A 100 g ball with a speed of 5 m/s hits a wall at an angle of 45 degrees. The ball then bounces off the wall at a speed of 5m/s at an angle of 45 degrees. What is the change in the momentum of the ball?

When I look at this problem it seems intuitive to me that the answer should be 0, as the ball's mass remains constant and the speed remains unchanged. However, this is wrong as the velocity changes in this problem. After some calculations, you get around 0.7 kg m/s.

I'm having trouble understanding what this value really represents. How might this answer be useful?

enter image description here

Consider the diagram above. A ball travelling at 5m/s hits the wall and then travels at 5m/s again. What's the change in velocity? Assuming the angle to the wall is 45 degrees.

The ball has the same magnitude of velocity before and after hitting the wall so it doesn't make sense to me to give a numerical value to the change in velocity. The only change is in the direction. However, when you solve for the change in velocity there is a magnitude. What is the representing? Why is there even a magnitude when the speed is the same?

Edit: The magnitude of the velocity and momentum remains the same after collision with the wall. Assuming this is true, then why is the change in momentum 0.7 kg m/s? To me, this is saying there is a change in the magnitude of the momentum which is not true. So what is the value representing? I understand the magnitude remained constant while the direction changed. So why is there a numerical value in the change of momentum?

Links and resources to learn more would be appreciated.

  • $\begingroup$ It may help to think about the components of the momentum. Specifically, the component parallel to the wall, and the component perpendicular to the wall. $\endgroup$
    – PM 2Ring
    Commented Mar 21, 2023 at 18:03
  • 2
    $\begingroup$ This is clearly a conceptual question, not a "check my work" question. $\endgroup$
    – PM 2Ring
    Commented Mar 21, 2023 at 20:34

6 Answers 6


Sounds to me, you are not familiar with vectors. So I highly suggest to learn about vectors quickly, which is a simple field of mathematics.

Let's have a look in some more detail. x-axis points to the right, y-axis downwards, as I was being lazy ... and it doesn't matter.

In point A we have $\vec v = (v_x, v_y) = (v_x, 0) + (0, v_y)$, which means:

  • there's a vector $\vec v$, i.e. having direction and magnitude $|\vec v|$ (length)
  • you can describe it by its two components $(v_x, v_y)$
  • which you can also understand as the sum of two vectors $(v_x, 0) + (0, v_y)$
  • and you can always decompose $\vec v$ into its components or reconstruct it from $(v_x, v_y)$ (red parallelogram)


Now, what happens at point B, the wall?

  • component $v_y$ is conserved, i.e. vector $(0, v_y)$ doesn't change, as there is no physical reason to do so
  • component $v_x$ is reflected, i.e. it turns into $\vec v_x* = - \vec v_x = (-v_x,0)$ (the blue vector)

Point C shows the result, which holds since it bounced back from the wall:

  • $\vec v* = (-v_x, v_y)$

The change in direction you can easily tell by looking at its components. But what's about its magintude?

  • incoming: $|\vec v|^2 = (v_x)^2 + (v_y)^2$ (according to Pythagoras)
  • reflected: $|\vec v*|^2 = (-v_x)^2 + (v_y)^2 = (v_x)^2 + (v_y)^2 =|\vec v|^2 $
  • so magnitude $|v|$ didn't change, while direction did
  • $\begingroup$ I understand the magnitude is the same. However, then why does the change in momentum or velocity have a magnitude after hitting the wall? $\endgroup$ Commented Mar 23, 2023 at 18:09
  • 2
    $\begingroup$ @QuinGardinerBax Because the change in a vector is itself a vector $\endgroup$ Commented Mar 23, 2023 at 18:30

One way to look at this is the following: for any system, you can relate the force the object experiences to its change in momentum by $$ \Delta \vec{p} = \int \vec{F} \, dt $$ In particular, if the force is constant throughout the period during which it acts, you have $$ \Delta \vec{p} = \vec{F} \Delta t. $$ This integral (or product, for constant force) is called the impulse delivered to that object. This vector equation works equally well if $\vec{p}$ changes direction, magnitude, or both.

This means that calculating the change in momentum $\Delta \vec{p}$ for an object tells us something valuable about the direction that a force acted on that object to effect that change. In your case, this means that the wall must have exerted a force directly to the left — any vertical components to the force between the wall and the ball must have been negligible, because $\Delta \vec{p}$ doesn't have a significant vertical component.


I'd recommend that you think first about a simpler case: you throw a ball straight at a wall (in the x-direction, let's say) and it bounces back with equal and opposite velocity. The wall has to exert an impulse in the –x direction on the ball to stop it and then an equal impulse to get it moving at equal speed in the –x direction. Twice as much impulse is needed as to stop it, because the momentum change is twice as much!

The other answer explains nicely what is meant by 'impulse'.


Momentum is a vector, so actually its change, being a vector sum, is itself a vector. Asking "what's the change in momentum" is a little misleading. If they asked you "what changes in the components of the momentum vector", it would have been helpful. However, being exposed to this type of questions, I tend to think that they want to "trick" more than to "point" you.

At any rate.

You already have the pieces of the puzzle, and I am answering just because I see you want "a more detailed answer".

MS-SPO has already beautifully depicted the components of the momentum vector during the three salient instants of your problem. Before, during and after collision.

From there, you see that the vertical component of the ball's momentum,

$$p_y = m v_y$$

is not affected by the collision as its direction is parallel to that of the wall.

On the other hand, the horizontal component does get affected, as it is perpendicular to the wall.

$$p_{x,after} = m v_{x,after} = - m v_{x,before}$$

So, what's the change? The change in $p_y$ is zero, the change in $p_x$ is

$$p_{x,after} - p_{x,before} = - 2 m v_{x,before}$$

And to calculate this, you just have to find out what $v_{x,before}$ was, which is simply your 5 m/s multiplied by $\cos (\pi/4)$, or $\dfrac{1}{\sqrt{2}}$, which gives me $v_{x,before} \approx 3.53 m/s$.

If you now plug the 100g mass into the equation above for the change in momentum, you get that rough 0.7 kg m/s you were looking for.

What does it mean? It means that the collision actually did change the momentum of the ball. You can see it in two ways:

  • it altered its direction, rotating it by 90 degrees, while leaving its modulo unaltered

  • it left the y component unchanged, while it flipped the x component 180 degrees ('reflected' one could say), without changing its modulo.


Think of a taking a slow drive forwards with $v=10\,\mathrm{m/s}$. Suddenly you are hit by a truck. You are now moving backwards with the truck with $v=-10\,\mathrm{m/s}$.

Did a numerical change happen? After all, the numerical values are the same, only the directions are different (the sign is a descriptor of direction for scalar values).

Yes, obviously a numerical change happened. In fact, that numerical change is $\Delta v=20 \,\mathrm {m/s}$ (there is a difference of 20 between the two speed values on the number line).

In general when not dealing in 1-dimensional scenarioes and using vectors, the difference between two is called a difference vector, because both a numerical change as above and a directional change happens. The numerical change you mention in the question would be the magnitude of this difference vector. And you can be certain that you can feel this numerical difference just as well as you would feel it when hitting a truck.


let us look at what happens at the wall: the ball has incoming momentum in the positive x-direction, and departing momentum in the negative x-direction ; the momentum for the y-direction is the same in both cases.

given that the velocity of the ball is 5m/s , it follows that we have v[y]=5 sin(45) in all cases concerning the y-direction. For the incoming case concerning the x-direction we have v{incoming}= 5 cos(45)=3.54, and for the departing case we have v{departing}=-5 cos(-45)=-3.54.

Since we are interested in the change in momentum, we have mass to consider. The change in momentum is p{change}=p{departing}-p{incoming}

=0.1kg (-3.54m/s)-0.1kg(3.54m/s) =-0.354-0.354=>-0.708kg m/s

For the y-axis the change in momentum is equal to zero.

The total change in momentum is given by the vector contributions from both directions . It follows that p{change}^2=p{change in x}^2 + p{change in y}^2 = (-0.708)^2 +(0)^2 => p{change}=0.708kg m/s.

*The negative sign indicates a change in direction.


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