How to Include Weight in a Vertical Impact-Force Calculation

A 0.20 kg ball hits a floor travelling downwards at 5.0 m s\(^{-1}\). It comes to rest in 0.025 s. If you calculate the average force using \(F=\Delta p/\Delta t\), you get 40 N. Is that the force exerted by the floor?

Not quite. That’s the average resultant force on the ball. The floor must also overcome the ball’s weight while bringing it to rest.

With upwards taken as positive, the average contact force is:

\[ F_{\text{contact,avg}} =\frac{m\Delta v}{\Delta t}+mg \]

\[ =40+(0.20)(9.8) =41.96\ \mathrm N \]

So the average force exerted by the floor is approximately 42.0 N upwards.

The key is understanding where that extra \(mg\) comes from. It is not an arbitrary correction. It comes directly from the impulse of gravity during the collision.

01Why you need to include weight

Imagine a ball hitting a horizontal floor.

There are two vertical forces acting on the ball during contact:

• The normal contact force \(N\), exerted by the floor, acts upwards.
• The weight \(mg\), caused by gravity, acts downwards.

The ball responds to the resultant of these forces, not to the contact force alone.

Taking upwards as the positive direction, Newton’s second law gives:

\[ F_{\text{net}}=N-mg=ma \]

Rearranging:

\[ N=ma+mg \]

This is the important distinction. The acceleration tells you about the resultant force. It does not directly tell you the force exerted by the floor.

For a collision, however, it is usually easier to work with momentum and impulse than acceleration.

Using impulse to include weight

The impulse-momentum theorem states:

\[ J_{\text{net}}=\Delta p \]

The net impulse is the sum of the impulses from all forces acting during contact.

For a ball striking a horizontal floor:

\[ J_N+J_W=\Delta p \]

Taking upwards as positive:

\[ J_N-mg\Delta t=\Delta p \]

Therefore:

\[ \boxed{J_N=\Delta p+mg\Delta t} \]

Here:

• \(J_N\) is the impulse from the floor, in N s.
• \(m\) is the object’s mass, in kg.
• \(g\) is gravitational acceleration, in m s\(^{-2}\).
• \(\Delta t\) is the contact time, in seconds.
• \(\Delta p=m(v_f-v_i)\) is the signed change in momentum.

To find the average contact force, divide the contact impulse by the contact time:

\[ \boxed{ N_{\text{avg}}= \frac{m(v_f-v_i)}{\Delta t}+mg } \]

This is the main equation for a vertical impact against a horizontal floor, with upwards positive.

What if the ball rebounds?

Suppose a ball approaches the floor at speed \(u\) downwards and rebounds at speed \(v\) upwards.

Its initial and final velocities are:

\[ v_i=-u,\qquad v_f=+v \]

Therefore:

\[ \Delta p=m(v-(-u)) \]

\[ \Delta p=m(u+v) \]

Substituting into the average contact force equation:

\[ \boxed{ N_{\text{avg}}=\frac{m(u+v)}{\Delta t}+mg } \]

Notice that you add the speeds when calculating the momentum change because the ball reverses direction.

If the ball simply stops, its final velocity is zero, so the equation becomes:

\[ \boxed{ N_{\text{avg}}=\frac{mu}{\Delta t}+mg } \]

These equations assume that the contact force from the floor acts upwards, gravity acts downwards, and the contact interval is correctly identified.

02Understanding force-time graphs

Force-time graphs are particularly useful in impact questions because they show how the force changes throughout a collision.

The most important feature is the area under the graph.

The area under a force-time graph represents impulse:

\[ J=\text{area under the force-time graph} \]

But there is a detail you need to get right: the area depends on which force the graph represents.

Contact force versus net force

Consider a graph showing the upward contact force \(N\) exerted by the floor.

The area under this graph represents the impulse from the floor:

\[ J_N=\text{area under the contact-force graph} \]

Gravity also acts during the collision. Its impulse is:

\[ J_W=-mg\Delta t \]

Therefore, the net impulse is:

\[ \boxed{\Delta p=J_N-mg\Delta t} \]

This means the area under the contact-force graph is not equal to the change in momentum by itself.

The relationship is:

There are two different situations you need to recognise in an exam:

Situation 1: The graph shows contact force.

You must account for the impulse of weight when relating the graph’s area to the change in momentum.

\[ J_N=\Delta p+mg\Delta t \]

Situation 2: The graph shows resultant force.

Its area already represents the change in momentum.

\[ J_{\text{net}}=\Delta p \]

You do not add weight again. Doing so would count gravity twice.

Always read the vertical axis before calculating the area.

03Average force versus peak force

Another common mistake is confusing average impact force with peak impact force.

They are not necessarily equal.

The average contact force is calculated from the total contact impulse:

\[ N_{\text{avg}}=\frac{J_N}{\Delta t} \]

The peak contact force is the greatest instantaneous force reached during the collision.

It depends on the shape of the force-time graph.

Consider two possible force-time graphs.

For a rectangular force-time graph, the force remains constant:

\[ N_{\text{peak}}=N_{\text{avg}} \]

For a triangular force-time graph that starts and ends at zero:

\[ J_N=\frac{1}{2}N_{\text{peak}}\Delta t \]

Rearranging:

\[ \boxed{N_{\text{peak}}=2N_{\text{avg}}} \]

This factor of two only applies to that triangular shape.

For a more complicated graph, you must use its actual geometry to calculate the peak. Knowing the average force alone does not tell you the peak force.

Important: If a question gives you momentum change and contact time but does not provide the force-time graph or a model for its shape, you can calculate the average force, but you cannot determine the peak force uniquely.

04How significant is the weight correction?

You might be wondering whether weight is large enough to make a noticeable difference during such a short collision.

Often, the correction is relatively small. However, it is still part of the correct force calculation.

For an object moving downwards and coming to rest, the momentum-change contribution to average force is:

\[ F_{\text{net,avg}}=\frac{mu}{\Delta t} \]

The ratio of the weight correction to this contribution is:

\[ \frac{mg}{mu/\Delta t} =\frac{g\Delta t}{u} \]

For example, consider an object travelling downwards at 5.0 m s\(^{-1}\), with a contact time of 0.020 s.

\[ \frac{g\Delta t}{u} = \frac{9.8(0.020)}{5.0} =0.0392 \]

The weight correction is approximately 3.9% of the momentum-change contribution.

That might seem small, but the distinction matters when a question specifically asks for the force exerted by the floor rather than the resultant force.

05Worked examples

06Questions and solutions

Question 1

A 0.40 kg training ball is travelling vertically downwards at 3.5 m s\(^{-1}\). It strikes a horizontal floor and comes to rest in 0.050 s.

Calculate the average force exerted by the floor on the ball.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 1

The average force exerted by the floor is 31.9 N upwards.

The first step is to calculate the change in momentum.

Take upwards as positive:

\[ v_i=-3.5\ \mathrm{m\,s^{-1}} \]

\[ v_f=0 \]

Therefore:

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

\[ \Delta p=0.40[0-(-3.5)] \]

\[ \Delta p=1.40\ \mathrm{N\,s} \]

Now calculate the weight impulse:

\[ J_W=-mg\Delta t \]

\[ J_W=-(0.40)(9.8)(0.050) \]

\[ J_W=-0.196\ \mathrm{N\,s} \]

The impulse-momentum theorem gives:

\[ J_N+J_W=\Delta p \]

Therefore:

\[ J_N=1.40-(-0.196) \]

\[ J_N=1.596\ \mathrm{N\,s} \]

Finally, calculate the average contact force:

\[ N_{\text{avg}}=\frac{1.596}{0.050} \]

\[ \boxed{N_{\text{avg}}=31.9\ \mathrm N} \]

The floor must provide a greater upward impulse than the ball’s momentum change because gravity continues to act downwards during contact.

Question 2

A 0.25 kg ball is travelling downwards at 4.0 m s\(^{-1}\) when it strikes a horizontal force plate.

The ball remains in contact with the plate for 0.060 s.

A force sensor records an upward contact-force impulse of 1.50 N s. This is the total area under the contact-force-time graph.

Calculate the ball’s velocity immediately after contact.

Take upwards as positive and \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 2

The ball leaves the force plate travelling 1.41 m s\(^{-1}\) upwards.

The important detail is that the given 1.50 N s is the contact impulse, not the net impulse.

Step 1: Identify the known values.

\[ m=0.25\ \mathrm{kg} \]

\[ v_i=-4.0\ \mathrm{m\,s^{-1}} \]

\[ J_N=1.50\ \mathrm{N\,s} \]

\[ \Delta t=0.060\ \mathrm s \]

Step 2: Calculate the impulse from weight.

\[ J_W=-mg\Delta t \]

\[ J_W=-(0.25)(9.8)(0.060) \]

\[ J_W=-0.147\ \mathrm{N\,s} \]

Step 3: Calculate the net impulse.

\[ J_{\text{net}}=J_N+J_W \]

\[ J_{\text{net}}=1.50-0.147 \]

\[ \boxed{J_{\text{net}}=1.353\ \mathrm{N\,s}} \]

This is equal to the change in momentum:

\[ \Delta p=1.353\ \mathrm{N\,s} \]

Step 4: Calculate the change in velocity.

Using:

\[ \Delta p=m\Delta v \]

\[ \Delta v=\frac{1.353}{0.25} \]

\[ \Delta v=5.412\ \mathrm{m\,s^{-1}} \]

Step 5: Calculate the final velocity.

\[ v_f=v_i+\Delta v \]

\[ v_f=-4.0+5.412 \]

\[ \boxed{v_f=+1.41\ \mathrm{m\,s^{-1}}} \]

The positive sign means the ball is travelling upwards.

A tempting mistake would be to use the full 1.50 N s as the net impulse. That would ignore the downward impulse of gravity and produce an incorrect final velocity of 2.0 m s\(^{-1}\) upwards.

Question 3

A 0.20 kg ball enters two different horizontal impact mats at 5.0 m s\(^{-1}\) downwards. It rebounds from each mat at 3.0 m s\(^{-1}\) upwards.

Both mats produce triangular contact-force-time graphs, starting and ending at zero force.

The contact durations are:

Calculate the average and peak contact forces for each mat.

Does doubling the contact time halve the peak force exactly?

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 3

Mat A produces an average force of 66.0 N and a peak force of 132 N. Mat B produces an average force of 34.0 N and a peak force of 67.9 N.

Doubling the contact time does not halve the peak force exactly because the impulse from weight also increases with contact time.

Step 1: Calculate the momentum change.

Take upwards as positive.

\[ v_i=-5.0\ \mathrm{m\,s^{-1}} \]

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

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

\[ \Delta p=0.20[3.0-(-5.0)] \]

\[ \boxed{\Delta p=1.60\ \mathrm{N\,s}} \]

Both mats produce the same momentum change.

Step 2: Calculate the contact impulse for Mat A.

\[ J_A=\Delta p+mg\Delta t_A \]

\[ J_A=1.60+(0.20)(9.8)(0.025) \]

\[ \boxed{J_A=1.649\ \mathrm{N\,s}} \]

Step 3: Calculate the average and peak forces for Mat A.

The average force is:

\[ N_{\text{avg,A}}=\frac{1.649}{0.025} \]

\[ \boxed{N_{\text{avg,A}}=65.96\ \mathrm N} \]

Because the graph is triangular:

\[ N_{\text{peak,A}}=2N_{\text{avg,A}} \]

\[ \boxed{N_{\text{peak,A}}=131.92\ \mathrm N} \]

Step 4: Calculate the contact impulse for Mat B.

\[ J_B=\Delta p+mg\Delta t_B \]

\[ J_B=1.60+(0.20)(9.8)(0.050) \]

\[ \boxed{J_B=1.698\ \mathrm{N\,s}} \]

Step 5: Calculate the average and peak forces for Mat B.

\[ N_{\text{avg,B}}=\frac{1.698}{0.050} \]

\[ \boxed{N_{\text{avg,B}}=33.96\ \mathrm N} \]

For the triangular graph:

\[ N_{\text{peak,B}}=2(33.96) \]

\[ \boxed{N_{\text{peak,B}}=67.92\ \mathrm N} \]

Step 6: Compare the results.

Mat B has approximately half the peak force of Mat A, but not exactly.

Why?

The momentum change is identical, but the ball experiences gravity for twice as long during Mat B’s collision.

Consequently, the contact impulse required from Mat B is slightly larger.

If you ignored weight, you would calculate peak forces of 128 N and 64 N, which are exactly half of each other. Including weight changes that relationship.

Question 4

A 0.30 kg ball is released from rest at a height of 1.25 m above a horizontal floor.

After striking the floor, it rebounds to a maximum height of 0.45 m.

The ball is in contact with the floor for 0.040 s.

Calculate the average upward contact force exerted by the floor during the collision.

Assume air resistance is negligible and take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 4

The average upward contact force is 62.3 N.

This question requires two stages of reasoning: first, determine the velocities immediately before and after impact; then use impulse to calculate the contact force.

Step 1: Calculate the speed immediately before impact.

The ball falls from rest through 1.25 m.

Using the vertical motion equation:

\[ v^2=u^2+2as \]

With \(u=0\) and \(a=g\):

\[ v^2=2gh \]

\[ v=\sqrt{2(9.8)(1.25)} \]

\[ \boxed{v_{\text{before}}=4.95\ \mathrm{m\,s^{-1}}} \]

The ball is travelling downwards.

Step 2: Calculate the speed immediately after impact.

The ball rises to a maximum height of 0.45 m.

At its highest point, its velocity is zero.

Using the same vertical motion equation:

\[ 0=u^2-2gh \]

Rearranging:

\[ u=\sqrt{2gh} \]

\[ u=\sqrt{2(9.8)(0.45)} \]

\[ \boxed{v_{\text{after}}=2.97\ \mathrm{m\,s^{-1}}} \]

The ball is travelling upwards immediately after contact.

Step 3: Calculate the change in momentum.

Take upwards as positive.

\[ v_i=-4.95\ \mathrm{m\,s^{-1}} \]

\[ v_f=+2.97\ \mathrm{m\,s^{-1}} \]

Therefore:

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

\[ \Delta p=0.30[2.97-(-4.95)] \]

\[ \boxed{\Delta p=2.376\ \mathrm{N\,s}} \]

Step 4: Calculate the impulse from weight.

\[ J_W=-mg\Delta t \]

\[ J_W=-(0.30)(9.8)(0.040) \]

\[ \boxed{J_W=-0.1176\ \mathrm{N\,s}} \]

Step 5: Calculate the contact impulse.

Using the impulse-momentum theorem:

\[ J_N+J_W=\Delta p \]

\[ J_N=2.376-(-0.1176) \]

\[ \boxed{J_N=2.4936\ \mathrm{N\,s}} \]

Step 6: Calculate the average contact force.

\[ N_{\text{avg}}=\frac{J_N}{\Delta t} \]

\[ N_{\text{avg}}=\frac{2.4936}{0.040} \]

\[ \boxed{N_{\text{avg}}=62.3\ \mathrm N} \]

The floor must provide enough upward impulse to reverse the ball’s momentum and account for the downward impulse of gravity during contact.

Question 5

A 0.20 kg rubber ball undergoes two different vertical collisions.

In Trial A, it strikes a horizontal floor while travelling downwards at 4.0 m s\(^{-1}\). It rebounds upwards at 3.0 m s\(^{-1}\).

In Trial B, the same ball is thrown vertically upwards and strikes a horizontal ceiling at 4.0 m s\(^{-1}\). It rebounds downwards at 3.0 m s\(^{-1}\).

Both collisions last 0.050 s.

Assume that the contact-force-time graph is triangular in both trials, starting and ending at zero contact force.

For each trial:

a) Calculate the magnitude of the average contact force.

b) Calculate the peak contact force.

c) Explain why the contact forces are different, even though the ball has the same mass, the same speed before impact, the same speed after impact, and the same contact time.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 5

The floor collision produces a peak contact force of 59.9 N, while the ceiling collision produces a peak contact force of 52.1 N.

The difference comes from the direction of gravity relative to the required change in momentum. Gravity opposes the momentum change during the floor collision but assists it during the ceiling collision.

This is why you cannot blindly add \(mg\) to every vertical impact calculation.

Step 1: Calculate the momentum change in both trials.

For Trial A, take upwards as positive.

The ball approaches the floor downwards:

\[ v_i=-4.0\ \mathrm{m\,s^{-1}} \]

It rebounds upwards:

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

Therefore:

\[ \Delta p_A=0.20[3.0-(-4.0)] \]

\[ \boxed{\Delta p_A=+1.40\ \mathrm{N\,s}} \]

For Trial B, use the same positive direction.

The ball approaches the ceiling upwards:

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

It rebounds downwards:

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

Therefore:

\[ \Delta p_B=0.20[-3.0-(4.0)] \]

\[ \boxed{\Delta p_B=-1.40\ \mathrm{N\,s}} \]

The momentum changes have equal magnitudes but opposite directions.

Step 2: Calculate the weight impulse.

Weight acts downwards in both trials.

\[ J_W=-mg\Delta t \]

\[ J_W=-(0.20)(9.8)(0.050) \]

\[ \boxed{J_W=-0.098\ \mathrm{N\,s}} \]

Step 3: Calculate the contact impulse for the floor collision.

For Trial A:

\[ J_N+J_W=\Delta p_A \]

Rearranging:

\[ J_N=\Delta p_A-J_W \]

\[ J_N=1.40-(-0.098) \]

\[ \boxed{J_N=1.498\ \mathrm{N\,s}} \]

The contact impulse from the floor acts upwards.

Step 4: Calculate the mean and peak floor forces.

The average contact force magnitude is:

\[ N_{\text{avg}}=\frac{1.498}{0.050} \]

\[ \boxed{N_{\text{avg}}=29.96\ \mathrm N} \]

The force-time graph is triangular, so:

\[ N_{\text{peak}}=2N_{\text{avg}} \]

\[ \boxed{N_{\text{peak}}=59.92\ \mathrm N} \]

Step 5: Calculate the contact impulse for the ceiling collision.

For Trial B:

\[ J_C+J_W=\Delta p_B \]

Rearranging:

\[ J_C=\Delta p_B-J_W \]

\[ J_C=-1.40-(-0.098) \]

\[ \boxed{J_C=-1.302\ \mathrm{N\,s}} \]

The negative sign indicates that the ceiling’s contact impulse acts downwards.

Its magnitude is:

\[ |J_C|=1.302\ \mathrm{N\,s} \]

Step 6: Calculate the mean and peak ceiling forces.

The average contact force magnitude is:

\[ C_{\text{avg}}=\frac{1.302}{0.050} \]

\[ \boxed{C_{\text{avg}}=26.04\ \mathrm N} \]

For the triangular force-time graph:

\[ C_{\text{peak}}=2C_{\text{avg}} \]

\[ \boxed{C_{\text{peak}}=52.08\ \mathrm N} \]

The comparison is:

The tempting approach is to argue that the two collisions must have equal average contact forces because their momentum changes have equal magnitudes and their contact times are identical.

That would be true if the contact force were the only force acting.

But gravity acts throughout both collisions.

• At the floor, gravity acts downwards while the ball’s momentum change is upwards. The floor must provide additional upward impulse to overcome gravity.
• At the ceiling, gravity acts downwards in the same direction as the ball’s momentum change. The ceiling therefore needs to provide less downward impulse.

The key principle is to use the signed impulse equation:

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

Do not memorise that weight must always be added. Its contribution depends on the direction of the momentum change and the direction of the contact force.

07Where to go next

When you encounter another vertical impact question, make your first decision about what the force-time graph represents: contact force or resultant force. Then choose a positive direction and account for every impulse during contact.

Once you can separate contact impulse from net impulse, the next useful skill is calculating impulse from different graph shapes. Practise with How to Calculate Impulse from a Triangular Force-Time Graph.

For the broader distinction between normal force and weight, see When Is the Normal Force Not Equal to Weight?.