Exploding Projectile: Momentum and Center of Mass Approach

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Күн бұрын

Physics Ninja looks at an exploding projectile problem. The explosion at the top of the projectile cause it to split into 2 pieces of equal mass. We use conservation of momentum at the top of the trajectory to find the velocity of the second fragment and it's total range. I also use center of mass to calculate the same distance.

Пікірлер: 33

@patrickzhao9072 3 жыл бұрын

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

@ArgentumFirst 4 ай бұрын

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 4 ай бұрын

Thank you so much

@mileslegend Жыл бұрын

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

@trishla. 3 жыл бұрын

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

@user-lo7pi5nc3c 6 ай бұрын

well explaining continue creating more and more videos

@mileslegend Жыл бұрын

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?

@ceeyar2003 8 ай бұрын

thank you. it was very helpful..

@sanjayshinde1468 3 ай бұрын

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

@nimoo1929 14 күн бұрын

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

@hueypeng8026 3 жыл бұрын

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

@utkarshshirbhate2938 2 жыл бұрын

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

@patrickzhao9072 3 жыл бұрын

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 7 ай бұрын

has anyone figured this out lol

@saptaswabasu4957 3 жыл бұрын

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 3 жыл бұрын

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

@neonskie7008 3 жыл бұрын

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

@ajithshaminda4827 2 жыл бұрын

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

@PoonamSingh-oi4zj Жыл бұрын

Vertical velocity at the top most point is 0

@Rizwan_Khan07 3 жыл бұрын

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 Жыл бұрын

Ke decreases?

@Rkplays567 11 ай бұрын

thanks bruh it helps a lot

@PhysicsNinja 11 ай бұрын

Happy to help

@umarruban5012 3 жыл бұрын

You look like abd villers 😂

@MEBMohdSaifuddinKhan 4 жыл бұрын

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 4 жыл бұрын

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 4 жыл бұрын

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 5 ай бұрын

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

@bluefire6470 5 ай бұрын

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.