Portal Physics

Classical Mechanics Level 4

The image above shows two circular planar portals: a blue portal, with its axis oriented in the vertical direction, and an orange portal, with its axis oriented in the horizontal direction. The blue portal is situated at y = 0 y=0 , and its x x -coordinate is irrelevant. The orange portal is situated at x = 0 x=0 , and its height ( y y -position) varies as described later.

The rules of portal physics are the following:

  • An object enters one portal and emerges from the other
  • Portals do not conserve vector momentum or total energy (kinetic + potential)
  • Portals do conserve the scalar speed of objects that enter / exit them
  • An object entering one portal at its center on a trajectory perpendicular to its axis, will emerge from the other portal in like manner
  • Portals do not impart their own velocity to objects which pass through them
  • Aside from their own strange properties, portals do not otherwise alter the physics of nearby objects and environments
  • It takes zero time to go through a pair of portals

At time t = 0 t = 0 , two things happen simultaneously:

  • A massive ball drops from its initial resting position at height h B h_B and falls under the influence of a uniform downward gravitational acceleration g g . It falls toward the center of the blue portal.

  • The orange portal (position defined by its center) begins to descend from its initial vertical position h P h_P at a constant speed v P v_P .

Consider the x x -coordinate of the ball at the instant at which it intersects the x x -axis after emerging from the orange portal. The value of h B h_B which maximizes this quantity can expressed as:

h B max = a b g ( h P v P ) 2 . {h_{B_\text{max}} = \frac{a}{b} g \left(\frac{h_P}{v_P}\right)^2}.

If a a and b b are coprime positive integers, determine the value of a + b a+b .


The answer is 11.

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Steven Chase by Steven Chase , 37, USA

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2 solutions

Rohit Gupta
Feb 2, 2017

Relevant wiki: Projectile motion - Medium

Let the time taken by the ball to reach the blue portal is t t and its speed there is v v . As it is a free fall where the acceleration remains constant equal to g g , we can calculate these values using the equations of motion. Applying these equations we will get,
t = 2 h B g , v = 2 g h B . t = \sqrt {\frac{2h_B}{g}} \quad, \quad v= \sqrt{2gh_B}.

In this time the orange portal will descend by a distance d = v P t = v P 2 h B g d = v_P t = v_P \sqrt {\frac{2h_B}{g}} .

Now, we know that for projectile thrown horizontally at speed u u , from a tower of height h h the horizontal distance traveled is x = u 2 h g . x= u \sqrt{\frac{2h}{g}}. Therefore, the ball will land at a distance

x = 2 g h B 2 ( h p − d ) g = 2 g h B 2 ( h p − v P 2 h B g ) g = 4 h P h B − 4 V P 2 g ( h B ) 3 2 . \begin{aligned} x &= \sqrt{2gh_B} \sqrt{\frac{2(h_p-d)}{g}} \\ &= \sqrt{2gh_B} \sqrt{\frac{2(h_p-v_P \sqrt {\frac{2h_B}{g}})}{g}} \\ &= \sqrt{4h_Ph_B - 4V_P \sqrt{\frac{2}{g}}(h_B)^{\frac{3}{2}}}. \end{aligned}

For x x to be maximum, the term in the square root needs to be maximum, then its derivative is zero, x ′ = 0 x' = 0 . Doing this will gives 4 h P − 6 v P 2 h B g = 0 ⇒ h B = 2 9 g ( h P v P ) 2 . 4h_P - 6v_P \sqrt{\frac{2h_B}{g}} = 0 \quad \Rightarrow \quad h_B = \frac{2}{9}g \left(\frac{h_P}{v_P}\right)^2.

Our answer is a = 2 , b = 9 ⇒ a + b = 11 a=2, b=9 \Rightarrow a+b=\boxed{11} .

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Assumption : If a particle moves with speed v v through blue portal, it will come out with speed v v through orange portal with respect to ground. If it comes out with speed v v w.r.t portal, that would be more realistic but answer would be totally different.

Let t 1 = ( 2 h B g ) t_1=\sqrt(\frac{2h_B}{g}) and v 1 = ( 2 g h B ) ) v_1=\sqrt(2gh_B))

So t 2 = ( 2 ( h p − v p t 1 ) g ) ) t_2=\sqrt(\frac{2(h_p-v_pt_1)}{g}))

We need to maximize R = v 1 t 2 R=v_1t_2

Differentiating and rearranging, we get h B m a x = 2 9 g ( h P v P ) 2 \large{h_{B_{max}} = \frac{2}{9} g \Big(\frac{h_P}{v_P}\Big)^2}

.

If we do take the other assumption, equation for t 2 t_2 would change to a quadratic equation

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for the solution. I have added the following note to remove the ambiguity.

"Portals do not impart their own velocity to objects which pass through them"

Steven Chase - 4 years, 4 months ago

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Oh I assumed that due to Galilean relativity the velocity of the object would be v b + v p v_b +v_p . I guess I should've read more clearly! Fun problem though.

Krishna Karthik - 1 year, 2 months ago

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