# A mass collides with an inclined plane

• Rikudo

#### Rikudo

Homework Statement
An object with mass m free fall and collides with an inclined plane that cannot move with velocity V. The angle between the vector of final velocity and a line that is perpendicular to the surface?
Relevant Equations
Newton third law
impulse-momentum theorem
In my textbook, it is stated that "if an object elastically hit an frictionless inclined surface with angle between the vector of initial velocity and an imaginary line that is perpendicular to the surface ##\alpha##,then the angle between the line and final velocity vector will also be ##\alpha##. Fig.1 illustrates this statement

However, I'm not sure that this is true,so I'm going to prove it.

Note:
1. At X axis, the final velocity is marked as ##V'_x## ,and ##V_x##means initial velocity
2. Fig.2 illustrates the force that works on the object

M's impulse at X-axis :
$$−N_x\,dt=dP_{x_M}$$$$∫−N_x\,dt=ΔP_{x_M}$$
since the velocity of the plane does not change, then ##\Delta P_{x_M}## is 0. As a result :$$∫N_x\,dt=0...(1)$$

Now, for the impulse works on ##m## at X-axis : $$N_x\,dt=dP_{x_m}$$$$∫N_x\,dt=ΔP_{x_m}...(2)$$ Substitute equation 1 into equation 2
$$0=m(V'_{x_m} − V_{x_m})$$ Since the object free fall,the initial velocity of ##m## in X axis is 0$$V'_{x_m}=0$$

This breaks my textbook's statement ,and this does not make any sense because as in fig.1 , we know exactly that ##m## has ##V'_x##

It can be concluded that my method is forbidden.
So,my question is : Is there any law that my method breaks?

#### Attachments

• IMG_20210619_182934.jpg
16.6 KB · Views: 58

Your assumption that the momentum of the plane is zero is incorrect. Look at the 1d collision of elastic balls where the stationary ball is much more massive than the moving ball to see how to think about this.

... momentum of an object is not conserved in a collision with the Earth, which is effectively what this is (the inclined plane is fixed to the Earth). E.g. a bouncing ball reverses the direction of momentum at every bounce.

Instead, in an elastic collision with the Earth it's the object's energy and not momentum that is conserved. As @caz says, you can see this from an elastic collision with a large mass, and taking the limit as that mass becomes infinite.

##\Delta P_{x_M}## is not zero and hence (1) isn't true either. There is some small change (in practice negligible) in the velocity of the plane, which when multiplied by its huge mass, gives a good non zero value for ##\Delta P_{x_M}## .

... momentum of an object is not conserved in a collision with the Earth, which is effectively what this is (the inclined plane is fixed to the Earth). E.g. a bouncing ball reverses the direction of momentum at every bounce.

Instead, in an elastic collision with the Earth it's the object's energy and not momentum that is conserved. As @caz says, you can see this from an elastic collision with a large mass, and taking the limit as that mass becomes infinite.
But, there is no external force.So ,isn't the momentum is supposed to be conserved?

But, there is no external force.So ,isn't the momentum is supposed to be conserved?
The momentum of the system (particle +earth )is conserved . The momentum of the particle (object) alone is not conserved, that's what @PeroK says.
But in order to apply conservation of momentum for the particle+earth system, you have to take into account the change in the momentum of the Earth which is of the form ##0\cdot\infty## because the change in velocity might be small (practically 0) but the mass of Earth is too big (in a way infinite).

But, there is no external force.So ,isn't the momentum is supposed to be conserved?
The external force is provided by the collision with the Earth.

And, in any case, gravity is an external force on the object, so there is no reason for conservation of momentum of the falling object and bouncing object.

The momentum of the system (particle +earth )is conserved . The momentum of the particle (object) alone is not conserved, that's what @PeroK says.
But in order to apply conservation of momentum for the particle+earth system, you have to take into account the change in the momentum of the Earth which is of the form ##0\cdot\infty## because the change in velocity might be small (practically 0) but the mass of Earth is too big (in a way infinite).
I followed your advice, but the result is that the final velocity of ##m## in X-axis has no solution

##M##'s impulse:
$$- \int N_x \,dt= M\Delta V_{x_M}$$ ##M## is so big so that it can be considered as ##\infty##
$$- \int N_x\,dt = \infty (0)$$
Infinity times 0 is undefined , and I will replace this result as "?"
$$- \int N_x \,dt = ?\,...(1)$$

##m##'s impulse :
$$∫N_x\,dt=ΔP_{x_m}...(2)$$ Substitute equation 1 to 2$$-? = m (V'_{x_m} - V_{x_m})$$
The initial velocity in X direction is 0,so :
$$-? = m (V'_{x_m})$$ $$\frac {-?} m = V'_{x_m}$$
Which means the final velocity of m in X-axis has no solution

• Delta2
Infinity times 0 is not always undefined it might be something finite and I think such is the case here, it is ##M\Delta V_{x_M}## where M is very big but ##\Delta V_{x_M}## is very small. When you multiply something very big by something very small you usually get something in between, that is something medium if I can put it this way.

Your final equation should be if i replace "?" by ##M\Delta V_{x_M}##
$$V'_{x_m}=-\frac{M}{m}\Delta V_{x_M}$$

@Rikudo you're making this way too complicated. First, consider an elastic collision between a mass ##m## with initial velocity ##u## and a stationary mass ##M##. After the collision we have:
$$v = \frac{m - M}{m + M}u, \ \ V = \frac{2m}{m + M}u$$ That's an exercise for you, using conservation of energy and momentum. Where ##v## is the velocity of ##m## and ##V## is the velocity of ##M## after the collision.

Note that if we take the limit as ##M \rightarrow \infty## (or ##M >> m##), then ##V \rightarrow 0## and ##v \rightarrow -u##. And we see that, as expected, the small mass changes direction in a collision with a large object.

Now, if the small mass collides elastically with a smooth, immovable surface at some incident angle, then we assume that there is no friction in the direction tangential to the surface and an elastic collision in the direction normal to the surface. E.g. if the surface is horizontal, then: $$v_x = u_x,\ v_y = -u_y$$ And this gives us directly that the angle of reflection is equal to the angle of incicence.

Also notice what MV goes to in the limit.

@Rikudo you're making this way too complicated. First, consider an elastic collision between a mass ##m## with initial velocity ##u## and a stationary mass ##M##. After the collision we have:
$$v = \frac{m - M}{m + M}u, \ \ V = \frac{2m}{m + M}u$$ That's an exercise for you, using conservation of energy and momentum. Where ##v## is the velocity of ##m## and ##V## is the velocity of ##M## after the collision.

Note that if we take the limit as ##M \rightarrow \infty## (or ##M >> m##), then ##V \rightarrow 0## and ##v \rightarrow -u##. And we see that, as expected, the small mass changes direction in a collision with a large object.

Now, if the small mass collides elastically with a smooth, immovable surface at some incident angle, then we assume that there is no friction in the direction tangential to the surface and an elastic collision in the direction normal to the surface. E.g. if the surface is horizontal, then: $$v_x = u_x,\ v_y = -u_y$$ And this gives us directly that the angle of reflection is equal to the angle of incicence.
Yes, I am also aware of this method. I notice that when using conservation of energy to create this formula, we always uses velocity that is normal to the surface (Obviously,the reason is that there is an action-reaction force that works on both masses in that direction).This reason also applies to the velocity in tangential direction (Because there is no friction,there is no action-reaction force working on both object in tangential direction. As a result, the velocity in that direction does not change after collision ).

I agree with you that my method is really complicated. one of the reasons I use the method is to prove my theory, that is using ##x## and ##y## direction of cartesian on the conservation of energy will yields the same result.

Infinity times 0 is not always undefined it might be something finite and I think such is the case here, it is ##M\Delta V_{x_M}## where M is very big but ##\Delta V_{x_M}## is very small. When you multiply something very big by something very small you usually get something in between, that is something medium if I can put it this way.

Your final equation should be if i replace "?" by ##M\Delta V_{x_M}##
$$V'_{x_m}=-\frac{M}{m}\Delta V_{x_M}$$

$$V'_{x_m}=-\frac{M}{m}\Delta V_{x_M}$$
Since the value of##\Delta V_{x_M}## is almost zero, it can be replaced by ## \sin(\frac{1}{x})## with ##x \rightarrow \infty## (because both ##\Delta V_{x_M}## and ## \sin(\frac{1}{\infty})## has the same value. Also, ##M## can be replaced by ##x##
$$V'_{x_m}=-\frac{1}{m}\lim_{x \rightarrow \infty} {x \sin(\frac{1}{x})}$$
$$V'_{x_m}=-\frac{1}{m}\lim_{x \rightarrow \infty} {\frac {\sin(\frac{1}{x})} {\frac {1} {x}}}$$
With using L' Hospital's rule, we get :
$$V'_{x_m}=-\frac{1}{m}\lim_{x \rightarrow \infty} {\frac {\cos(\frac{1}{x}) \frac {-1} {x^2}} {\frac {-1} {x^2}}}$$
$$V'_{x_m}=-\frac{1}{m}\lim_{x \rightarrow \infty} { \cos(\frac{1}{x})}$$
$$V'_{x_m}=-\frac{1}{m}\cos(\frac{1}{\infty})$$
$$V'_{x_m}=-\frac{1}{m}\cos(0)$$
$$V'_{x_m}=-\frac{1}{m}$$

But, doesn't this answer is wrong (because ##V'_{x_m} ##depends on the value of mass ##m##)?

Since the value ofΔVxM is almost zero, it can be replaced by sin⁡(1x) with x→∞ (because both ΔVxM and sin⁡(1∞) has the same value. Also, M can be replaced by
Sorry these assumptions are simply too arbitrary. Why ##\Delta V_M## can't be ##\frac{1}{x^2}## (this also is zero when ##x\to\infty##) and why M not be ##e^{x}## (this also goes to ##\infty## for ##x\to\infty##).

I think my guidance on considering the product ##M\Delta V_M## as sort of ##0\cdot\infty## is causing you to take wrong turns in your thought process. I suggest to better follow @PeroK guidance, I judge that his guidance at post #11 is really good.