the momentum changes will differ.
• Sensor sampling rate: If the logger samples too slowly, it might miss a brief peak force.
• Apparatus stiffness: The padded surface and sensor do not perfectly reproduce the behaviour of human hands.

There is also an important distinction between contact force and net force.

The force-time area gives the impulse from the net force acting on the ball. In this horizontal experiment, gravitational force acts vertically, so it does not contribute to the horizontal momentum change. If other horizontal forces are negligible, the measured horizontal contact force is approximately the net horizontal force.

This assumption becomes important in vertical catching problems, where the ball’s weight also contributes.

04Worked examples

05A reliable approach for HSC questions

When a question involves catching a ball with moving hands, work through these decisions in order.

1. Identify the initial and final velocities. Do not assume that the ball is stationary simply because it has been caught.
2. Calculate the change in momentum. This determines the required net impulse.
3. Determine the collision duration. A longer stopping time generally means a smaller average force for the same impulse.
4. Read the force-time graph carefully. The area gives impulse, the average height gives average force, and the highest point gives peak force.
5. Check the forces acting on the ball. For a horizontal catch, weight is perpendicular to the motion. For a vertical catch, weight must be included.

For a more detailed treatment of calculating collision forces directly from velocity changes, see How to Calculate Average Collision Force from Velocity Change.

06Questions and solutions

Question 1

A 0.12 kg ball travels horizontally towards a student at \(+10\text{ m s}^{-1}\). The student catches it and brings it to rest over 0.040 s by moving their hands backwards.

Take the positive direction as the ball’s initial direction of motion.

Calculate:

(a) The impulse delivered to the ball.

(b) The average net force acting on the ball.

(c) The direction of the average force.

Assume other horizontal forces are negligible.

Solution 1

The impulse is \(-1.2\text{ N s}\), and the average net force is \(-30\text{ N}\), acting opposite to the ball’s initial motion.

Part (a): Calculate the impulse.

Use the change in momentum:

\[ J=m(v_f-v_i) \]

Substitute the given values:

\[ \begin{aligned} J&=0.12(0-10)\\ &=-1.2\text{ N s} \end{aligned} \]

Part (b): Calculate the average net force.

Use:

\[ F_{\mathrm{net,avg}}=\frac{J}{\Delta t} \]

Substitute:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{-1.2}{0.040}\\ &=-30\text{ N} \end{aligned} \]

Part (c): Interpret the direction.

The negative sign indicates that the average net force acts opposite to the ball’s initial direction.

The student’s hands exert a backward force on the ball, reducing its forward momentum until it reaches rest.

Question 2

A force sensor records the magnitude of the backward net force acting on a 0.14 kg ball during a horizontal catch.

The ball initially travels at \(+10\text{ m s}^{-1}\) and is brought to rest.

The sensor produces the following readings. Assume that the force changes linearly between consecutive measurements.

Calculate:

(a) The impulse magnitude using the trapezium rule.

(b) The average net force acting on the ball, including its direction.

(c) The peak force magnitude.

(d) Verify your impulse calculation using the change in momentum.

Solution 2

The impulse magnitude is \(1.40\text{ N s}\), the average net force is \(-35\text{ N}\), and the peak force magnitude is 60 N.

The impulse agrees with the ball’s calculated change in momentum.

Part (a): Calculate the impulse from the force-time readings.

First, convert the time intervals into seconds.

\[ 10\text{ ms}=0.010\text{ s} \]

Each interval forms a trapezium.

The area of a trapezium is:

\[ A=\frac12(F_1+F_2)\Delta t \]

Calculate the four areas separately.

From 0 to 10 ms:

\[ A_1=\frac12(0+40)(0.010) \]

\[ A_1=0.20\text{ N s} \]

From 10 to 20 ms:

\[ A_2=\frac12(40+60)(0.010) \]

\[ A_2=0.50\text{ N s} \]

From 20 to 30 ms:

\[ A_3=\frac12(60+40)(0.010) \]

\[ A_3=0.50\text{ N s} \]

From 30 to 40 ms:

\[ A_4=\frac12(40+0)(0.010) \]

\[ A_4=0.20\text{ N s} \]

Add the areas:

\[ \begin{aligned} |J|&=A_1+A_2+A_3+A_4\\ &=0.20+0.50+0.50+0.20\\ &=1.40\text{ N s} \end{aligned} \]

The force acts backwards, so the signed impulse is \(-1.40\text{ N s}\).

Part (b): Calculate the average net force.

The total collision duration is:

\[ \Delta t=0.040\text{ s} \]

Therefore:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{-1.40}{0.040}\\ &=-35\text{ N} \end{aligned} \]

The average net force is 35 N opposite to the ball’s initial direction.

Part (c): Find the peak force.

The peak force is the largest recorded force magnitude.

\[ F_{\mathrm{peak}}=60\text{ N} \]

It is greater than the average force magnitude of 35 N.

Part (d): Check using momentum.

The ball is brought to rest, so:

\[ v_i=+10\text{ m s}^{-1} \]

\[ v_f=0 \]

Calculate the change in momentum:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.14(0-10)\\ &=-1.40\text{ N s} \end{aligned} \]

This matches the signed impulse calculated from the force-time graph.

The two methods give the same result, as required by the impulse-momentum theorem.

Question 3

A 0.16 kg ball approaches a student’s hands horizontally at \(+15\text{ m s}^{-1}\).

The student performs two different catches.

• Catch A: The ball is brought to rest in 0.080 s.
• Catch B: The ball rebounds from the hands at \(-3.0\text{ m s}^{-1}\), with the collision lasting 0.060 s.

Calculate the impulse and average net force for each catch.

Determine which catch produces the greater average force magnitude.

Explain why the rebound in Catch B changes the impulse, even though the ball’s rebound speed is smaller than its initial speed.

Solution 3

Catch A produces an impulse of \(-2.40\text{ N s}\) and an average force of \(-30\text{ N}\).

Catch B produces an impulse of \(-2.88\text{ N s}\) and an average force of \(-48\text{ N}\).

Catch B produces the greater average force magnitude.

Catch A: Ball brought to rest

Calculate the impulse:

\[ \begin{aligned} J_A&=m(v_f-v_i)\\ &=0.16(0-15)\\ &=-2.40\text{ N s} \end{aligned} \]

Calculate the average force:

\[ \begin{aligned} F_{A,\mathrm{avg}} &=\frac{-2.40}{0.080}\\ &=-30\text{ N} \end{aligned} \]

Catch B: Ball rebounds

This is where sign conventions become important.

The final velocity is negative because the ball moves in the opposite direction after rebounding.

Therefore:

\[ v_i=+15\text{ m s}^{-1} \]

\[ v_f=-3.0\text{ m s}^{-1} \]

Calculate the impulse:

\[ \begin{aligned} J_B&=m(v_f-v_i)\\ &=0.16(-3.0-15)\\ &=0.16(-18)\\ &=-2.88\text{ N s} \end{aligned} \]

Calculate the average force:

\[ \begin{aligned} F_{B,\mathrm{avg}} &=\frac{-2.88}{0.060}\\ &=-48\text{ N} \end{aligned} \]

Compare the two catches.

Catch B produces a larger impulse because the ball reverses direction.

A common mistake is to calculate the change in speed as \(15-3=12\text{ m s}^{-1}\).

That ignores the reversal of direction.

The actual change in velocity is:

\[ \Delta v=-3.0-(+15)=-18\text{ m s}^{-1} \]

The ball therefore experiences a larger momentum change in Catch B.

Its shorter collision duration also increases the average force.

Both effects contribute to the larger average force magnitude.

Question 4

A 0.25 kg ball falls vertically downwards at \(8.0\text{ m s}^{-1}\) and is caught by a student.

The ball comes to rest over 0.10 s.

Take upwards as the positive direction and use \(g=9.8\text{ m s}^{-2}\).

(a) Calculate the impulse and average net force acting on the ball.

(b) Calculate the average upward force exerted by the student’s hands.

(c) The student repeats the catch but moves their hands downwards, increasing the stopping time to 0.20 s. The ball still finishes at rest. Calculate the new average net force and average force exerted by the hands.

Explain why the hand force is not equal to the average net force.

Solution 4

For the original catch, the impulse is \(+2.0\text{ N s}\), the average net force is 20 N upwards, and the average force from the hands is approximately 22.45 N upwards.

When the stopping time doubles, the average net force falls to 10 N upwards, and the average hand force falls to approximately 12.45 N upwards.

Part (a): Calculate the impulse and average net force.

Upwards is positive, so the initial velocity is negative.

\[ v_i=-8.0\text{ m s}^{-1} \]

The final velocity is zero.

Calculate the impulse:

\[ \begin{aligned} J&=m(v_f-v_i)\\ &=0.25(0-(-8.0))\\ &=+2.0\text{ N s} \end{aligned} \]

Calculate the average net force:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{2.0}{0.10}\\ &=+20\text{ N} \end{aligned} \]

The average net force acts upwards.

Part (b): Calculate the average force from the hands.

This question requires you to account for gravity.

Two vertical forces act on the ball:

• The upward force from the student’s hands.
• The downward gravitational force, equal to \(mg\).

Using Newton’s second law for the average forces:

\[ F_{\mathrm{net,avg}}=F_{\mathrm{hands,avg}}-mg \]

Rearrange:

\[ F_{\mathrm{hands,avg}} =F_{\mathrm{net,avg}}+mg \]

Substitute:

\[ \begin{aligned} F_{\mathrm{hands,avg}} &=20+(0.25\times9.8)\\ &=20+2.45\\ &=22.45\text{ N} \end{aligned} \]

The average upward force from the hands is 22.45 N.

The net force is smaller because gravity acts in the opposite direction.

Part (c): Increase the stopping time to 0.20 s.

The ball has the same initial and final velocities, so the impulse remains unchanged:

\[ J=+2.0\text{ N s} \]

Calculate the new average net force:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{2.0}{0.20}\\ &=+10\text{ N} \end{aligned} \]

Now calculate the average force from the hands:

\[ \begin{aligned} F_{\mathrm{hands,avg}} &=F_{\mathrm{net,avg}}+mg\\ &=10+2.45\\ &=12.45\text{ N} \end{aligned} \]

The average upward hand force has decreased from 22.45 N to 12.45 N.

The important distinction is that the impulse-momentum theorem relates the change in momentum to the net impulse, not just the impulse from one individual force.

Gravity continues to act throughout the collision.

Question 5

A 0.20 kg ball travels horizontally towards a catcher at \(+12\text{ m s}^{-1}\).

The catcher moves their hands backwards as the ball makes contact.

After the first 0.10 s, the ball is stationary relative to the catcher’s hands. However, both the ball and hands are still moving together at \(+4.0\text{ m s}^{-1}\) relative to the floor.

There are now two possible outcomes.

Outcome A: The catcher continues moving forwards at a constant \(4.0\text{ m s}^{-1}\), holding the ball securely.

Outcome B: Instead, the catcher decelerates and brings both the ball and hands to rest over a further 0.20 s.

Assume:

• The ball remains in contact with the hands throughout the interaction.
• Horizontal forces other than the hand-ball contact force are negligible.
• All velocities are measured relative to the floor.
• The positive direction is the ball’s original direction of motion.

Answer the following:

(a) Calculate the impulse and average net force acting on the ball during the first 0.10 s.

(b) For Outcome A, calculate the total impulse received by the ball from initial contact until it is held securely. Explain whether the ball has actually stopped.

(c) For Outcome B, calculate the impulse and average net force during the second stage.

(d) Calculate the total impulse, total interaction time, and overall average net force for Outcome B.

(e) A student claims that the impulse during the first 0.10 s must be \(-2.4\text{ N s}\), because the ball has been caught. Explain the exact error in this reasoning.

(f) Compare Outcome B with a rigid catch that brings the same ball directly from \(+12\text{ m s}^{-1}\) to rest in 0.040 s. Explain why the overall average forces are different even though the ball eventually reaches the same final velocity.

Solution 5

The first stage delivers an impulse of \(-1.6\text{ N s}\), not \(-2.4\text{ N s}\).

If the catcher continues moving at \(+4.0\text{ m s}^{-1}\), that is the total impulse received by the ball.

If the catcher subsequently brakes to rest, an additional impulse of \(-0.8\text{ N s}\) is delivered. The total impulse then becomes \(-2.4\text{ N s}\), with an overall average net force of \(-8.0\text{ N}\).

The central issue is the difference between being stationary relative to the hands and being stationary relative to the floor.

Part (a): Calculate the impulse during the first stage.

The ball initially travels at:

\[ v_i=+12\text{ m s}^{-1} \]

At the end of the first stage, it travels at:

\[ v_f=+4.0\text{ m s}^{-1} \]

It has stopped moving relative to the hands, but it is still moving relative to the floor.

Calculate the change in momentum:

\[ \begin{aligned} J_1&=m(v_f-v_i)\\ &=0.20(4.0-12)\\ &=0.20(-8.0)\\ &=-1.6\text{ N s} \end{aligned} \]

Calculate the average net force:

\[ \begin{aligned} F_{\mathrm{net,avg,1}} &=\frac{-1.6}{0.10}\\ &=-16\text{ N} \end{aligned} \]

The average net force during the first stage is 16 N opposite to the ball’s original direction.

Part (b): Analyse Outcome A.

The catcher continues moving at a constant velocity of \(+4.0\text{ m s}^{-1}\).

The ball is already moving at this velocity.

Therefore, no further horizontal impulse is required to keep the ball moving with the catcher.

The total impulse received by the ball is:

\[ J_{\mathrm{total}}=-1.6\text{ N s} \]

The ball has been caught in the everyday sense because there is no longer any motion between the ball and the hands.

However, it has not stopped relative to the floor.

Its remaining momentum is:

\[ \begin{aligned} p&=mv\\ &=0.20(4.0)\\ &=0.80\text{ kg m s}^{-1} \end{aligned} \]

This is a crucial distinction.

Being caught does not necessarily mean having zero momentum in the ground frame.

Part (c): Calculate the impulse and average force during the second stage of Outcome B.

In Outcome B, the catcher now brings the ball and hands to rest.

The initial velocity for this second stage is:

\[ v_i=+4.0\text{ m s}^{-1} \]

The final velocity is:

\[ v_f=0 \]

Calculate the additional impulse:

\[ \begin{aligned} J_2&=m(v_f-v_i)\\ &=0.20(0-4.0)\\ &=-0.80\text{ N s} \end{aligned} \]

The second stage lasts 0.20 s.

Therefore:

\[ \begin{aligned} F_{\mathrm{net,avg,2}} &=\frac{-0.80}{0.20}\\ &=-4.0\text{ N} \end{aligned} \]

The average net force during the second stage is 4.0 N backwards.

Part (d): Calculate the total impulse and average force for Outcome B.

The total impulse is the sum of the impulses from both stages.

\[ \begin{aligned} J_{\mathrm{total}}&=J_1+J_2\\ &=-1.6+(-0.80)\\ &=-2.4\text{ N s} \end{aligned} \]

This also agrees with the overall change in momentum:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.20(0-12)\\ &=-2.4\text{ N s} \end{aligned} \]

The total interaction time is:

\[ \Delta t_{\mathrm{total}}=0.10+0.20 \]

\[ \Delta t_{\mathrm{total}}=0.30\text{ s} \]

Calculate the overall average net force:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{J_{\mathrm{total}}}{\Delta t_{\mathrm{total}}}\\ &=\frac{-2.4}{0.30}\\ &=-8.0\text{ N} \end{aligned} \]

The overall average net force is 8.0 N opposite to the original direction of motion.

Notice that you cannot simply take the ordinary arithmetic average of the two stage-average forces.

The overall average must account for both the impulse and the duration of each stage.

Part (e): Identify the student’s mistake.

The student’s error is assuming that the ball’s final velocity is zero after the first 0.10 s.

That is not true in the floor frame.

The ball is stationary relative to the hands, but both are still travelling at \(+4.0\text{ m s}^{-1}\).

The correct impulse for the first stage is therefore:

\[ J_1=0.20(4.0-12) \]

\[ J_1=-1.6\text{ N s} \]

The remaining \(-0.80\text{ N s}\) is delivered when the catcher subsequently brakes.

This distinction matters whenever a question involves moving objects or a moving reference point. Always identify the reference frame before deciding that an object has stopped.

Part (f): Compare with a rigid catch.

For the rigid catch, the ball is brought directly from \(+12\text{ m s}^{-1}\) to zero in 0.040 s.

The impulse is still:

\[ J=-2.4\text{ N s} \]

The average net force is:

\[ \begin{aligned} F_{\mathrm{net,avg}} &=\frac{-2.4}{0.040}\\ &=-60\text{ N} \end{aligned} \]

Compare the results:

Both catches produce exactly the same overall momentum change.

However, the drawn-back catch spreads that change over a much longer time.

Its average net force magnitude is 7.5 times smaller.

The important lesson is that you must compare the complete interaction, not just the moment when the ball first becomes stationary relative to the catcher’s hands.

The same reasoning is useful when evaluating airbags, crumple zones, and protective padding. In each case, the momentum change determines the required impulse, while the duration and shape of the force-time interaction determine the average and peak forces.

For the next step, explore How to Compare Airbag Designs Using Force-Time Graphs, where the same principles are applied to comparing different impact-protection systems.