When a bomb explodes, does it momentum remain same?

Question

If, from an aircraft, a bomb is thrown to an object placed at ground and bomb explodes before it hits the object, i.e if it explodes in the middle of its path, does it momentum remain same?

I knew that law of conservation of momentum can be used when no other forces or agents, except action and reaction, do not act between and among objects. Since in this case many kinds of chemical reactions involve, can we apply law of conservation of momentum in this case? If law of conservation of momentum cannot be applied here, what would happen? Either momentum will increase or it will decrease. But I'm not sure which may occur. Please tell me what may happen and what are the reasons behind it.

At first I thought momentum will decrease because the mass of particles created after explode of bomb will be near to zero, and the product of velocity with a large value and mass with very small value will give another value which is also near to zero. Later I started to think that my idea is not correct, because the product can be higher if the value of velocity of very small particle becomes so much higher.

Answer 1

UPDATE : The explosion itself conserves linear momentum, regardless of how small the fragments are. If we ignore gravity and air resistance and all other external forces, there is no change in total momentum. This is because the internal forces all occur in equal and opposite pairs (Newton's 3rd Law).

If we take the external forces into account, then momentum is not conserved. Even during the brief explosion, the bomb and its fragments are being accelerated downwards by gravity. Air resistance will also affect different fragments unequally because it depends on the size and speed of the fragments, which are very unlikely to be equal.

However, if we extend "the system" to include the Earth and its atmosphere as well as the bomb and its fragments, then we can again say that momentum is conserved - before, during and after the explosion. All of the forces which we are taking account of (gravity, air resistance, buoyancy, and the bomb blast pushing fragments apart) are now internal forces, so Newton's 3rd Law again applies.

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ORIGINAL ANSWER :

Neglecting air resistance, and until any of the fragments of the bomb reach the ground, the centre of mass of the bomb follows the same trajectory as it would if the bomb did not explode - ie part of a parabola.

Conservation of linear momentum does not apply here because there is an outside force (gravity) which changes the magnitude and direction of the total momentum. However, all of the forces in the explosion are internal forces (action/ reaction) which do not alter the motion of the centre of mass and do not affect total momentum.

The explosion does not change the total mass of the bomb.

Answer 2

It will stay the same, if we neglect the variation due to gravity (every external force is going to change the momentum).

If we assume a uniform distribution of the shrapnels' mass (same size for all shrapnels), the shrapnels going in the direction the bomb was originally going will have, on average, higher velocity.

With a great simplification, we can say that, on explosion, the bomb splits in two identical parts of mass $M/2$ upon explosion. Assuming no gravity, the motion will be 1-dimensional. Since total momentum is conserved, we will have:

$$M V = \frac M 2 v_1 + \frac M 2 v_2$$

That is to say,

$$V = \frac{v_1 + v_2}{2}$$

Notice that kinetic energy won't be conserved, because we have to take into account some kind of chemical energy that triggers the explosion, which will be partly converted in heath, sound and radiation.

Anyway, remember that in the absence of external forces total momentum is always conserved.

Edit (clarification):

Yes, gravity is going to increase the total momentum of the system: $\frac{ d\vec q}{dt} = m \vec g$. The momentum of the whole system increases constantly in time (the bomb -before the explosion- and the shrapnels -after the explosion- accelerate with constant acceleration g towards the ground).

Anyway, your question if I got it correctly was more if the process triggering the explosion (chemical reaction etc.) would change the momentum. That's not the case: only an external force will change the total momentum of any system.

So the momentum of the system as a whole is increasing constantly due to gravity; but, the total momentum immediately before the explosion is exactly the same a the total momentum immediately after

Answer 3

That the bomb breaks apart due to the explosive forces which are internal to the system, has nothing to do with the trajectory of the center of mass. As user Sammy Gerbil points out correctly, the initial trajectory of the bomb was parabolic and even if the bomb exploded into million fragments of different masses, the center of mass would continue on it's path the same way as if nothing has happened. Explosion doesn't change anything for the COM. The only external force is the gravitational force (we are neglecting any drag forces due to air and the wind direction) and it's direction is straight down, so the COM's vertical component of the velocity is increasing. Component of linear momentum along the y-axis through out the parabolic trajectory of the COM continues to increase. While the horizontal veloctiy of COM and hence the horizontal component of linear momentum of the system remains unchanged because there is no net external force acting in that direction. Mass of the bomb remains unchanged. This scenario includes mass being converted to energy.

Momentum is not conserved in this situation because an external force, gravity, is acting on your system, the bomb.

Remember that analyzing each fragment of the bomb individually would require you to take into account all the forces that act on it. This includes the explosive forces whose magnitude is impossible to know. Also, for individual fragments, vertical and horizontal linear momentum are changing in a way different from the COM, but when you vector add them over all the individual pieces, you would once again find the same vertical and horizontal values of the COM as a function of time, as if it were the only particle moving in a parabolic path.