# Does the theory of the Conservation of motion apply to horizontal-vertical systems?

Suppose, a trolley is moving along a frictionless surface with a constant velocity. After some time a mass is added to the trolley. I read somewhere that the theory of Conservation of motion will apply to the system. As in the momentum of the trolley will be the same even after adding the mass. But how's that possible? Because I thought the theory only applied to horizontal-horizontal or vertical-vertical systems. But in the system above, the trolley was moving along the horizon and the mass was added on top of it aka vertically. So how does the theory apply here?

• Please give a link for this"conservation of motion". I only know of conservation of momentum and energy as far as kinematics goes. Sep 12 at 5:30
• @annav some books define momentum as the quantity of motion of a body. Never really understood this definition myself. Sep 12 at 5:31
• What did you mean by “… only applies to horizontal-horizontal or vertical-vertical systems…”? Conservation of momentum applies to every system, moving in every direction. Sep 12 at 15:30

You can consider the addition of mass to be an inelastic collision where kinetic energy is not conserved by the momentum is conserved. The linear momentum in the horizontal direction of motion should remain the same before and after the collision.

$$mu = (m+M)v$$

where $$m$$ is the mass of the trolley, $$M$$ is the added mass, u is the initial velocity of the trolley, and $$v$$ is the velocity of the trolley + added mass system

Momentum is always conserved.

Perhaps you are thinking of a system where the mass is brought up to speed with the trolley, then gently attached to it?

In that case, then the trolley will, indeed, have more momentum than before the addition -- precisely as much as the mass had by itself!

If the trolley crashes into the mass, then you're talking about the "sticking" collision that Joseph described.

If you drop the mass onto the trolley as it passes, and it doesn't slip off, the result should be very similar to having the mass stick to a trolley that crashes into it.

The only difference is that the mass had some vertical momentum before the collision. This will cause the entire system -- the trolley, the mass, and the ground -- to start moving slightly downward. Of course, in practice, this doesn't matter, since the Earth weighs so much more than everything else.

When you say "conservation of motion" you are referring to the conservation of momentum. Quantity of motion refers to momentum. Note that conservation of momentum not only works in horizontal and vertical directions, but in all three dimensions, provided net forces in all directions are zero. For example, consider projectile motion (with zero air resistance). Momentum will be conserved in the horizontal direction because the horizontal forces are zero, but along the vertical direction, their is a non-zero net vertical force (gravity) and so the vertical momentum of the projectile will not be conserved.

Your system is the mass and trolley, so that before the mass is added, the total momentum is that of the trolley only since the mass is presumably not moving.

Now when the mass is added (gently, so that there is no net downward force), the trolley will move with a different velocity, but the total momentum of the trolley and mass system will be conserved even though kinetic energy may not be. That is $$m_1v_i +m_2\times 0=(m_1+m_2)v_f$$ $$\rightarrow m_1v_i=(m_1+m_2)v_f$$ where $$m_1$$ and $$m_2$$ are the masses of the trolley and the mass respectively, and $$v_i$$, $$v_f$$ are the velocities of the trolley and then trolley+mass respectively.

This is analogous to a "sticking" collision where two objects combine. But for any collision occurring in an isolated system, the momentum is conserved so that the total momentum of all of the objects making up the system, is the same before and after the collision.

This is actually a common physics experiment for first year students where one places a weight on a moving cart measuring the velocities before and after, confirming that momentum is conserved.

To understand the conservation of momentum here, it is necessary to think about the forces involved, because the principle of conservation of momentum has a dependent clause:

• If the net external force on a system is zero, then the momentum is conserved.

Because momenta and forces are vector quantities, this actually holds true for each component of an orthonormal coordinate system as well, i.e.,

• If the component $$F_x$$ of the net force along the $$x$$-direction is zero, then the $$x$$-component $$p_x$$ of the momentum is conserved.

In the case in the OP, the trolley is moving along the $$x$$-direction at constant velocity because the net force is zero. Now, consider the mass and trolley as our system. If the mass is dropped onto the trolley, then the net external force on the system is the gravitational force exerted on the mass as it is dropping toward the trolley (because the net vertical force on the trolley is zero). Thus, the $$x$$-component of the net force on the system during this process is always zero. As a consequence, the $$x$$-component of the momentum is conserved, and you can set up, as usual $$p_{\textrm{trolley},x,i} + p_{\textrm{block},x,i} = p_{\textrm{trolley},x,f} + p_{\textrm{block},x,f}.$$

Note, of course, that the $$y$$-component of the momentum of the system is not conserved! It goes from non-zero due to the downward momentum of the block to zero after the block has hit the trolley and come to rest. This is due to the non-zero forces exerted before and during the collision: the gravitational force on the block before the collision and the increased normal force exerted upward on the trolley during the collision.