Exploding Projectile: Momentum and Center of Mass Approach

[@ArgentumFirst]

Thank you very much Sir. Truly, what you are doing is a *Great Service* to the student science community of the whole globe. Keep it up! The Karma returns!👍

[@PhysicsNinja]

Thank you so much

[@patrickzhao9072]

thanks for the video, this is the exact same question I'm searching for. My teacher never explained clearly. Greate

[@mileslegend]

Amazing.....🔥🔥🔥🔥🔥🔥 I have understood it immediately Thanks for the work

[@trishla.]

Thanku ninja.... I was searching for this concept...

[@EsauBanda-x6j]

well explaining continue creating more and more videos

[@ceeyar2003]

thank you. it was very helpful..

[@andrewjustin256]

Would you please delineate why the time for the object with mass m should take as much time to trace the new projectile as it took for the mass 2m to approach to the top?

[@sanjayshinde1468]

Please can you tell me the final velocity of mass 2 , how to calculate it ?

[@nimoo1929]

when explosion is in any point of the path, after explosion the path of the center of mass will be same or not?

[@Rizwan_Khan07]

A projectile on its path gets divided into two parts at its highest point which is true? A)k.e increase B)k.e decrease C) momentum increase D)momentum decrease

[@bhavitbansal6502]

Ke decreases?

[@hueypeng8026]

May I know why does the vertical components for the first, second fragments and centre of mass will be the same?

[@utkarshshirbhate2938]

Laws are motion are true for point masses. I case of big bodies, we consider centre of mass, a point where we suppose whole mass is focused or concentrated. Thus velocity of COM is the velocity of body as a whole

[@mileslegend]

An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is x0. The projectile breaks into two pieces that fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. Neglect air resistance and effects due to the earth’s curvature. How far away from the original launching point does the larger piece land?

[@patrickzhao9072]

Why is the time to got top and bottom are the same, will the distance to go bottom longer since the explosion happened so the time being longer?

[@randomstuff535]

has anyone figured this out lol

[@JovialJolaChip]

In such questions we neglect air resistance, so there is no opposing force expect gravity that affects the time it takes to reach the top. Since gravity is a conservative force, the time taken to the reach from the top to the bottom or from bottom to top remains the same regardless of the path taken.

[@saptaswabasu4957]

What if the 2 masses didnt break uniformly ??then the formula will not work right?Then we have to do by full method like writingequations and all

[@neonskie7008]

U can substitute the different masses in the law of conservation of momentum equation mv1=mv2

[@neonskie7008]

U can substitute the different masses in the law of conservation of momentum equation mv1=mv2

[@Rkplays567]

thanks bruh it helps a lot

[@PhysicsNinja]

Happy to help

[@MEBMohdSaifuddinKhan]

But isn't gravity acting? And gravity would be an external force. And it's effect cannot be neglected since it's acting for sufficiently large time.

[@PhysicsNinja]

Gravity is acting in the vertical direction only so no impact in the horizontal. Also assume the the explosion is fast and the impulse due to gravity in that short period of time is small compared to the explosion.

[@MEBMohdSaifuddinKhan]

Thanks a lot for replying. I think the only reason we can conserve momentum is because the impulse of gravity can be neglected since time for which mg acts is very small. Even if the broken part had a velocity in y we could have used momentum conservation to find it (obviously just after the explosion)

[@bluefire6470]

​@@MEBMohdSaifuddinKhan I have been looking for an answer to this question for so long. I think you are right. Thank you.

[@bluefire6470]

It might also be due to the fact that Gravity is much, much smaller than the internal forces of the explosion, so you just ignore it.

[@ajithshaminda4827]

is the vertical velocity or the horizontal component of it equal to zero?

[@PoonamSingh-oi4zj]

Vertical velocity at the top most point is 0

[@umarruban5012]

You look like abd villers 😂

[@gladycherian]

Thank you so much.... 🙏

[@PhysicsNinja]

You're most welcome

[@nkaur8362]

this question came in jee advanced 2021 doing some modifications