How to Solve Elastic Collisions Between Equal Masses

Two identical carts collide on a low-friction track. Cart A is moving to the right at \(4.0\text{ m s}^{-1}\), while cart B is stationary.

Predict what happens after a perfectly elastic, head-on collision. Do both carts move away at \(2.0\text{ m s}^{-1}\), or does one cart stop completely?

The correct result is that cart A stops and cart B moves to the right at \(4.0\text{ m s}^{-1}\).

The prediction that both carts move at \(2.0\text{ m s}^{-1}\) conserves momentum, but it does not conserve kinetic energy. That’s the important difference.

Equal masses give us some particularly useful shortcuts when solving elastic collisions. In one dimension, the objects exchange velocities. In two dimensions, if one object starts stationary, their final velocity vectors are perpendicular.

However, these shortcuts have conditions. Applying the wrong one can give you an answer that looks mathematically reasonable but describes an impossible collision.

01The method: what changes when the masses are equal?

Start by treating both colliding objects as one system.

During the collision, each object exerts a force on the other. These forces are internal to the two-object system and cancel when considering its total momentum.

Provided the external impulse is negligible during the collision, total momentum is conserved.

For an elastic collision, total kinetic energy is also conserved.

The equal-mass condition makes both equations simpler because the common mass can be cancelled.

There are three situations to recognise:

The third situation is where students often apply the wrong shortcut. Equal masses do not automatically mean that the final velocities are perpendicular or that the objects exchange their velocity vectors.

Assume throughout this article that:

• Both objects have equal masses.
• External impulses during the collision are negligible.
• The collision is perfectly elastic.
• Translational kinetic energy is the relevant kinetic energy, with rotational motion ignored.

For a fuller explanation of why elastic collisions conserve both quantities, see Elastic Collisions: Momentum and Kinetic Energy for HSC Physics.

02How an experiment tests the equal-mass prediction

Before relying on the mathematical shortcuts, consider how you could test them.

Imagine two identical pucks colliding on a horizontal air table.

First, use a balance to check that the pucks have equal masses. The air table blows air through tiny holes in its surface, creating a thin cushion that reduces friction.

Place a calibrated grid underneath the transparent table surface. Position a camera directly above the apparatus.

Put puck B near a marked collision point. Place puck A to its west and give it an initial velocity towards B.

For a head-on collision, align the centres of both pucks along the same straight line. For a glancing collision, offset B slightly from A’s original path.

What should the experiment show?

For a head-on collision, the prediction is straightforward:

• Puck A should leave with B’s original velocity.
• Puck B should leave with A’s original velocity.

For a glancing collision with B initially stationary, the two final velocity vectors should be perpendicular.

To test this, track the centre of each puck across several video frames before and after contact.

The gradient of a position-time graph gives the velocity component:

• The gradient of an \(x\)-position graph gives horizontal velocity.
• The gradient of a \(y\)-position graph gives vertical velocity.

Avoid using frames during the actual collision. The pucks accelerate rapidly during contact, so their velocities are not approximately constant over that interval.

A single photogate would be a poor choice for the glancing collision. It can measure speed, but it does not provide the complete direction of travel. An overhead camera is a better arrangement because it captures both velocity components.

A quick measurement calculation

Suppose a puck moves \(15\text{ mm}\) across four frame intervals in a camera recording at 200 frames per second.

The elapsed time is:

\[
\Delta t=\frac{4}{200}
=0.020\text{ s}
\]

Convert the displacement into metres:

\[
\Delta x=0.015\text{ m}
\]

The average velocity component is:

\[
v_x=\frac{\Delta x}{\Delta t}
\]

\[
v_x=\frac{0.015}{0.020}
\]

\[
v_x=0.75\text{ m s}^{-1}
\]

The same calculation can be performed vertically to obtain \(v_y\).

In a real investigation, use several frames to estimate the position-time gradient more reliably.

What would convincing results look like?

Consider the following illustrative idealised data for a glancing collision.

Two equal-mass pucks are involved. A initially moves east at \(4.0\text{ m s}^{-1}\), while B is stationary.

All components are in \(\text{m s}^{-1}\). These are illustrative values, not measurements from a recorded experiment.

The final momentum components add to the original velocity components:

\[
v_{Ax}+v_{Bx}=1.44+2.56=4.00
\]

\[
v_{Ay}+v_{By}=1.92-1.92=0
\]

The final squared speeds also add correctly:

\[
v_A^2+v_B^2
=2.4^2+3.2^2
\]

\[
=5.76+10.24=16
\]

The initial squared speed was also \(4.0^2=16\).

These results are consistent with an elastic collision between equal masses.

In an actual experiment, small differences would be expected because of friction, imperfect mass matching, and measurement uncertainty. The purpose is to test whether the results agree with the model, not to force every measurement to match it exactly.

03Solving a one-dimensional elastic collision

A one-dimensional collision occurs when both objects move along the same straight line.

This is where the equal-mass shortcut is most powerful.

Let:

• \(m\) be the mass of each object, in kilograms.
• \(u_A\) and \(u_B\) be their initial signed velocities.
• \(v_A\) and \(v_B\) be their final signed velocities.

Choose one direction as positive. An object travelling in the opposite direction has a negative velocity.

Step 1: Apply conservation of momentum

For two equal masses:

\[
mu_A+mu_B=mv_A+mv_B
\]

Cancel the common mass:

\[
u_A+u_B=v_A+v_B
\]

Step 2: Apply conservation of kinetic energy

For an elastic collision:

\[
\frac12 mu_A^2+\frac12 mu_B^2
=
\frac12 mv_A^2+\frac12 mv_B^2
\]

Cancel the common factor:

\[
u_A^2+u_B^2=v_A^2+v_B^2
\]

Combining these two equations gives an important result.

The initial and final velocities have the same sum and the same product:

\[
u_A+u_B=v_A+v_B
\]

\[
u_Au_B=v_Av_B
\]

Therefore, the pair of final velocities must be the same pair of values as the initial velocities.

For an actual head-on collision, the non-trivial result is:

\[
\boxed{v_A=u_B}
\]

\[
\boxed{v_B=u_A}
\]

In other words, the objects exchange their signed velocities.

There is a mathematical solution in which both objects simply retain their original velocities, but that represents no change in their motion. For an actual collision, it is the exchange solution that applies.

Notice that we exchange velocities, not just speeds. The signs matter.

04Solving a two-dimensional collision

Now consider a glancing collision.

Puck A moves east towards stationary puck B. The collision is not head-on, so the pucks leave along different directions.

You cannot simply exchange their velocities as you did in one dimension.

Instead, conserve momentum separately in the horizontal and vertical directions.

Why are the final velocities perpendicular?

Let:

• \(\vec u_A\) be A’s initial velocity vector.
• \(\vec v_A\) be A’s final velocity vector.
• \(\vec v_B\) be B’s final velocity vector.

Puck B initially has zero velocity.

Because the masses are equal, momentum conservation gives:

\[
m\vec u_A=m\vec v_A+m\vec v_B
\]

Cancel the mass:

\[
\boxed{\vec u_A=\vec v_A+\vec v_B}
\]

This is a vector equation. It must hold horizontally and vertically.

Now apply conservation of kinetic energy:

\[
\frac12 mu_A^2
=
\frac12 mv_A^2+\frac12 mv_B^2
\]

Cancel the common factors:

\[
u_A^2=v_A^2+v_B^2
\]

Here’s the important step.

Square the magnitude of the momentum equation:

\[
u_A^2=
v_A^2+v_B^2+
2\vec v_A\cdot\vec v_B
\]

But kinetic energy conservation already tells us:

\[
u_A^2=v_A^2+v_B^2
\]

Comparing the equations:

\[
\vec v_A\cdot\vec v_B=0
\]

A zero dot product means the velocity vectors are perpendicular.

Therefore:

\[
\boxed{\theta=90^\circ}
\]

This is the origin of the famous right-angle result for an elastic collision between equal masses when one object is initially stationary.

It does not mean the final speeds are equal. One puck can leave much faster than the other.

05What if the equal-mass objects stick together?

Here’s a common misconception: equal masses always exchange velocities.

They don’t.

The exchange rule depends on the collision being elastic. If the objects stick together, the collision is perfectly inelastic.

Consider two identical \(0.50\text{ kg}\) carts. Cart A moves right at \(4.0\text{ m s}^{-1}\), and cart B is stationary. They collide and stick together.

Both carts now have the same final velocity.

Apply conservation of momentum:

\[
mu_A+mu_B=2mv_f
\]

Cancel the mass:

\[
u_A+u_B=2v_f
\]

Substitute:

\[
4.0+0=2v_f
\]

\[
v_f=2.0\text{ m s}^{-1}
\]

The joined carts move right at \(2.0\text{ m s}^{-1}\).

Now check kinetic energy.

Before:

\[
K_i=\frac12(0.50)(4.0^2)
\]

\[
K_i=4.0\text{ J}
\]

After:

\[
K_f=\frac12(1.00)(2.0^2)
\]

\[
K_f=2.0\text{ J}
\]

The final kinetic energy is only half the initial kinetic energy.

The other \(2.0\text{ J}\) has been transferred into other forms of energy, including deformation, heating, and sound.

Momentum is conserved, but kinetic energy is not.

This is why you must establish that a collision is elastic before using the equal-mass exchange rule.

06What if both objects are initially moving in two dimensions?

There is one more important distinction.

The perpendicular-velocity result is often remembered as a rule for equal masses. However, it is more specific than that.

It follows when the initial velocity vectors have a zero dot product. Having one object initially stationary is one way this happens.

For a general two-dimensional elastic collision between equal masses, momentum conservation gives:

\[
\vec u_A+\vec u_B
=
\vec v_A+\vec v_B
\]

Kinetic energy conservation gives:

\[
u_A^2+u_B^2=v_A^2+v_B^2
\]

Combining these equations gives:

\[
\boxed{
\vec u_A\cdot\vec u_B
=
\vec v_A\cdot\vec v_B
}
\]

The dot product of the two velocity vectors is unchanged.

This means:

• If the initial velocity vectors are perpendicular, the final velocity vectors are also perpendicular.
• If the initial velocity vectors are not perpendicular, the final velocity vectors do not have to be perpendicular.
• If one object starts stationary, the initial dot product is automatically zero.

There is another practical consequence. In a general two-dimensional collision, the horizontal and vertical momentum equations plus the kinetic energy equation may not provide enough information to determine all four final velocity components. You may need an additional measured direction or information about the collision geometry.

For a broader treatment of this situation, see How to Solve Collisions When Both Objects Are Moving.

07Questions and solutions

Try each question before reading its solution. The difficulty increases as you progress.

Question 1

Two identical carts, each with a mass of \(0.30\text{ kg}\), collide head-on on a low-friction track.

Initially, cart A travels right at \(3.0\text{ m s}^{-1}\), while cart B is stationary.

The collision is perfectly elastic.

Calculate the final velocity of each cart.

Solution 1

Cart A stops, and cart B moves right at \(3.0\text{ m s}^{-1}\).

Because the collision is one-dimensional, elastic, and involves equal masses, the carts exchange their signed velocities.

Therefore:

\[
v_A=0
\]

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

Check momentum.

Before:

\[
p_i=0.30(3.0)=0.90
\]

After:

\[
p_f=0.30(0)+0.30(3.0)
\]

\[
p_f=0.90\text{ kg m s}^{-1}
\]

Check kinetic energy.

Before:

\[
K_i=\frac12(0.30)(3.0^2)
\]

\[
K_i=1.35\text{ J}
\]

After:

\[
K_f=\frac12(0.30)(3.0^2)
\]

\[
K_f=1.35\text{ J}
\]

Both conservation laws are satisfied.

Question 2

Two identical carts, each with a mass of \(0.40\text{ kg}\), approach one another along a straight track.

Take right as positive.

Initially:

\[
u_A=+6.0\text{ m s}^{-1}
\]

\[
u_B=-2.0\text{ m s}^{-1}
\]

The carts collide head-on and elastically.

Determine:

(a) The final velocity of each cart.

(b) The total momentum before and after the collision.

(c) The total kinetic energy before and after the collision.

(d) Which cart or carts reverse direction?

Solution 2

Cart A finishes moving left at \(2.0\text{ m s}^{-1}\), while cart B finishes moving right at \(6.0\text{ m s}^{-1}\). Both carts reverse direction.

(a) Final velocities

The equal-mass exchange rule gives:

\[
v_A=u_B=-2.0\text{ m s}^{-1}
\]

\[
v_B=u_A=+6.0\text{ m s}^{-1}
\]

The negative sign for A indicates leftward motion.

(b) Momentum

Before:

\[
p_i=0.40(6.0)+0.40(-2.0)
\]

\[
p_i=2.4-0.8
\]

\[
p_i=1.6\text{ kg m s}^{-1}
\]

After:

\[
p_f=0.40(-2.0)+0.40(6.0)
\]

\[
p_f=-0.8+2.4
\]

\[
p_f=1.6\text{ kg m s}^{-1}
\]

Momentum is conserved.

(c) Kinetic energy

Before:

\[
K_i=\frac12(0.40)(6.0^2+2.0^2)
\]

\[
K_i=0.20(36+4)
\]

\[
K_i=8.0\text{ J}
\]

After:

\[
K_f=\frac12(0.40)(2.0^2+6.0^2)
\]

\[
K_f=0.20(4+36)
\]

\[
K_f=8.0\text{ J}
\]

Kinetic energy is conserved.

(d) Direction changes

A initially travels right and finishes travelling left.

B initially travels left and finishes travelling right.

Both carts reverse direction.

The important detail is that the velocities are exchanged with their signs. If you had treated both initial speeds as positive, you would have missed the direction changes.

Question 3

Two identical pucks collide on a frictionless horizontal air table.

Puck B is initially stationary.

After a glancing elastic collision:

• Puck A travels at \(4.0\text{ m s}^{-1}\), at \(30^\circ\) north of east.
• Puck B travels at \(3.0\text{ m s}^{-1}\), at \(60^\circ\) south of east.

Determine the initial velocity of puck A, including its speed and direction.

Verify whether the given velocities are consistent with an elastic collision.

Solution 3

Puck A initially travels at \(5.0\text{ m s}^{-1}\), approximately \(6.9^\circ\) south of east. The given final velocities are consistent with an elastic collision.

Since B was initially stationary, momentum conservation gives:

\[
\vec u_A=\vec v_A+\vec v_B
\]

Step 1: Resolve A’s final velocity into components.

\[
v_{Ax}=4.0\cos30^\circ
\]

\[
v_{Ax}=3.464\text{ m s}^{-1}
\]

\[
v_{Ay}=4.0\sin30^\circ
\]

\[
v_{Ay}=2.00\text{ m s}^{-1}
\]

Step 2: Resolve B’s final velocity into components.

\[
v_{Bx}=3.0\cos60^\circ
\]

\[
v_{Bx}=1.50\text{ m s}^{-1}
\]

Since B travels south:

\[
v_{By}=-3.0\sin60^\circ
\]

\[
v_{By}=-2.598\text{ m s}^{-1}
\]

Step 3: Calculate A’s initial velocity components.

Horizontally:

\[
u_{Ax}=3.464+1.50
\]

\[
u_{Ax}=4.964\text{ m s}^{-1}
\]

Vertically:

\[
u_{Ay}=2.00-2.598
\]

\[
u_{Ay}=-0.598\text{ m s}^{-1}
\]

The negative vertical component means the initial direction is slightly south of east.

Step 4: Calculate the initial speed.

\[
u_A=\sqrt{4.964^2+(-0.598)^2}
\]

\[
u_A\approx5.00\text{ m s}^{-1}
\]

Step 5: Calculate the initial direction.

\[
\theta=\tan^{-1}
\left(\frac{0.598}{4.964}\right)
\]

\[
\theta\approx6.9^\circ
\]

Therefore, A initially travels at:

\[
\boxed{5.0\text{ m s}^{-1}}
\]

\[
\boxed{6.9^\circ\text{ south of east}}
\]

Step 6: Check kinetic energy.

Because the masses are equal, compare the sums of the squared speeds.

Before:

\[
u_A^2=5.0^2=25
\]

After:

\[
v_A^2+v_B^2=4.0^2+3.0^2
\]

\[
=16+9=25
\]

The kinetic energy condition is satisfied.

The outgoing vectors are also perpendicular because their directions differ by \(90^\circ\).

The result is consistent with an elastic collision.

Question 4

Two identical pucks, each with a mass of \(0.20\text{ kg}\), undergo a glancing collision on a frictionless air table.

Their measured velocity components are:

All components are in \(\text{m s}^{-1}\).

A student argues that these measurements cannot describe an elastic collision because the final velocity vectors are not perpendicular.

(a) Determine whether momentum is conserved in both directions.

(b) Determine whether kinetic energy is conserved.

(c) Evaluate the student’s claim.

Solution 4

The measurements are consistent with an elastic collision. The student’s claim is incorrect because the two objects were both initially moving, and their initial velocity vectors were not perpendicular.

(a) Momentum conservation

The masses are equal, so we can compare the sums of the velocity components.

Horizontally, before:

\[
u_{Ax}+u_{Bx}=4.0+1.0=5.0
\]

Horizontally, after:

\[
v_{Ax}+v_{Bx}=3.5+1.5=5.0
\]

Therefore, horizontal momentum is conserved.

Vertically, before:

\[
u_{Ay}+u_{By}=0+2.0=2.0
\]

Vertically, after:

\[
v_{Ay}+v_{By}=2.5-0.5=2.0
\]

Vertical momentum is also conserved.

The actual momentum components are:

\[
p_x=0.20(5.0)=1.0\text{ kg m s}^{-1}
\]

\[
p_y=0.20(2.0)=0.40\text{ kg m s}^{-1}
\]

Both are unchanged.

(b) Kinetic energy conservation

Before the collision:

\[
K_i=\frac12(0.20)
(4.0^2+1.0^2+2.0^2)
\]

\[
K_i=0.10(16+1+4)
\]

\[
K_i=2.10\text{ J}
\]

After the collision:

\[
K_f=\frac12(0.20)
(3.5^2+2.5^2+1.5^2+0.5^2)
\]

\[
K_f=0.10(12.25+6.25+2.25+0.25)
\]

\[
K_f=2.10\text{ J}
\]

Kinetic energy is conserved.

(c) Why the student’s claim is wrong

The right-angle result comes from the initial velocity vectors having a zero dot product.

Here, the initial dot product is:

\[
(4.0)(1.0)+(0)(2.0)=4.0
\]

The final dot product is:

\[
(3.5)(1.5)+(2.5)(-0.5)
\]

\[
=5.25-1.25=4.0
\]

The dot product is unchanged, as expected for an equal-mass elastic collision.

The final vectors are therefore not required to be perpendicular.

The student has incorrectly extended a special result from the stationary-target case to every two-dimensional equal-mass collision.

Question 5

Two identical pucks, each with a mass of \(0.20\text{ kg}\), collide on a frictionless horizontal air table.

Initially:

• Puck A travels east at \(5.0\text{ m s}^{-1}\).
• Puck B is stationary.

A tracking program reports that after the collision, puck A travels at \(4.0\text{ m s}^{-1}\), at \(60^\circ\) north of east.

A student calculates puck B’s velocity using conservation of momentum and argues that the collision must be elastic because total momentum is unchanged.

Investigate this claim.

(a) Use momentum conservation to calculate B’s reported final velocity components and speed.

(b) Calculate the total kinetic energy before and after the collision. Decide whether both reported measurements can describe a perfectly elastic collision.

(c) Starting with momentum and kinetic energy conservation, derive a relationship between A’s final speed and its deflection angle \(\theta\), measured from its original direction.

Use this relationship to determine:

• The required speed of A if its reported \(60^\circ\) direction is correct.
• The required deflection angle of A if its reported \(4.0\text{ m s}^{-1}\) speed is correct.

In each case, calculate B’s final speed.

(d) What is the greatest possible deflection angle of A if it continues moving after the collision?

Solution 5

The reported measurements cannot both be correct for a perfectly elastic collision.

Momentum conservation gives B a final speed of approximately \(4.58\text{ m s}^{-1}\), but this would increase the total kinetic energy from \(2.50\text{ J}\) to \(3.70\text{ J}\).

The deeper issue is that the final speed and deflection angle of A are linked. They cannot be chosen independently.

(a) Calculate B’s final velocity using momentum.

Take east as positive \(x\) and north as positive \(y\).

A’s reported final velocity components are:

\[
v_{Ax}=4.0\cos60^\circ
\]

\[
v_{Ax}=2.0\text{ m s}^{-1}
\]

\[
v_{Ay}=4.0\sin60^\circ
\]

\[
v_{Ay}=3.464\text{ m s}^{-1}
\]

Because B was initially stationary, the equal-mass momentum equations become:

\[
u_{Ax}=v_{Ax}+v_{Bx}
\]

\[
u_{Ay}=v_{Ay}+v_{By}
\]

Horizontally:

\[
5.0=2.0+v_{Bx}
\]

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

Vertically:

\[
0=3.464+v_{By}
\]

\[
v_{By}=-3.464\text{ m s}^{-1}
\]

Therefore:

\[
\vec v_B=(3.0,-3.464)
\]

Calculate B’s speed:

\[
v_B=\sqrt{3.0^2+(-3.464)^2}
\]

\[
v_B=\sqrt{9+12}
\]

\[
v_B\approx4.58\text{ m s}^{-1}
\]

Momentum is conserved in both directions.

So far, the student’s calculation looks correct.

(b) Check kinetic energy.

Initial kinetic energy:

\[
K_i=\frac12(0.20)(5.0^2)
\]

\[
K_i=2.50\text{ J}
\]

Final kinetic energy:

\[
K_f=\frac12(0.20)(4.0^2+4.58^2)
\]

\[
K_f\approx0.10(16+21)
\]

\[
K_f=3.70\text{ J}
\]

The final kinetic energy is greater than the initial kinetic energy.

The difference is:

\[
\Delta K=3.70-2.50
\]

\[
\Delta K=1.20\text{ J}
\]

This is impossible for the stated isolated, perfectly elastic collision.

The student has demonstrated momentum conservation, but that alone does not establish that the collision is elastic.

The reported speed and direction of A are inconsistent with the elastic collision model.

(c) Derive the relationship between final speed and deflection angle.

This is the most important part of the problem.

Let:

• \(u\) be A’s initial speed.
• \(v_A\) be A’s final speed.
• \(v_B\) be B’s final speed.
• \(\theta\) be the angle between A’s initial and final velocity vectors.

Since B initially rests and the masses are equal, momentum conservation gives:

\[
\vec u=\vec v_A+\vec v_B
\]

For an elastic collision, kinetic energy conservation gives:

\[
u^2=v_A^2+v_B^2
\]

Now square the magnitude of the momentum equation:

\[
u^2=v_A^2+v_B^2+
2\vec v_A\cdot\vec v_B
\]

Comparing this with the kinetic energy equation:

\[
\vec v_A\cdot\vec v_B=0
\]

The two final velocity vectors must be perpendicular.

But we can go further.

Rearrange the momentum equation:

\[
\vec v_B=\vec u-\vec v_A
\]

Take the dot product with \(\vec v_A\):

\[
\vec v_A\cdot\vec v_B=0
\]

Therefore:

\[
\vec v_A\cdot\vec u-v_A^2=0
\]

Using the definition of the angle between the initial and final velocities:

\[
uv_A\cos\theta=v_A^2
\]

For a non-zero final speed, divide by \(v_A\):

\[
\boxed{v_A=u\cos\theta}
\]

This is a particularly useful result.

The final speed of A is determined by its initial speed and its deflection angle.

Because the outgoing velocities are perpendicular, we also have:

\[
u^2=v_A^2+v_B^2
\]

Substituting \(v_A=u\cos\theta\):

\[
v_B^2=u^2-u^2\cos^2\theta
\]

Using the trigonometric identity:

\[
1-\cos^2\theta=\sin^2\theta
\]

We obtain:

\[
\boxed{v_B=u\sin\theta}
\]

These two relationships allow us to solve either version of the measurement problem.

Case 1: The \(60^\circ\) direction is correct.

Given:

\[
u=5.0\text{ m s}^{-1}
\]

\[
\theta=60^\circ
\]

A’s required final speed is:

\[
v_A=5.0\cos60^\circ
\]

\[
\boxed{v_A=2.50\text{ m s}^{-1}}
\]

B’s required final speed is:

\[
v_B=5.0\sin60^\circ
\]

\[
\boxed{v_B=4.33\text{ m s}^{-1}}
\]

B must travel at \(30^\circ\) south of east so that the two velocity vectors add to the original eastward velocity.

Case 2: The \(4.0\text{ m s}^{-1}\) speed is correct.

Given:

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

\[
u=5.0\text{ m s}^{-1}
\]

Rearrange:

\[
\cos\theta=\frac{v_A}{u}
\]

\[
\cos\theta=\frac{4.0}{5.0}
\]

\[
\theta=\cos^{-1}(0.80)
\]

\[
\boxed{\theta=36.9^\circ}
\]

B’s final speed is:

\[
v_B=\sqrt{u^2-v_A^2}
\]

\[
v_B=\sqrt{5.0^2-4.0^2}
\]

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

In this case, B travels at \(53.1^\circ\) south of east.

(d) Find the greatest possible deflection angle.

The relationship we derived was:

\[
v_A=u\cos\theta
\]

The final speed cannot be negative.

Therefore:

\[
\cos\theta\geq0
\]

This means that A cannot deflect into the backward half-plane.

If A continues moving with a non-zero final speed, its deflection angle must be less than \(90^\circ\).

As A’s final speed approaches zero, its deflection angle approaches \(90^\circ\). At exactly \(90^\circ\), A stops and B takes all of the original velocity.

This is a useful physical constraint that you would not discover by applying momentum conservation alone.

08A final decision process for HSC questions

When you see an elastic collision involving equal masses, work through these decisions before substituting numbers.

The most useful shortcuts to remember are:

One-dimensional elastic collision, equal masses:

\[
v_A=u_B,\qquad v_B=u_A
\]

Two-dimensional elastic collision, equal masses, one initially stationary:

\[
\vec u_A=\vec v_A+\vec v_B
\]

\[
\vec v_A\perp\vec v_B
\]

Two-dimensional elastic collision, both initially moving:

\[
\vec u_A+\vec u_B
=
\vec v_A+\vec v_B
\]

\[
u_A^2+u_B^2=v_A^2+v_B^2
\]

The key is knowing when each result applies.

Once you can recognise the equal-mass shortcuts, the next useful step is solving collisions where both objects are initially moving in two dimensions, particularly when their final directions are known but their speeds are not. That’s where the full component method becomes essential.