# How is the linear momentum conserved in this system?

Let there be a block on a frictionless surface. Let an agent constantly exert an invariable force, say $$F$$ on the block and does positive work by displacing the block by $$d$$ units. Here, the system is constituted of the agent and the block. By Newton's third law of motion, the block will exert force $$-F$$ on the agent and does negative work by displacing the agent by $$d$$ units. Thus work done on the block $$W = F\cdot{d}$$ and the work done on the agent is $$W' = -F\cdot{d}$$. Thus, the energy of the system is conserved.

Now, since it is an isolated system (agent + block) and no external force acts on the system, the linear momentum of the system must be conserved.

Hmm... Here I got blanked & baffled and couldn't point out how the linear momentum of this system is conserved. The initial velocity of both the components of the system is zero and hence their sum of the momentum is zero initially. Therefore at any moment after that the their sum of the momentum must be zero and since mass cannot be negative, their velocities must be opposite to each other. But they are moving in the same direction. So how can one's velocity is positive and other's is negative though they are moving in the same direction?

[Note: Here sign only denotes direction. Positive velocity means the body is moving in the positive direction & vice-versa.]

• "[T]hey are moving in the same direction"? If the agent pushes on the block, they will drift apart, in opposite directions. – J-T Nov 1 '14 at 12:10

Let an agent constantly exert an invariable force , say F on the block and does positive work by displacing the block by d unit. Here, the system is constituted of the agent and the block. By Newton's third law of motion, the block will exert force -F on the agent and does negative work by displacing the agent by d unit.

This is not physically possible for an isolated agent+block system. This setup is the horizontal equivalent of lifting oneself by ones bootstraps.

Suppose the agent and block have masses M and m, respectively. The easiest way to see what will happen is to analyze things from the perspective of a frame in which the agent and the block are initially at rest. Assume the agent exerts some force F on the block in the +x direction, for some time t, moving the block through some distance d in the +x direction. If the force is constant, the distance d is given by $d=\frac 1 2 \frac F m t^2$. Since the block moves in line with the force, the work on the block is $Fd = \frac 1 2 \frac {F^2} m t^2$. The block's velocity after applying this force is $v=\frac F m t$. The change in the block's kinetic energy is $\frac 1 2 m v^2 = \frac 1 2 \frac {F^2} m t^2$, which is equal to the work done on the block.

What about the agent? The block exerts an equal but opposite force on the agent. This force pushes the agent in the -x direction. The agent travels a distance $D=d \frac m M$ during the time $t$ during which the force is applied, but in the -x direction. The work on the agent is positive as the agent moves with rather than against the force applied by the block: $FD = \frac 1 2 \frac{F^2} M t^2$. The agent's velocity after applying this force is $V=- \frac F M t = -\frac m M v$. The change in the agent's kinetic energy is $\frac 1 2 M V^2 = \frac 1 2 \frac {F^2} M t^2$, once again equal to the work done on the agent. The work done on the agent and on the block is $W_{\text{tot}} = F (d+D) = Fd(1+\frac m M)$. This is of course equal to the total change in kinetic energy.

What about linear momentum? The linear momentum after the interaction is complete is $mv + MV = mv - M\frac m M v = 0$. Linear momentum is conserved.

Note that kinetic energy is not conserved. There is no such thing as conservation of kinetic energy, or conservation of work. The concept is conservation of energy. Some form of energy was needed to generate that push. For example, that push might have come from releasing a compressed spring or using a human arm. Releasing the spring or pushing with an arm converts potential energy into kinetic energy.

• +1 . If my hand pushes the block, the block exerts force on my hand. My hand would have gone in the opposite direction in response to the force,but my hands have done work against the force applied by the block and hence hands and the block always go together... Thanks sir, I will bother you for any query later! – user36790 Nov 1 '14 at 18:34
• @user36790 - Your hand by itself is not the agent. Your whole body is. Your hand stays in contact with the block because you are extending your arm ever further to keep your hand in contact with the block. Your body as a whole gains momentum opposite that of the block while you push on the block. In the end, your body and the block have equal but opposite momenta. If the block is small, it will have a much greater speed and greater kinetic energy than you will. It's the other way around if the block is huge (e.g., a ten ton block of granite). – David Hammen Nov 1 '14 at 19:03
• Sir, one thing I am not understanding is that though the velocity of the agent (my hand) at time $t$ has negative velocity, how can it move together with the block in the positive direction?? – user36790 Nov 2 '14 at 1:49
• You are constructing a system (block+hand) that has external forces acting on it. The conservation laws in general do not apply to such systems. Then to make matters worse you are ignoring those external forces. – David Hammen Nov 2 '14 at 13:37
• Sir, I think I got my answer. My body will move in $-x$ direction but the velocity will be negligible to that of block as in $\frac{m}{M} . v$ , $M > m$ . Thanks! – user36790 Nov 2 '14 at 17:58

What you described is completely impossible. Your agent must use some sort of engine to generate the force, this engine has to interact with the ground, or the atmosphere, or something else which lies outside your 'block + agent' system. Therefore, your system is not isolated.

A quick example: suppose the agent uses a jet engine. Then the true isolated system is: 'block + agent + jet fuel'. Jet fuel flies backwards with enormous speed, preserving the total linear momentum.