# Mechanics 7: Conservation of Momentum and Collisions

This chapter, we introduce a new quantity, called *momentum.* It is defined as \({\bf p} = m{\bf v}.\) Since velocity is a vector, so is momentum — it is a scalar (\(m\)) times a vector, which is a vector. Momentum is a way of quantifying the force and impact of a moving object in a collision. Think of it this way: If you were standing on a baseball field, would you rather be hit by a baseball moving 100 mph (45 m/s) or a car? Even though these objects would be moving at the same speed, the car would have 5000 times the momentum of the baseball due to having 5000 times more mass. That’s momentum!

### Momentum and Newton’s Second Law

We learned previously that Newton’s second law tells us that the sum of the forces on an object equals it mass times acceleration, or

$$\sum {\bf F} = m {\bf a}$$

If we do a little math, we can re-write this another way. If we take the derivative of the momentum equation, we get

$$\frac d{dt} {\bf p} = \frac d{dt} (m {\bf v})$$

since mass is a constant, we can re-write this as

$$\frac{d{\bf p}}{dt} = m \frac{d{\bf v}}{dt}$$

and since we know that the time-derivative of velocity is acceleration, or \(dv/dt = a,\) we get

$$\frac{d{\bf p}}{dt} = m {\bf a}$$

and setting this equal to the first equation, we have another version of Newton’s second law,

$$\sum {\bf F} = \frac{d{\bf p}}{dt} \; .$$

To summarize, the first version tells us that the sum of the forces on a object are proportional to its acceleration. This second form tells us that the sum of the forces on a object tell us how its momentum changes with time. Same thing, just two different ways of seeing/thinking about it.

### Collisions and Conservation of Momentum

Conservation of momentum means that

*total*momentum does not change throughout an interaction (\(\Delta\bf{p} = 0\)). This is usually some type of collision. The two types of collisions are

*elastic*and

*inelastic.*For both types of collision, momentum is conserved. For an

*elastic*collision, energy is also conserved. Energy is not conserved for the

*inelastic*collision because work is done by one particle on the other, like crumpling or deformation, and probably heating.

For an *elastic* collision, where the masses do not change and energy and momentum are conserved, we can write (subscript \(i\) for initial, \(f\) for final)

$$m_1 {\bf v}_{1i} + m_2 {\bf v}_{2i} = m_1 {\bf v}_{1f} + m_2 {\bf v}_{2f}$$

and the energy (kinetic) is also conserved, written as

$$\frac12 m_1 {\bf v}_{1i}^2 + \frac12 m_2 {\bf v}_{2i}^2 = \frac12 m_1 {\bf v}_{1f}^2 + \frac12 m_2 {\bf v}_{2f}^2$$

Some algebra with these two equations gives

$${\bf v}_{1f} = \left (\frac{m_1 – m_2}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{2 m_2}{m_1 + m_2} \right) {\bf v}_{2i}$$

$${\bf v}_{2f} = \left (\frac{2m_1}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{m_2 – m_1}{m_1 + m_2} \right) {\bf v}_{2i}$$

#### Examples

- Say we have mass 1 of \(m_1 = 2\,\)kg traveling to the right at \(v_{1i} = 6\,\)m/s, and a mass 2 of \(m_2 = 4\,\)kg traveling left at \(v_{2i} = -2\,\)m/s. What are their final velocities?
Using the above equations, we get

$${\bf v}_{1f} = \left (\frac{2 – 4}{2 + 4}\right ) {\bf v}_{1i} + \left( \frac{2 \cdot 4}{2 + 4} \right) {\bf v}_{2i} =

\frac{-1}3 {\bf v}_{1i} + \frac 4 3 {\bf v}_{2i}$$

$${\bf v}_{2f} = \left (\frac{2 \cdot 2}{2 + 4}\right ) {\bf v}_{1i} + \left( \frac{4 – 2}{2 + 4} \right) {\bf v}_{2i} =

\frac 2 3 {\bf v}_{1i} + \frac 1 3 {\bf v}_{2i}$$

now insert the values for initial velocities

$${\bf v}_{1f} = \frac{-1}3 (6) + \frac 4 3 (-2) = – \frac 6 3 – \frac 8 3 = -\frac{14}3$$

$${\bf v}_{2f} = \frac 2 3 (6) + \frac 1 3 (-2) = \frac{12}3 – \frac 2 3 = \frac{10} 3 $$So what are these numbers telling us? Mass 1 enters the collision headed right (positive velocity) and leaves headed left (negative velocity). Vice versa for mass 2 — enters headed left and leaves headed right. Also, \(m_1\) is half the mass of \(m_2\), and it enters with \(3 \times\) the velocity of \(m_2\), and leaves only \(1.4 \times\) as fast.

Finally, we check the conservation of momentum. The initial momentum is \(m_1 {\bf v}_{1i} + m_2 {\bf v}_{2i} = 2\cdot 6 + 4(-2) = 4\) and the final momentum is

\(m_1 {\bf v}_{1f} + m_2 {\bf v}_{2f} = 2\left ( \frac {-14} 3 \right ) + 4\left( \frac {10} 3 \right) = 4,\) so momentum is conserved. - What if the masses and velocities were equal? Let \(m_1 = m_2 = m\) and \(v_{1i} = v,\) and \(v_{2i} = -v,\) so we get

$${\bf v}_{1f} = \left (\frac{m_1 – m_2}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{2 m_2}{m_1 + m_2} \right) {\bf v}_{2i} = 0 + \frac{2m}{2m} (-v) = -v$$

$${\bf v}_{2f} = \left (\frac{2m_1}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{m_2 – m_1}{m_1 + m_2} \right) {\bf v}_{2i} = \frac{2m}{2m} v + 0 = v$$

So the final velocities are equal and opposite from the initial velocities. In other words, the masses leave with the same velocity they started with, but headed in the opposite direction. Intuitively, this makes sense. - Now, let the masses and velocities of one be \(2 \times\) the other, like so: \(m_1 = m,\) \(m_2 = 2m,\) \(v_{1i} = v,\) \(v_{2i} = -2v.\) This gives us

$${\bf v}_{1f} = \left (\frac{m_1 – m_2}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{2 m_2}{m_1 + m_2} \right) {\bf v}_{2i} = \left(\frac{-m}{3m}\right) v + \left(\frac{2m}{3m}\right) (-2v)$$

$${\bf v}_{2f} = \left (\frac{2m_1}{m_1 + m_2}\right ) {\bf v}_{1i} + \left( \frac{m_2 – m_1}{m_1 + m_2} \right) {\bf v}_{2i} = \left(\frac{2m}{3m}\right) v + \left(\frac{m}{3m}\right) (-2v)$$

which simplifies to

$${\bf v}_{1f} = \left(\frac{-1}{3}\right) v + \left(\frac{2}{3}\right) (-2v) = -\frac{5}3 v$$

$${\bf v}_{2f} = \left(\frac{2}{3}\right) v + \left(\frac{1}{3}\right) (-2v) = 0$$

So this tells us that the first mass leaves at a velocity 5/3 larger than it started with, and headed in the opposite direction. The second mass comes to a dead stop! - Now, let’s do an example of a 2-dimensional elastic collision. That means we will need to work out the \(x\) and \(y\) coordinates separately, and this will be an opportunity to use the vectors that we talked about previously. Let mass one be \(m_1 = 6\,\)kg with an initial velocity of \({\bf v}_{1i} = 3\hat\imath + 3\hat\jmath\,\)m/s, and mass two is \(m_2 = 3\,\)kg with initial velocity of \({\bf v}_{2i} = -6\hat\jmath\,\)m/s. Now, we do the same calculation as before, but we need one for \(x\) and one for \(y.\)
\(x\) coordinate

$${\bf v}_{1f_x} = \left (\frac{m_1 – m_2}{m_1 + m_2}\right ) {\bf v}_{1i_x} + \left( \frac{2 m_2}{m_1 + m_2} \right) {\bf v}_{2i_x} =

\left ( \frac 3 9 \right) \cdot 3 + \left( \frac{2 \cdot 3}{9}\right) \cdot 0 = 1$$$${\bf v}_{2f_x} = \left (\frac{2m_1}{m_1 + m_2}\right ) {\bf v}_{1i_x} + \left( \frac{m_2 – m_1}{m_1 + m_2} \right) {\bf v}_{2i_x} =

\left( \frac{2 \cdot 6}9 \right) \cdot 3 + \left( \frac {-3} 9 \right) \cdot 0 = 4$$\(y\) coordinate

$${\bf v}_{1f_y} = \left (\frac{m_1 – m_2}{m_1 + m_2}\right ) {\bf v}_{1i_y} + \left( \frac{2 m_2}{m_1 + m_2} \right) {\bf v}_{2i_y} =

\left ( \frac 3 9\right) \cdot 3 + \left( \frac{2 \cdot 3}{9}\right) \cdot (-6) = -3$$$${\bf v}_{2f_y} = \left (\frac{2m_1}{m_1 + m_2}\right ) {\bf v}_{1i_y} + \left( \frac{m_2 – m_1}{m_1 + m_2} \right) {\bf v}_{2i_y} =

\left( \frac{2 \cdot 6} 9 \right) \cdot 3 + \left( \frac {-3} 9 \right) \cdot (-6) = 6$$So in summary, we get

$${\bf v_{1i}} = 3\hat\imath + 3\hat\jmath \qquad \qquad {\bf v_{2i}} = -6\hat\jmath $$

$${\bf v_{1f}} = 1\hat\imath + -3\hat\jmath \qquad {\bf v_{2f}} = 4\hat\imath + 6\hat\jmath $$Let’s check conservation of momentum. The initial momentum is

\(x\) direction

$$ m_1 {\bf v}_{1ix} + m_2 {\bf v}_{2ix} = 6 \cdot 3 + 3 \cdot 0 = 18$$

$$ m_1 {\bf v}_{1fx} + m_2 {\bf v}_{2fx} = 6 \cdot 1 + 3 \cdot 4 = 18$$\(y\) direction

$$ m_1 {\bf v}_{1iy} + m_2 {\bf v}_{2iy} = 6 \cdot 3 + 3 (-6) = 0$$

$$ m_1 {\bf v}_{1fy} + m_2 {\bf v}_{2fy} = 6 (-3) + 3 \cdot 6 = 0$$

which shows that momentum is conserved.Finally, we’ll finish with a diagram. Remember, vectors have a magnitude and a direction, which we can calculate. The magnitude formula is just the distance formula from geometry, \(d = \sqrt{x^2 + y^2}\) and the direction (angle) is from trig, or \(\theta = \arctan(y/x).\) So, we get

$$\| {\bf v_{1i}} \| = \sqrt{3^2 + 3^2} = 4.24 \qquad \theta_{1i} = \arctan(3/3) = 45^\circ$$

$$\| {\bf v_{2i}} \| = \sqrt{(-6)^2} = 6 \qquad \qquad \theta_{2i} = \arctan(-6/0) = -90^\circ$$$$\| {\bf v_{1f}} \| = \sqrt{1^2 + (-3)^2} = 3.16 \qquad \theta_{1f} = \arctan(-3/1) = -71.6^\circ$$

$$\| {\bf v_{2f}} \| = \sqrt{4^2 + 6^2} = 7.21 \qquad \theta_{2f} = \arctan(6/4) = 56.3^\circ$$Using these magnitudes (lengths) and direction (angles), we get this diagram:

Which shows us that mass \(m_1[.latex] comes in at a [latex]45^\circ\) angle and leaves slightly slower and at a little sharper angle, but mass \(m_2\) comes in headed straight “south” and leaves moving faster and headed “northeast.” You can use this technique to calculate the vectors (magnitude and direction) for any elastic collision. I’ll let you try a few in the homework.

### Inelastic Collisions

The second type of collision is

*inelastic,*where the masses stick together and there is a single final velocity; here momentum is conserved but energy is not conserved

$$m_1 {\bf v}_{1i} + m_2 {\bf v}_{2i} = (m_1 + m_2) {\bf v}_{f} $$

which can be re-written as

$${\bf v}_f = \frac{m_1{\bf v}_{1i} + m_2{\bf v}_{2i}}{m_1 + m_2}$$

### Examples

**One dimensional:**Say we have two masses that collide inelastically. Mass 1 is \(m_1=4\) kg with an initial velocity of \( {\bf v_{1i}} = 4\) m/s to the right. Mass 2 is \(m_2=2\) kg and is sitting still, so \( {\bf v_{2i}} = 0 \)m/s. Find the final velocity of the combined masses after the collision.Use the formula above to get

$${\bf v}_f = \frac{4 (3) + 2 (0)}{4 + 2} = \frac {12}6 = 2\,{\rm m/s}$$

So the final velocity is 1/3 less that the initial, also moving to the right.**Two dimensional:**Suppose we have two masses moving toward each other and they collide inelastically. Again, two dimensions means that we will have to use vectors. Mass 1 is \(m_1=3 \) kg with an initial velocity of \( v_{1i} = 2\hat\imath + 2\hat\jmath\) m/s. Mass 2 is \(m_2=2 \)kg with velocity \( v_{2i} = 2\hat\imath – 2\hat\jmath\) m/s. Find the final velocity of the combined masses after the collision.We can do this just as before, but we have to do each direction (\(x\) and \(y\)) separately.

\(x\) direction

$${\bf v_{f_x}} = \frac{3 (2) + 2 (2)}{5} = \frac {10}5 = 2\,{\rm m/s}$$

\(y\) direction

$${\bf v_{f_y}} = \frac{3 (2) + 2 (-2)}{5} = \frac {2}5\,{\rm m/s}$$

So the final velocity vector is

$${\bf v_f} = 2 \hat\imath + \frac2 5 \hat\jmath\;{\rm m/s}$$And the figure looks like this:

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Hello Doc. This is a timely entry, thanks!

We are teaching momentum right now, and ran into something interesting. In the case of a ball hitting a stationary wall (one-dimensional, perfectly elastic collision), the conservation of momentum equation ALONE does not work. One needs to add conservation of energy, in order for things to make sense. But the kid is not there yet: we have not talked about kinetic energy formula — we proceeded directly to momentum because we had covered m and v, touched upon vectors, so we were ready to mesh them together and introduce momentum, p.

So we are here:

m1v1 (initial) + m2v2 (initial) = m1v1 (final) + m2v2 (final)

But the wall (2) is always stationary, so its v2 = 0, both in the initial and final states.

It follows that

m1v1 (initial) = m1v1 (final)

Which only works in the scalar sense, as the velocity vector has changed direction.

Yes, yes: Conservation of energy makes sense of it. But the child is not there yet, and we are looking for a way to make sense of momentum WITHOUT introducing energy into the picture. Is there a way to explain what happens in this perfectly elastic, one-dimensional, collision of a ball vs. stationary wall, so that it makes intuitive sense to the kid while demonstrating the conservation of momentum?

Thanks,

Syd

Syd — thanks for the comment. Glad you are finding my material useful.

Sorry for the long delay in replying — I had to think about this question for a few days.

The problem is that a ball hitting a wall is “strange math.” It works, but you have to “interpret” the results.

Here’s the problem: It’s true that the wall has velocity v=0, but it’s also true that (compared to the ball) it has infinite mass. So, when you calculate its momentum, you have (infinity x 0). That’s troublesome for a direct calculation. The “full” answer requires calculus and limits, but we can approximate it.

Let the wall’s mass be very large — say 1000 kg. Then let its velocity be something very small — say 0.001 m/s. Now, do the elastic collision calculation with these numbers, but interpret anything that comes out very small as being 0, and anything that comes out very large as being infinity. I did it, and I get just the answer you’re looking for. I let the ball have a mass of 3 kg and velocity 5 m/s. I got a final velocity for the ball of -4.97 and the wall 0.031.

Let me know if you want further clarification.

Hope this helps! Thanks for the questions —

Doc