This has made front page before na

No one dies, the rolling ball isn't heavier than the one on the lever, so it can't lift it up, also the opening on the rolling stone will save B , so no one dies. Hope am right

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Incomplete information.
If Mass A and B are of equal weight (completing the information), then Mass A will not tip the see-saw, hence C will not die. If Mass A = Mass B, then MAss A is invariably lesser than Mass C in weight. If C does not die, then B and A cannot die.

We are left with E and D. Obviously, E cannot die since Mass B cannot be flung over to E's position by a lesser Mass A.
Now will D die? Most likely.

Taking each square as a unit.
Take circumference of ball and you get pi*d = pi*2. For ball to travel down that slope with length sqrt(3^2 + 4^2) = 5.
It will make 5/(pi*2) = effectively covering a distance the length of it's own circumference in one rotation.
That's = 0.795 which is not up to one rotation but close. Half rotation brings the curved part (absent part) of the ball to D's head, just escaping him. Since you won't be getting half rotation, but you will be past it and just getting close to full rotation, you might just get that guys head before curved part faces up again.

D will therefore mostly likely die in this scenario.


In other scenarios (difference in ball weights, for example), there are other possibilities. Hence, the incompelte information!

"D" dies first

B will die

Its all about probability. If you want the probability, inbox me.

E would die..

Codec:
[img][/img]

none

If they are all alive right now, something has to happen to make one or all of them die. OP should tell us. All the replies are based on the assumption that the cut ball is pushed but there's nowhere in the post to indicate that the ball will be pushed.

they r all cartoons, so no one dies...

None of them will die
Ask that guy to roll the ball
And you will see that they will all suvive

THEY ALL DIE!!

Its important knowing the mass of the 2 balls. But, if we
assume that they are of the same size, then
the ball rolls and does not kill D due to the
hole, it lands on the ramp, but, since the hole is
cut into the rock and would therefore have
presumably less mass that the other rock, it can
not fully move the ramp, therefore,
No one dies.

THEN, since all of the other's survive, they get
together and lynch E.

E dies.

tempem:
Simply put, Nobody.

i say too...nobody....

BabaO2:

Won't he dunk ?

u mean duck?

I was the first to actually post this here on NL now someone copied it... Op be sincere and tell us where you got the pic from.
https://www.nairaland.com/2240417/give-me-answers-these-puzzles

goofyone:
Incomplete information.
If Mass A and B are of equal weight (completing the information), then Mass A will not tip the see-saw, hence C will not die. If Mass A = Mass B, then MAss A is invariably lesser than Mass C in weight. If C does not die, then B and A cannot die.

We are left with E and D. Obviously, E cannot die since Mass B cannot be flung over to E's position by a lesser Mass A.
Now will D die? Most likely.

Taking each square as a unit.
Take circumference of ball and you get pi*d = pi*2. For ball to travel down that slope with length sqrt(3^2 + 4^2) = 5.
It will make 5/(pi*2) = effectively covering a distance the length of it's own circumference in one rotation.
That's = 0.795 which is not up to one rotation but close. Half rotation brings the curved part (absent part) of the ball to D's head, just escaping him. Since you won't be getting half rotation, but you will be past it and just getting close to full rotation, you might just get that guys head before curved part faces up again.

D will therefore mostly likely die in this scenario.


In other scenarios (difference in ball weights, for example), there are other possibilities. Hence, the incompelte information!

simply put, pi(D)

goofyone:
Incomplete information.
If Mass A and B are of equal weight (completing the information), then Mass A will not tip the see-saw, hence C will not die. If Mass A = Mass B, then MAss A is invariably lesser than Mass C in weight. If C does not die, then B and A cannot die.

We are left with E and D. Obviously, E cannot die since Mass B cannot be flung over to E's position by a lesser Mass A.
Now will D die? Most likely.

Taking each square as a unit.
Take circumference of ball and you get pi*d = pi*2. For ball to travel down that slope with length sqrt(3^2 + 4^2) = 5.
It will make 5/(pi*2) = effectively covering a distance the length of it's own circumference in one rotation.
That's = 0.795 which is not up to one rotation but close. Half rotation brings the curved part (absent part) of the ball to D's head, just escaping him. Since you won't be getting half rotation, but you will be past it and just getting close to full rotation, you might just get that guys head before curved part faces up again.

D will therefore mostly likely die in this scenario.


In other scenarios (difference in ball weights, for example), there are other possibilities. Hence, the incompelte information!

simply put, pi(D)

If you notice the animated gif, you know that D, C and B are going to die. D dies directly after coming in contact with the object, C dies after being thrust by the spikes of the see-saw and B dies with the blue ball.

But consider the situation that the person at E stands for Ebola. Eventually A dies as well.

ahhhh...i have other things to do...than thinking who is going to die..stvpid puzzle...

Chelcy10:
Seriously D is already a dead man cos when E pushes the ball, it will grind his head.

D won't die. Look at the ball very well. The ball would have made a revolution before getting to D.

Codec:
[img][/img]

D is the only person I see that is at risk of dying.

The rolling ball will eventually crush him going by the metric indicators.

It will land on the seesaw and give the other ball a shunt but not enough to push it over the parapet. there also wont be enough power to deal a death crush to the guy under the spike. Ultimately the ball will rest side by side with the other ball on the othe end and case closed

C and D but if D is a born again Christian he will not die lol

E

Codec:
@young sage it cannot be see cos the rolling ball has a lesser weight to the resting ball

d c and b re dead men

It is graphical. 20cm in length, both of the ball block are 2 cm each in diameter, the pushed block seems to be thicker than the ball on the balance, and also the pushed ball or block is graphical too, every 90 degrees covers one centimetre. D is save, no matter what.
So C dies, because the pushed block is thicker and heavier than the block on the ledge, and if not heavier, it is definitely equal in size, the light black will simply roll towards the thick block because of the force.

It now depends on the force at which A pushes the thicker block. The higher the force from E, The smasher A. Na make A just beg E. C na sacrifice.

c will die if d stone is being rolld over his head

It all depends on the weight of the rolling stone and the speed. Only D could die, C and D could die, B, C, D could die, C, D and E could die.

As a matter of fact the only person that is completely safe is A.


https://www.youtube.com/watch?v=LKLzv8INA3E

B dies there is a ramp right above C on the see saw that would propel the rock E is pushing. C wouldn't die because the rock E is pushing is lighter than the rock near B

D is the only death


The ball descends with a total of 3 cu sq of acceleration, the small plane at the foot of the descend cancels out 2 cu sq of force but an additional 2.5 cu sq (additional .5 to compensate for hollowed out portion) is needed to kill C and also have enough force accumulated to lift the lever. It doesn't.

Its E dat would die

REASON

By the time E pushes the stone D wud squat in order not to get squashed, the spikes won't touch C because of the lightness of the rolling ball (this is where it gets interesting).
Angered by the actions of E (their fellow brother from sambisa forest), A,B and D ganged up to kill E and save C from the trap.


Next Question please.