An elastic cube sliding without friction along a horizontal floor hits a vertical wall with one of its faces parallel to the wall. The coefficient of friction between the wall and the cube is $\mu$. The angle between the direction of the velocity $\mathbf{v}$ of the cube and the wall is $\alpha$. What will this angle be after the collision (see Figure for a bird's-eye view of the collision)? 1

Disclaimer: Before this gets tagged "Homework-like", I'd like to clarify that I'm not just looking for the answer. I'd like to get clarifications on how exactly this object interacts with the surface during collision.

Approach: Suppose the components of the velocity after collision are $v_{x}$ and $v_{y}$, in the $-x$ and $+y$ directions. The frictional and Normal impulses act along $-x$ and $-y$ directions. Keeping that in mind, we can write the equations for the momentum change(s): $$m(v_{x}+v\sin(\alpha))=\int_{t_{i}}^{t_{f}}N\text{d}t$$ $$m(-v_{y}+v\cos(\alpha))=\int_{t_{i}}^{t_{f}}\mu N\text{d}t$$

We can the divide the two equations to get: $$\mu=\frac{v\cos(\alpha)-v_{y}}{v_{x}+v\sin(\alpha)}$$

Now, from here on, you can take 2 approaches:

  1. The word "elastic cube" seems to hint at an elastic collision going on, which I initially treated as implying the total KE is constant, which would mean the speed is the same after collision, i.e $v$, and if we designate the desired angle as $\theta$ (with the upward vertical), then we can write $v_{x}$ and $v_{y}$ in terms of $v$ and $\theta$, and solve for $\theta$ easily, using basic trig identities. The expression comes out to be $\theta= \alpha +2 \arctan(\mu)$.

  2. However, the presence of friction challenges the assumption of constant KE. I tried to investigate whether or not friction does work:

The cube is mentioned to be elastic. Elasticity means the tendency to regain its shape. Based on this, I initially tried to reason that initially the surface gets deformed, but then the original shape is regained. I thought that this implies the net "displacement" of the surface is $0$, hence friction does no work.

However, this assumption is also wrong, now that I think of it. That's simply how non-conservative forces work: bringing a ball from point A to B, then back to A, on a rough surface, does not mean that the net work by friction is 0. Friction does negative work in both trips. Something similar happens here too, I presume. Besides, I believe the boundary of the surface too, moves vertically, (albeit slightly)?

To calculate the angle, we need the ratio $v_x/v_y$. The above argument suggests it's difficult to explicitly know what $v_{y}$ is, hence we somehow need to extract the value of $v_x/v_y$ from that equation.

Now, there has to be some significance of the cube being elastic. In most collision problems, Newton's coefficient of restitution $e$ is used to account for how elastic the collision is. And, when the collision isn't head on, many sources state the equation $e=1$ (for elastic collisions) is applied along the line of impact. In this case, this means $\dfrac{|v_{x,\text{after collision}}|}{|v_{x,\text{before collision}}|}=1$, i.e $v_{x}=v\sin(\alpha)$. With this, we can easily solve for the desired ratio. We get: $$\theta= \arctan{\dfrac{\tan(\alpha)}{1-2\mu \tan(\alpha)}}$$

However, the problem with this is that in my opinion, $e=1$ along the line of impact, is not some independent law of nature. It has to be derivable using energy and momentum considerations, and this is precisely what I wish to know primarily.

How can we, purely using energy and momentum considerations, and the fact that the cube is "elastic", deduce that the horizontal velocity magnitude is the same as that before the collision?

I just wish to know what the "elasticity" of the cube translates to, in terms of energy and momentum. Alternatively, I want to know why is it that we can apply $e=1$ "only along the line of impact". The answer to this must be based on how $N$ operates during collision.

Moreover, whatever that explanation may be, will directly imply that there is some loss of energy, since the answer using approach 2 differs from approach 1 (which assumed energy conservation). Which leads me to my second question:

What exactly happens during the collision which results in the loss of energy?

At the start of approach 2, I tried to come up with some sort of theory, however I need some kind of verification whether that's the correct reasoning or not.

Lastly, the answer in the book, states (apart from the expression given in approach 2), that the angle will be $\pi /2$ if $\tan(\alpha)>1/2 \mu$.

Why will be this the case?

It seems that I have posed a lot of questions, but in my opinion they are all very closely related, with the correct understanding of how the cube interacts with the wall during the collision.

  • 4
    $\begingroup$ If anyone is voting to close this question, please do consider mentioning the reason so that I can edit it appropriately. IMO, the question is not homework like, asks valid questions, and is definitely mainstream. To me, it seems like a perfectly valid question to ask on this site. $\endgroup$
    – satan 29
    Jul 25, 2021 at 17:19
  • $\begingroup$ Is your cube sliding on a horizontal floor, or moving up at a small angle from the vertical? $\endgroup$
    – R.W. Bird
    Jul 25, 2021 at 17:39
  • $\begingroup$ It's sliding along a horizontal floor, yes, and the velocity direction is shown in the image, at the moment of impact $\endgroup$
    – satan 29
    Jul 25, 2021 at 17:56
  • $\begingroup$ Is your horizontal surface moving up? $\endgroup$
    – R.W. Bird
    Jul 25, 2021 at 19:07

1 Answer 1


How can we, purely using energy and momentum considerations, and the fact that the cube is "elastic", deduce that the horizontal velocity magnitude is the same as that before the collision?

To understand this, it might be helpful to simplify the problem. We can first remove friction everywhere, since friction only acts in the $y$-direction and does not influence motion in the $x$-direction, which is what we are interested in. We can then perform a Galilean boost of $v\cos\alpha$ in the $y$-direction, so that the block moves only in the $x$-direction. In this frame, the floor and wall are moving backwards ($-y$-direction), but because we have removed friction, this movement is irrelevant and can be set to zero (assuming that they do not deform).

Therefore, we have reduced the situation to a perfectly elastic cube moving straight towards a wall, hitting and then moving back along the same direction from which it came. Because the cube is elastic, it can be replaced by a spring of equal mass oriented perpendicular to the wall.

There are two ways to deduce that the speeds of the spring before and after the collision must be equal:

  1. Energy argument: This is the easy way. The spring starts off with only kinetic energy. At the midpoint of the collision (where the spring is instantaneously still and its compression is maximum), all of this kinetic energy becomes elastic potential energy in the spring. All of this elastic potential energy must be converted back to kinetic energy when the spring moves off again. In the non-ideal case, some of the initial kinetic energy is lost to permanent deformation of the spring/surface.

  2. Force argument: This is slightly harder to see. The spring force is what brings the spring to a stop. This force is a function of the compression $x$ of the spring only, i.e. $F=F(x)$. During any infinitesimal span of time $\text{d}t$ during the collision, the change in momentum is $F(x) \text{d}t = F(x) \text{d}x/\dot{x}$. The total change in momentum from the start till the midpoint (incoming) is then $-\Delta p = \int_0^{x_0} F(x) \text{d}x/\dot{x}$. When the spring moves off (outgoing) again, the change in momentum is simply the same integral with the limits reversed. This must be equal to $\Delta p$ because both $F(x)$ and $\dot{x}$ are state functions of $x$ only. In other words, the outgoing movement is the incoming movement played in reverse. In the non-ideal case, the $F$ for the outgoing portion is not the same as the $F$ for the incoming portion, because the spring has been deformed and exerts less force when outgoing. Therefore the outgoing integral is not the same as the incoming integral for the momentum.

What exactly happens during the collision which results in the loss of energy?

When we go back to the original problem, the force argument (in the $x$-direction) is the one that applies directly, because the kinetic energy $$\frac{1}{2} m\left({v_x}^2+{v_y}^2\right)$$ is not conserved because of the friction on ${v_y}$. It might appear that the energy argument does not apply, but that is not true. The quantity that is conserved is $$\frac{1}{2} m{v_x}^2$$ except that it is not called the kinetic energy. This boils down to the fact that the $x$ and $y$ axes are orthogonal to each other. This is a subtle point, because the question did not say that the collision itself is elastic; rather only the cube is.

  • $\begingroup$ Thanks for the answer, this is almost exactly along the lines of what I wanted . However, just how appropriate is it to treat the cube as a spring? $\endgroup$
    – satan 29
    Jul 26, 2021 at 6:44
  • $\begingroup$ @satan29 Every object is compressible, and the ratio of force per unit cross-sectional area to the (fractional) change in length is known as the Young's modulus of the object (this is where the spring constant $k$ comes from). The word "elastic" means that when you let go, the object returns completely to its original shape, and all elastic potential energy is returned. $\endgroup$ Jul 26, 2021 at 6:49
  • $\begingroup$ That makes sense then, to model the ball as a spring, hmm $\endgroup$
    – satan 29
    Jul 26, 2021 at 6:50

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