How to Find the Speed of Two Objects That Stick Together

A \(0.60\ \mathrm{kg}\) cart moving right at \(1.50\ \mathrm{m\,s^{-1}}\) collides with a stationary \(0.40\ \mathrm{kg}\) cart. Their Velcro bumpers lock together, so they move as one object after the collision.

What is their new speed?

The answer is \(0.90\ \mathrm{m\,s^{-1}}\) to the right. You find the speed of two objects that stick together by adding their initial momenta and dividing by their combined mass.

Before working through the calculation, make a prediction. If the stationary cart had a mass of \(0.80\ \mathrm{kg}\) instead, would the combined carts move faster or slower?

They would move slower. The initial momentum would stay the same, but there would be more mass moving after the collision. The same momentum would have to carry a heavier combined object.

01The one-dimensional method

The key is to treat both objects as a single system.

Momentum is calculated using:

\[ p=mv \]

where:

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

Momentum has direction, so you must use signed velocities rather than simply adding speeds.

The conservation of momentum equation

If the net external impulse on the two-object system is negligible during the collision, its total momentum is conserved.

This means:

\[ \text{Initial momentum}=\text{Final momentum} \]

Before the collision, the objects have separate velocities. After they stick together, they share one common velocity.

For a one-dimensional collision:

\[ m_1u_1+m_2u_2=(m_1+m_2)v \]

Rearranging gives the formula you need:

\[ \boxed{ v=\frac{m_1u_1+m_2u_2}{m_1+m_2} } \]

Here:

• \(m_1\) and \(m_2\) are the masses of the two objects.
• \(u_1\) and \(u_2\) are their initial signed velocities.
• \(v\) is their common final velocity.

The formula works for objects moving in the same direction, opposite directions, or with one object initially stationary.

If an object is stationary, its initial velocity is zero.

Why you cannot simply average the speeds

A common mistake is to calculate the average of the two initial speeds.

For example, suppose a \(0.80\ \mathrm{kg}\) cart moves at \(3.0\ \mathrm{m\,s^{-1}}\) and collides with a stationary \(0.20\ \mathrm{kg}\) cart.

The ordinary average of their initial speeds is:

\[ \frac{3.0+0}{2}=1.5\ \mathrm{m\,s^{-1}} \]

But the actual final speed is:

\[ v=\frac{(0.80)(3.0)+(0.20)(0)}{0.80+0.20} \]

\[ v=2.4\ \mathrm{m\,s^{-1}} \]

The lighter cart does not have the same influence on the final motion as the heavier cart.

The correct calculation is a mass-weighted average of the initial velocities. A heavier object contributes more momentum at the same speed.

For objects initially moving in the same direction, the final velocity will lie between their initial velocities. If they move in opposite directions, their momenta can partially or completely cancel.

Getting the direction right

Choose a positive direction before substituting numbers.

For example, if right is positive:

Suppose a cart moves left at \(2.0\ \mathrm{m\,s^{-1}}\). Its velocity is:

\[ u=-2.0\ \mathrm{m\,s^{-1}} \]

The negative sign does not mean the cart has negative speed. It identifies its direction.

Your momentum calculation gives a final velocity. The speed is the magnitude of that velocity.

For example, a final velocity of \(-0.75\ \mathrm{m\,s^{-1}}\) means a speed of \(0.75\ \mathrm{m\,s^{-1}}\) to the left.

02What changes when the objects stick together?

When two objects collide and remain attached, the collision is called a perfectly inelastic collision.

There are two important things to understand:

1. Total momentum is conserved if the net external impulse is negligible.
2. Total kinetic energy is generally not conserved.

These are not contradictory.

Momentum and kinetic energy are different quantities

Momentum depends on mass and velocity:

\[ p=mv \]

Kinetic energy depends on mass and speed squared:

\[ KE=\frac12 mv^2 \]

During a sticking collision, some of the initial kinetic energy is transformed into other forms of energy, including:

• Thermal energy.
• Sound.
• Deformation of the objects.
• Internal energy associated with the materials.

The energy has not disappeared. It has changed form.

This distinction matters when checking your answer. You should use momentum to calculate the common final velocity, then calculate kinetic energy separately if the question asks how much energy was transferred.

For a broader method of testing whether collisions are elastic, see How to Check Whether a Collision Is Elastic from Measured Speeds.

03How an experiment tests the prediction

The momentum equation is a model. We can test its prediction by measuring the motion of two objects before and after they stick together.

Consider a practical collision experiment using two dynamics carts.

Setting up the experiment

Place a low-friction track horizontally on a bench.

Position cart A on the left, moving towards stationary cart B on the right. Attach Velcro bumpers to their front faces so that they connect during the collision.

Place a photogate before the collision and another just beyond the collision point. Attach a small flag of known length to cart A. The flag passes through the first photogate before impact and the second photogate after the carts have joined.

Measure both cart masses using a balance.

What should the experiment show?

Suppose the measured masses are:

• Cart A: \(0.60\ \mathrm{kg}\).
• Cart B: \(0.40\ \mathrm{kg}\).

The first photogate measures cart A’s initial speed as \(1.50\ \mathrm{m\,s^{-1}}\).

Cart B is stationary.

Our momentum calculation predicts:

\[ v=\frac{(0.60)(1.50)+(0.40)(0)} {0.60+0.40} \]

\[ v=0.90\ \mathrm{m\,s^{-1}} \]

Therefore, the joined carts should travel at approximately \(0.90\ \mathrm{m\,s^{-1}}\).

Measuring the actual speed

A photogate measures how long an object blocks its light beam.

If the flag has a known length, its speed can be calculated using:

\[ s=\frac{L}{\Delta t} \]

where:

• \(s\) is speed in \(\mathrm{m\,s^{-1}}\).
• \(L\) is the flag length in metres.
• \(\Delta t\) is the time taken for the flag to pass through the beam, in seconds.

Consider the following illustrative measurements.

The measured final speed is:

\[ s=\frac{0.050}{0.0568} \]

\[ s=0.88\ \mathrm{m\,s^{-1}} \]

The predicted speed was \(0.90\ \mathrm{m\,s^{-1}}\), while the measured speed was \(0.88\ \mathrm{m\,s^{-1}}\).

The difference is approximately \(2.2\%\) of the predicted value.

This is reasonably close, but it is not an exact match. Whether the difference is consistent with momentum conservation depends on the experiment’s measurement uncertainty and any external impulse.

What could cause the difference?

A first attempt might use a stopwatch to measure how long the carts take to travel a relatively long distance after the collision.

The problem is that this gives an average speed over that distance, rather than the speed immediately after impact. Friction can also change the carts’ speed while the measurement is being made.

Photogates improve the measurement because they record the time taken for a short flag to pass through a beam.

The experiment can be improved further by:

• Checking that the track is level.
• Reducing friction as much as possible.
• Positioning the second photogate close to, but beyond, the collision.
• Repeating the experiment several times.
• Using two photogates in sequence, or video analysis, when the direction of motion also needs to be established.

There is an important limitation here. A single photogate measures speed, not direction. A cart moving backwards can produce the same blocking time as one moving forwards.

The experiment supports the momentum model when the measured results agree within the uncertainty of the measurements and external impulses are sufficiently small.

04Choosing the correct method in an exam

When you see a question involving two objects that stick together, work through these decisions before substituting values.

1. Choose your system.

Usually, select both colliding objects as one system. The forces they exert on each other are internal forces and cancel when considering the total momentum of the system.

2. Check for external impulses.

Momentum conservation requires the net external impulse during the collision to be negligible.

Friction, a collision with a wall, or an external force from a spring can change the system’s momentum.

If an external impulse is significant, use the impulse-momentum relationship:

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

Here, \(J_{\mathrm{ext}}\) is the net external impulse in the chosen direction.

3. Choose a positive direction.

Assign positive and negative signs to the initial velocities.

4. Calculate the total initial momentum.

Add the momentum of both objects, including their directions.

5. Calculate the combined mass.

Because the objects stick, their final mass is:

\[ m_{\mathrm{total}}=m_1+m_2 \]

6. Find the common final velocity.

Divide the total final momentum by the combined mass.

7. Check kinetic energy separately.

Only calculate or conserve kinetic energy according to the physical conditions of the problem. Sticking together does not mean kinetic energy is conserved.

What if the objects move in two dimensions?

The same principle applies, but you must conserve momentum separately in each perpendicular direction.

For example, if two objects stick together after approaching at different angles:

\[ v_x= \frac{m_1u_{1x}+m_2u_{2x}} {m_1+m_2} \]

\[ v_y= \frac{m_1u_{1y}+m_2u_{2y}} {m_1+m_2} \]

Here, \(u_{1x}\) and \(u_{1y}\) are the initial velocity components of object 1, with equivalent definitions for object 2.

Once you have the final velocity components, calculate the common speed using:

\[ s=\sqrt{v_x^2+v_y^2} \]

The full process of resolving vectors and conserving momentum in perpendicular directions is explained in Conservation of Momentum in Two Dimensions for HSC Physics.

05Worked examples

06Questions and solutions

The following questions progress from straightforward momentum calculations to problems involving external impulses, two-dimensional motion, and multiple stages of energy transfer.

Question 1

A \(0.25\ \mathrm{kg}\) puck travels east at \(4.0\ \mathrm{m\,s^{-1}}\). It collides with a stationary \(0.75\ \mathrm{kg}\) puck, and the two pucks stick together.

Calculate their common final speed and direction.

Assume negligible external impulse during the collision.

Solution 1

The pucks move together at \(1.0\ \mathrm{m\,s^{-1}}\) east.

Take east as positive.

The initial velocities are:

\[ u_1=+4.0\ \mathrm{m\,s^{-1}} \]

\[ u_2=0 \]

Using conservation of momentum:

\[ m_1u_1+m_2u_2=(m_1+m_2)v \]

Substitute:

\[ (0.25)(4.0)+(0.75)(0) =(0.25+0.75)v \]

\[ 1.0=1.0v \]

Therefore:

\[ v=1.0\ \mathrm{m\,s^{-1}} \]

The velocity is positive, so the pucks move east.

The important point is that the stationary puck contributes mass but no initial momentum. Its mass still needs to be included when calculating the combined final velocity.

Question 2

A \(0.30\ \mathrm{kg}\) trolley moves east at \(5.0\ \mathrm{m\,s^{-1}}\). It catches up with a \(0.70\ \mathrm{kg}\) trolley moving east at \(2.0\ \mathrm{m\,s^{-1}}\).

The trolleys collide and stick together.

a) Calculate their common final velocity.

b) Calculate the total kinetic energy before and after the collision.

c) A student claims that the final speed must be the arithmetic mean of the initial speeds. Evaluate this claim.

Assume negligible external impulse.

Solution 2

The trolleys move together at \(2.9\ \mathrm{m\,s^{-1}}\) east, and their total kinetic energy decreases by \(0.95\ \mathrm{J}\).

Part a: Common final velocity

Take east as positive.

Both initial velocities are positive:

\[ u_1=+5.0\ \mathrm{m\,s^{-1}} \]

\[ u_2=+2.0\ \mathrm{m\,s^{-1}} \]

Apply conservation of momentum:

\[ (0.30)(5.0)+(0.70)(2.0) =(0.30+0.70)v \]

\[ 1.50+1.40=1.00v \]

\[ v=2.90\ \mathrm{m\,s^{-1}} \]

The final velocity is \(2.9\ \mathrm{m\,s^{-1}}\) east.

Part b: Change in kinetic energy

Calculate the initial kinetic energy:

\[ KE_i= \frac12(0.30)(5.0)^2 + \frac12(0.70)(2.0)^2 \]

\[ KE_i=3.75+1.40 \]

\[ KE_i=5.15\ \mathrm{J} \]

Now calculate the final kinetic energy:

\[ KE_f=\frac12(1.00)(2.90)^2 \]

\[ KE_f=4.205\ \mathrm{J} \]

The decrease is:

\[ \Delta KE=5.15-4.205 \]

\[ \Delta KE=0.945\ \mathrm{J} \]

To two significant figures:

\[ \Delta KE=0.95\ \mathrm{J} \]

This energy has been transferred into other forms during the collision.

Part c: Evaluating the student’s claim

The student’s claim is incorrect.

The arithmetic mean would be:

\[ v_{\mathrm{mean}}=\frac{5.0+2.0}{2} \]

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

But the actual final velocity is \(2.9\ \mathrm{m\,s^{-1}}\).

The problem is that the two trolleys have different masses. The heavier trolley contributes more initial momentum, so the final velocity is weighted towards its initial velocity.

An arithmetic mean would only give the correct result in this particular same-direction situation if the two masses were equal.

Question 3

Two carts approach each other on a horizontal track.

• Cart A has a mass of \(0.50\ \mathrm{kg}\) and moves right at \(2.0\ \mathrm{m\,s^{-1}}\).
• Cart B has a mass of \(0.70\ \mathrm{kg}\) and moves left at \(0.40\ \mathrm{m\,s^{-1}}\).

The carts stick together during a collision lasting \(0.10\ \mathrm{s}\).

A force sensor measures an average net external horizontal force of \(0.60\ \mathrm{N}\) to the left on the two-cart system during the collision.

Calculate the common final velocity, taking the external impulse into account.

Compare your answer with the result you would obtain by assuming momentum was conserved, and explain the difference.

Solution 3

The joined carts move right at \(0.55\ \mathrm{m\,s^{-1}}\). Ignoring the external impulse would incorrectly predict \(0.60\ \mathrm{m\,s^{-1}}\).

Step 1: Calculate the initial momentum.

Take right as positive.

\[ u_A=+2.0\ \mathrm{m\,s^{-1}} \]

\[ u_B=-0.40\ \mathrm{m\,s^{-1}} \]

The initial momentum is:

\[ p_i=(0.50)(2.0)+(0.70)(-0.40) \]

\[ p_i=1.00-0.28 \]

\[ p_i=0.72\ \mathrm{kg\,m\,s^{-1}} \]

Step 2: Calculate the external impulse.

Impulse is calculated using:

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

The external force acts left, so it is negative:

\[ J_{\mathrm{ext}}=(-0.60)(0.10) \]

\[ J_{\mathrm{ext}}=-0.060\ \mathrm{N\,s} \]

Step 3: Calculate the final momentum.

The impulse-momentum relationship is:

\[ p_f-p_i=J_{\mathrm{ext}} \]

Rearranging:

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

Substitute:

\[ p_f=0.72-0.060 \]

\[ p_f=0.66\ \mathrm{kg\,m\,s^{-1}} \]

Step 4: Calculate the common final velocity.

The combined mass is:

\[ m_{\mathrm{total}}=0.50+0.70 \]

\[ m_{\mathrm{total}}=1.20\ \mathrm{kg} \]

Therefore:

\[ v=\frac{0.66}{1.20} \]

\[ v=0.55\ \mathrm{m\,s^{-1}} \]

The positive sign means the joined carts move right.

Step 5: Compare with the isolated-system prediction.

If the external impulse were ignored, conservation of momentum would give:

\[ v=\frac{0.72}{1.20} \]

\[ v=0.60\ \mathrm{m\,s^{-1}} \]

This is higher than the actual result.

The difference occurs because the external force acts to the left and reduces the total momentum of the two-cart system.

The important distinction is that the contact forces between the carts are internal, but the force from outside the system is external. You cannot ignore a significant external impulse just because the objects stick together.

Question 4

Two pucks collide on a smooth horizontal air table and stick together.

Puck A has a mass of \(0.40\ \mathrm{kg}\) and moves at \(5.0\ \mathrm{m\,s^{-1}}\), at an angle of \(30^\circ\) north of east.

Puck B has a mass of \(0.60\ \mathrm{kg}\) and moves west at \(2.0\ \mathrm{m\,s^{-1}}\).

Calculate the speed and direction of the joined pucks immediately after the collision.

Assume negligible external horizontal impulse.

Solution 4

The joined pucks move at approximately \(1.13\ \mathrm{m\,s^{-1}}\), at \(62^\circ\) north of east.

Because the pucks approach at different angles, we must conserve momentum separately in the horizontal and vertical directions.

Step 1: Establish the coordinate system.

Take:

• East as positive \(x\).
• North as positive \(y\).

Puck A has both horizontal and vertical velocity components.

Puck B moves only horizontally.

Step 2: Calculate the initial horizontal momentum.

For puck A:

\[ p_{Ax}=(0.40)(5.0\cos30^\circ) \]

\[ p_{Ax}=1.732\ \mathrm{kg\,m\,s^{-1}} \]

For puck B:

\[ p_{Bx}=(0.60)(-2.0) \]

\[ p_{Bx}=-1.20\ \mathrm{kg\,m\,s^{-1}} \]

Therefore, total initial horizontal momentum is:

\[ p_x=1.732-1.20 \]

\[ p_x=0.532\ \mathrm{kg\,m\,s^{-1}} \]

Step 3: Calculate the initial vertical momentum.

Puck A contributes:

\[ p_{Ay}=(0.40)(5.0\sin30^\circ) \]

\[ p_{Ay}=1.00\ \mathrm{kg\,m\,s^{-1}} \]

Puck B has no vertical velocity, so:

\[ p_{By}=0 \]

Therefore:

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

Step 4: Calculate the combined mass and final velocity components.

The combined mass is:

\[ m_{\mathrm{total}}=0.40+0.60 \]

\[ m_{\mathrm{total}}=1.00\ \mathrm{kg} \]

Conserving horizontal momentum:

\[ v_x=\frac{0.532}{1.00} \]

\[ v_x=0.532\ \mathrm{m\,s^{-1}} \]

Conserving vertical momentum:

\[ v_y=\frac{1.00}{1.00} \]

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

Step 5: Calculate the final speed.

The final velocity has two perpendicular components.

Using Pythagoras:

\[ v=\sqrt{v_x^2+v_y^2} \]

\[ v=\sqrt{(0.532)^2+(1.00)^2} \]

\[ v=1.13\ \mathrm{m\,s^{-1}} \]

Step 6: Calculate the direction.

Using trigonometry:

\[ \tan\theta=\frac{v_y}{v_x} \]

\[ \theta=\tan^{-1}\left(\frac{1.00}{0.532}\right) \]

\[ \theta\approx62^\circ \]

The final velocity is therefore:

\[ v=1.13\ \mathrm{m\,s^{-1}} \]

at \(62^\circ\) north of east.

Notice that we did not add the two initial speeds. We resolved their momentum into perpendicular components and conserved each component separately.

Question 5

A \(0.25\ \mathrm{kg}\) block slides right along a horizontal, low-friction surface towards a stationary \(0.75\ \mathrm{kg}\) block. The blocks collide and stick together.

The joined blocks then move up a rough ramp inclined at \(30^\circ\) to the horizontal. They travel \(0.40\ \mathrm{m}\) along the ramp before coming momentarily to rest.

The coefficient of kinetic friction between the joined blocks and the ramp is \(0.10\).

Assume:

• The collision occurs on the horizontal section, where the external horizontal impulse is negligible.
• Friction acts only after the joined blocks reach the ramp.
• The ramp angle remains constant.
• The blocks slide without rotating.
• Gravitational acceleration is \(9.8\ \mathrm{m\,s^{-2}}\).

Calculate:

a) The initial speed of the moving block.

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

c) The work done against friction as the joined blocks travel up the ramp.

A student argues that the initial kinetic energy of the moving block must equal the gravitational potential energy gained by the joined blocks. Evaluate this argument.

Solution 5

The initial block speed is approximately \(8.58\ \mathrm{m\,s^{-1}}\). During the collision, approximately \(6.90\ \mathrm{J}\) of kinetic energy is converted into other forms, and a further \(0.34\ \mathrm{J}\) is transferred by friction during the upward motion.

This problem requires two different methods because two different physical processes occur:

1. The blocks collide and stick together.
2. The joined blocks travel up a rough ramp and stop.

We must not treat these as one continuous, energy-conserving collision.

Step 1: Calculate the total mass after the collision.

\[ m_{\mathrm{total}}=0.25+0.75 \]

\[ m_{\mathrm{total}}=1.00\ \mathrm{kg} \]

Step 2: Calculate the vertical height gained on the ramp.

The blocks travel \(0.40\ \mathrm{m}\) along the ramp, but this is not their vertical displacement.

Using trigonometry:

\[ h=s\sin\theta \]

\[ h=0.40\sin30^\circ \]

\[ h=0.20\ \mathrm{m} \]

This is the vertical height gained.

Step 3: Calculate the work done against gravity.

The gravitational potential energy gained is:

\[ \Delta GPE=m_{\mathrm{total}}gh \]

Substitute:

\[ \Delta GPE=(1.00)(9.8)(0.20) \]

\[ \Delta GPE=1.96\ \mathrm{J} \]

Step 4: Calculate the work done against friction.

First, calculate the normal force on the ramp.

\[ N=m_{\mathrm{total}}g\cos\theta \]

\[ N=(1.00)(9.8)\cos30^\circ \]

\[ N=8.49\ \mathrm{N} \]

The kinetic friction force is:

\[ F_f=\mu N \]

where \(\mu=0.10\) is the coefficient of kinetic friction.

\[ F_f=(0.10)(8.49) \]

\[ F_f=0.849\ \mathrm{N} \]

The work done against friction is:

\[ W_f=F_fs \]

\[ W_f=(0.849)(0.40) \]

\[ W_f=0.340\ \mathrm{J} \]

Step 5: Determine the kinetic energy immediately after the collision.

The joined blocks stop at the top of the ramp.

Their initial kinetic energy on the ramp must therefore provide both:

• The gravitational potential energy gained.
• The energy transferred through friction.

Using the work-energy principle:

\[ KE_{\mathrm{after}} = \Delta GPE+W_f \]

\[ KE_{\mathrm{after}}=1.96+0.340 \]

\[ KE_{\mathrm{after}}=2.30\ \mathrm{J} \]

This is the kinetic energy of the joined blocks immediately after the collision.

Step 6: Calculate their common speed immediately after the collision.

Using:

\[ KE=\frac12mv^2 \]

Rearrange:

\[ v=\sqrt{\frac{2KE}{m}} \]

Substitute:

\[ v=\sqrt{\frac{2(2.30)}{1.00}} \]

\[ v=2.14\ \mathrm{m\,s^{-1}} \]

This is the speed of the joined blocks as they begin travelling up the ramp.

Step 7: Work backwards to find the initial block speed.

Now return to the collision itself.

The initial \(0.75\ \mathrm{kg}\) block was stationary, and the external horizontal impulse during the collision is negligible.

Take right as positive.

Using conservation of momentum:

\[ m_1u_1+m_2u_2 = (m_1+m_2)v \]

Substitute:

\[ (0.25)u_1+(0.75)(0) = (1.00)(2.14) \]

\[ 0.25u_1=2.14 \]

\[ u_1=\frac{2.14}{0.25} \]

\[ u_1=8.56\ \mathrm{m\,s^{-1}} \]

Using the unrounded intermediate values gives:

\[ u_1\approx8.58\ \mathrm{m\,s^{-1}} \]

This is the original speed of the moving block.

Step 8: Calculate the kinetic energy converted during the collision.

Before the collision, only the \(0.25\ \mathrm{kg}\) block has kinetic energy.

\[ KE_i=\frac12m_1u_1^2 \]

\[ KE_i=\frac12(0.25)(8.58)^2 \]

\[ KE_i\approx9.20\ \mathrm{J} \]

Immediately after the collision:

\[ KE_f=2.30\ \mathrm{J} \]

Therefore, the kinetic energy decrease during the collision is:

\[ \Delta KE=KE_i-KE_f \]

\[ \Delta KE=9.20-2.30 \]

\[ \Delta KE=6.90\ \mathrm{J} \]

Approximately \(6.90\ \mathrm{J}\) has been converted into other forms of energy during the sticking collision.

Part c: Work done against friction

We calculated this in Step 4:

\[ W_f=0.340\ \mathrm{J} \]

This energy transfer happens during the motion up the ramp, not during the collision.

Evaluating the student’s argument

The student’s argument is incorrect because it treats the entire process as though mechanical energy were conserved.

There are two separate energy changes.

First, during the collision, the blocks stick together. This is an inelastic collision, so kinetic energy decreases.

Second, as the joined blocks move up the ramp, their remaining kinetic energy is transferred into gravitational potential energy and thermal energy through friction.

If the student incorrectly equates the initial kinetic energy directly to the gravitational potential energy gained, they would write:

\[ \frac12(0.25)u^2=1.96 \]

This gives:

\[ u=3.96\ \mathrm{m\,s^{-1}} \]

That is not the correct initial speed.

The calculation ignores both the kinetic energy converted during the collision and the work done against friction on the ramp.

We can check the correct energy accounting:

\[ 9.20=6.90+1.96+0.34 \]

The initial kinetic energy is accounted for across all three energy transfers.

This is the key lesson: momentum conservation applies to the collision, while the work-energy principle describes the subsequent motion up the ramp.

07What to remember

When two objects stick together, their final velocity is determined by their total initial momentum and combined mass:

\[ \boxed{ v=\frac{m_1u_1+m_2u_2}{m_1+m_2} } \]

Always remember:

• Include both objects when calculating the final mass.
• Use signed velocities to account for direction.
• Check that external impulses are negligible before conserving momentum.
• Report the magnitude of the final velocity as the speed.
• Do not assume kinetic energy is conserved just because momentum is conserved.
• For two-dimensional collisions, conserve momentum separately in each direction.

Once you can calculate the common velocity, the next useful decision is whether a collision is elastic or inelastic. Momentum conservation alone cannot answer that. You need to compare the total kinetic energy before and after the collision.