Collision of point mass and sphere in particular fashion

[timetraveller123]

1. Homework Statement
a point mass is projected at an angle of 60^° from the horizontal. it collides with the sphere at a maximum height of trajectory with a sphere of radius of 72.5cm such that it is projected off the sphere once again at an angle of 60^° from the horizontal.The sphere, due to the collision, is bounced off the floor. The point mass, in its descent, collides with the sphere again, through the same horizontal of the previous point-mass-sphere collision, and is projected at an angle of 30^° below the horizontal . if the density of the sphere is 50kg/m³ and uniform mass distribution then the mass of point mass,m and the initial velocity of the point mass is u. find floor(m + u) . Assume all the collisions are elastic, and the sphere is rigid.

Homework Equations

mv_i = mv_final
x_f = x_o + v_ot + 0.5 a t²

The Attempt at a Solution

couldn't try because i don't know why does the sphere bounces

someone please tell me why does the sphere bounce then i could probably start as in i know it is because the mass hit it but what exactly is the reason?[/B]

 

[.Scott]

Imagine that the sphere is actually a tiny height (0mm) above the ground - and at the time of the collision, motionless. Then the collision with the point will cause the sphere to move with a downward component, strike the surface, and bounce elastically.

 

[BvU]

I'll do as you ask and then you can start as you promised.
Drat -- Scotty was faster! Anyway: point mass m acquires vertical momentum upwards, so sphere gets downward momentum. F (change in momentum per unit time) downward gets inverteed by the ground and results in upward bounce.

 

[timetraveller123]

so say mass goes up with momentum x then sphere goes down with momentum x then rebound with momentum x also ? is it correct

 

[BvU]

Assume it is.

 

[timetraveller123]

wait based on conservation of momentum at point of first collision i am getting mass of point mass is same as mass of ball is that correct?

so let ucos60 be initial velocity of point mass before collision
then P be momentum of ball after collision
and h be velocity of ball after collision
m be mass of point mass and M be mass of ball
so
m ucos60 = mhcos60 + P_x
mhsin60 = P_y
time of flight for point mass and ball must be same after collision
hence
for point mass: 2hsin60/g = t
for ball : V_y = mhsin60/M
so for the two times to be same mustn't m/M be 1

am i right if wrong where did i go wrong

 

[BvU]

wait based on conservation of momentum at point of first collision i am getting mass of point mass is same as mass of ball is that correct?

Don't know until you post your work
[edit] ah, the work is following after this question mark. Will look, but I think we need an expert opinion (@haruspex for example...)

My hunch is that the point mass should be lighter than the big ball, but... who knows what evil lurks in the hearts of exercise composers (*)(*) The Shadow knows

 

[timetraveller123]

that was my work until now just what want to know if i am on right track

 

[BvU]

Is there a basis for your assumption that point mass hits the ball at the very moment the ball hits the ground again ? (In other words: did you render the complete problem statement ?)

 

[timetraveller123]

ooh so that isn't the case?because if it didn't the ball would keep bouncing until it hits the mass

 

[timetraveller123]

ok then this is how i want to approach the problem tell me if anything wrong with it

upon first collision momentum initial in x = momentum final in x
momentum initial in y = momentum final in y (momentum initial in y = 0 )
kinetic energy of the mass initially = kinetic energy of the mass finally + kinetic energy of the ball
do i need anymore equations

 

[timetraveller123]

Is there a basis for your assumption that point mass hits the ball at the very moment the ball hits the ground again ? (In other words: did you render the complete problem statement ?)

i did that because the sphere only bounced one time according to the picture

 

[BvU]

ooh so that isn't the case?because if it didn't the ball would keep bouncing until it hits the mass

I don't know if that is the case or not -- still trying to formulate a complete problem statement.

i did that because the sphere only bounced one time according to the picture

Is THAT picture in post #1 an original part of the problem statement ? Or did you sketch it yourself ?

 

[timetraveller123]

it is from the question and by the way the question i got it from brilliant.org https://brilliant.org/problems/projec-ception/

 

[BvU]

Goodness ! They target a brilliant audience, but their illustrating skills are rather limited -- and adding a mass and a velocity is close to complete heresy to ordinary physicists like me .

But now I understand your assumption and I agree.

 

[timetraveller123]

so it is correct ? and by the way what does the last part about the mass bouncing off at 30 degrees got to do with the question
on a side note i regard you way better than me at physics

 

[BvU]

so it is correct ?

ok then this is how i want to approach the problem tell me if anything wrong with it

upon first collision momentum initial in x = momentum final in x $\quad\quad\quad$ agree

momentum initial in y = momentum final in y (momentum initial in y = 0 )$\quad\quad\quad$ agree that initially it is 0.
But the floor changes the sign of the y momentum of the ball so then the total y momentum isn't 0 any more. Then gravity goes to work and changes the total vertical momentum. The second time the point mass hits the ball, the y-momentum of the point mass has changed sign. The y-momentum of the ball is ? (in the process of hitting the floor it seems to be rather undetermined to me).

kinetic energy of the mass initially = kinetic energy of the mass finally + kinetic energy of the ball$\quad\quad\quad$ agree -- if you include possible rotational kinetic energy:

do i need anymore equations$\quad\quad\quad$ unclear to me. I don't know if the ball starts spinning after the first collision.

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and by the way what does the last part about the mass bouncing off at 30 degrees got to do with the question

unclear to me too. probably Nanayaranaraknas has a purpose for it, but it eludes me ! Must be some smart insight that reveals the path to the solution !

on a side note i regard you way better than me at physics

make no mistake there: a PhD (which I do have) or a professorship (which I don't) are no guarantee whatsoever against stupidities of all kinds, including the worst. In this exercise, to me you are the leading problemsolver -- I just try to help by asking some questions...

 

[timetraveller123]

ok this is all i got after this i am stuck i need help! this problem is driving me crazy

mu² = mh² + MV² ---1
mu = mh + 2MV_x ---2
hsin60 = V_y ----3
V² = V_x² + V_y²
hence since M = m
u² = h² + V²
u² = h² + (u² + 4h² - 2uh)/4M

EDIT: i need one more equation

wait you need to include rotational motion that is going to make the problem so much more harder !

and when i mean total momentum in y to be zero i mean right after the collision before hitting the ground. after hitting the ground the velocity in y will just become negative as before and x-velocity remains unchanged

 

[haruspex]

I am inclined to dismiss this question.

The m+u in the target makes no sense unless it is specified what units m and u are to be expressed in. Asking for m+u is anyway bizarre. If it is possible to find that then it is possible to find m and u separately. (The same need not be true if asked for mu.). So why not ask for m and u?

In terms of m and u, we know everything about the initial impact: the velocity, location and angle of incidence. (You can use u and the launch angle to find the height of the impact.). Thereafter, we can know the position of each body in terms of m, u and t. That they should come together in that specific relationship at the same moment that the sphere hits the ground again gives three equations, so in principle we can find m, u and t.

It is not possible to use the final piece of information, the angle of m's path after second impact, without making some assumption about whether the ball bounces just before or just after the second impact with m. Either way, it now makes the problem overspecified. It almost certainly will lead to a contradiction.

 

[timetraveller123]

sir @haruspex me and @BvU believe mass of point mass and ball are the same if you may please take a look at my calculations in post #6 and tell us if we are correct thanks

and i so far only managed to get 3 equations namely momentum in x and momentum in y after first collision and finally conservation energy all these three collectively could only eliminate one variable please tell me what is the other equation i need to use
thanks!

 

[haruspex]

sir @haruspex me and @BvU believe mass of point mass and ball are the same if you may please take a look at my calculations in post #6 and tell us if we are correct thanks

and i so far only managed to get 3 equations namely momentum in x and momentum in y after first collision and finally conservation energy all these three collectively could only eliminate one variable please tell me what is the other equation i need to use
thanks!

Yes,you make a useful argument that since the flight times are the same they must have the same vertical velocity after the first collision (and the ball's immediate bounce). But that means the top of the ball is always higher than the mass, so the trajectories described in the problem are not physically possible.

 

[timetraveller123]

wait so are you saying i should just do the math without worrying about logic or are you saying this problem is impossible both ways and i should ditch it because i have been stuck on this problem for a quite a while now thanks for the insight didn't realize that

 

[BvU]

sir @haruspex me and @BvU believe mass of point mass and ball are the same

I'd like to go on record as NOT believing that at all -- until I'm convinced otherwise.

But I do see a road ahead: you may want to join 'Brilliant' to get a peek at the intended 'solution' and then we can hopefully say a bit more.

Like haru, I want to dismiss this exercise as a physical problem, but my natural curiosity makes me wonder what mathematical ingenuity is hidden inside this contraption.

 

[haruspex]

wait so are you saying i should just do the math without worrying about logic or are you saying this problem is impossible both ways and i should ditch it because i have been stuck on this problem for a quite a while now thanks for the insight didn't realize that

You could overlook the physical impossibility and pretend that the mass can magically pass through the sphere until the second impact.

 

[haruspex]

until I'm convinced otherwise.

Since the flight times are the same, the vertical velocities after the first impact must have the same magnitude.
Since the mass m strikes horizontally, the vertical momenta must have the same magnitude.
Ergo, the masses are the same.

 

[BvU]

I'm convinced. This makes the exercise physically despicable and the accompanying picture (purposely?) misleading. It's a mathematical figment of imagination -- and yet I'm still curious to know the trick hidden deep inside there. Almost enough to join 'Brilliant' to find out ...

 

[timetraveller123]

@BvU people have solved this problem in fact only 5 have but none have posted solution

 

[timetraveller123]

@BvU are you still trying that question i am getting nowhere i decided to give up on that sorry
meanwhile i have another question
why must a ball drop when drop it from a certain height as it is not going to get to lower energy level then why the motion of falling
also i am stuck on another similar but slightly easier problem should i post it here or separate thread?

 

[BvU]

post it here or separate thread

Separate thread

 

[haruspex]

As already deduced, the two masses are the same and they have equal and opposite vertical velocity components immediately after first impact.

When two equal masses collide elastically, the velocity components normal to the plane of contact simply swap over. The point mass therefore comes to rest in that direction, moving off tangentially to the sphere. Since that is 60 degrees above the horizontal, the line of centres must be 30 degrees below it.

It follows from the above that the sphere acquires a greater horizontal velocity than the point mass retains. It is therefore impossible for the point mass to overtake the sphere in the manner of the diagram.

Perhaps treating it as two collisions, point mass with sphere then sphere with ground, is wrong.

 

[timetraveller123]

As already deduced, the two masses are the same and they have equal and opposite vertical velocity components immediately after first impact.

When two equal masses collide elastically, the velocity components normal to the plane of contact simply swap over. The point mass therefore comes to rest in that direction, moving off tangentially to the sphere. Since that is 60 degrees above the horizontal, the line of centres must be 30 degrees below it.

It follows from the above that the sphere acquires a greater horizontal velocity than the point mass retains. It is therefore impossible for the point mass to overtake the sphere in the manner of the diagram.

Perhaps treating it as two collisions, point mass with sphere then sphere with ground, is wrong.

if that is not the case then why does the sphere bounce ?

 

[haruspex]

if that is not the case then why does the sphere bounce ?

It will bounce, for sure, but there really is no justification for treating it as two impacts in series, other than to make it simpler.
In reality, it would depend on the relative stiffness of the bodies, including the ground. If the point mass and the sphere are highly rigid (think, spring with very high k), but the ground is a softer spring, then it might approach such a serialisation. More likely, the ground would be the most rigid of the three.

A more accurate view would consider the two impacts occurring in parallel, but that gets quite hard to analyse. In principle, one of them would transit from compression phase to expansion phase at different times. A reasonable model would be to take the point mass and the ground as completely rigid and the sphere as the source of all elasticity.

One thing that makes it defy intuition somewhat is the assumption of no friction between the surfaces. We need to do that or we get tangled up with rotational energy too.