3
$\begingroup$

Suppose a box of mass $m$ is thrown at angle $\theta$ with velocity $v$. At the topmost point,the box explodes into two identical smaller boxes. One of the boxes falls straight to the ground at that point.

I have seen that conservation of momentum is used to calculate the velocity of other box after the explosion. But gravity is always acting downwards. So how can momentum be conserved?enter image description here

$\endgroup$
1

4 Answers 4

6
$\begingroup$

The horizontal component of momentum is conserved since there are no forces that act in this direction.

The same cannot be said in the vertical direction since gravity is acting on the boxes.

So there will be conservation of momentum in the x-direction. You can see this mathematically in the equation $$2mv_0\cos(\theta)=0\times m + mv_2$$ which relates the horizontal momentum before the box separates to the horizontal momentum after it separates. Note how the first term on the RHS vanishes because this corresponds to the part of the box that moves vertically downward with no component of velocity in the x-direction.

$\endgroup$
4
$\begingroup$

Momentum in the x direction will be conserved when looking at the two boxes alone.

Momentum in the y direction will be conserved when looking at the two boxes in the earth as one system, since it's the earth that imparts momentum to the boxes. This is easy to manage because it's well known that the earth will impart $mg\Delta t$ momentum to an object with mass $m$ that's in free fall for time $t$.

$\endgroup$
1
  • $\begingroup$ +1 Yes, momentum is conserved in the y-direction, but you don't have to include Earth in the "system". Conservation laws always include a current (impulse for momentum, work and heat for energy). Sometimes the current is zero, and the quantity is ALSO constant. Earth's gravitational pull produces an impulse (flow of momentum) into the system of the boxes. $\endgroup$
    – Bill N
    Nov 4, 2021 at 2:25
3
$\begingroup$

It is true that momemtnum is conserved only if external force is zero. But here as in written in photo: $$2m(v_0 \cos\theta)=m(0)+mv_2$$

This equation represents momentum conservation only in horizontal/x direction, where there is no external force (if mentioned no air resistance or wind). Gravity acts only in downward/y direction so in that direction, we can't conserve momentum.

$\endgroup$
5
  • 1
    $\begingroup$ Momentum is conserved in the vertical direction as well. You just need to remember that the Earth is also part of the system, and gets accelerated upwards. $\endgroup$
    – Mark
    Oct 31, 2021 at 1:44
  • $\begingroup$ Yes, but only if we take earth and mass as a system whole. Then the gravity between two will be counted as internal force, and will not affect momentum coservation equation. Here however, only $m$ mass is taken as sytem, so field force of gravity between the two by earth is counted as external force. $\endgroup$ Oct 31, 2021 at 2:56
  • $\begingroup$ Momentum is always conserved, but not always constant within the system. Impulses due to forces external to the system are part of the proper concept of conservation: $\vec{p}_{final}=\vec{p}_{initial}+\vec{J}$. Conservation always involves a current, and impulse is the current for momentum. "Conservation" does NOT equal "constant". $\endgroup$
    – Bill N
    Nov 4, 2021 at 2:22
  • $\begingroup$ @BillN It would needlessly, increase difficulties by adding impulse terms. Yes its true that I exchanged conservation for constant which is wrong. But we can safely say that vertical momentum is not constant. $\endgroup$ Nov 7, 2021 at 17:08
  • $\begingroup$ @KshitijKumar Pedagogically, I like to include the impuslese term to force the students to define their system and determine whether there are is a net force from outside the system. If there isn't, then they have a zero term. If it's not consciously examined, they assume it will never be there. $\endgroup$
    – Bill N
    Nov 7, 2021 at 19:01
0
$\begingroup$

Conservation of momentum us used here because the explosion forces are assumed to be much larger than the gravitational force, so that on the time scale of the explosion, the effects of gravity can be ignored.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.