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Carl throws 100 g of Bohrium-262 (Half life: 0.10 seconds, undergoes alpha decay) to Sukhdeep with a initial horizontal velocity of 20.0 m/s at an initial height of 2.00 m above the ground. What is the mass and final energy of the ball of Bohrium if Sukhdeep is only 5.00 m away from Carl?

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2 Answers

Conceptually, you can figure out how long the ball will take to reach Carl, by recognising that it is going to move at a (for practical purposes) constant horizontal velocity. This will allow you to figure out how much it has decayed in that time. Lay out your equations:

$${x \over t} = v$$

$$m_{final} = m_{initial} \left({1 \over2}\right)^{t \over t_{halflife}}$$

where $x$ = distance, $t$ = time, and $v$ = initial velocity

Just scanning those equations, already you're given all but two of them directly in the question:

Two equations and two unknown variables means you can solve for all variables --

Calculate the final mass and, use that to get its final energy.

(note that the final energy will have both horizontal and vertical components, so you'll need to figure out how much it's accelerated towards the ground due to gravity.)

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I'm not allowed to solve using Kinematics equations, I have to some how solve this with E = (mv^2)/2 and E = mgh –  SyedKamran Nov 23 '12 at 4:22
@SyedKamran well, you can use those equations to calculate the final kinetic energy (kinetic energy horizontally, plus change in potential energy vertically). You must need some other equations, right? Otherwise how could you calculate the decay due to half life. –  McGarnagle Nov 23 '12 at 4:27
@dbaseman When the atoms decay, they don't vanish into thin air. Your expression for mass vanishes for t tending to infinity, which is wrong, since we still have a considerable mass of the ball as "radioactive waste". See my answer below. –  Benji Remez Nov 23 '12 at 6:21
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Since the halflife is short enough and the ball is massive enough, we can assume that many decays take place, and the produced alpha particles are emitted in a spherically symmetric pattern away from the ball. This means that the recoil from the decays averages to zero, and the ball moves at constant velocity (i.e. momentum per unit of mass is conserved), so $$t = \frac{x}{v_x}$$ Note that this is not even an approximation.

Now, the number of Bohrium atoms as a function of time is $N=N_0 * 2^{-t/t_{half}}$. Each successive decay makes a single Bohrium atom lose the mass of about 4 proton masses, so it retains only 258/262 of its original mass. Thus, the mass at any time $t$ would be: $$m(t) = m_0 * (2^{-t/t_{half}}+\frac{258}{262}(1-2^{-t/t_{half}}))$$

Plugging in the time of flight above yields the final mass. Since mechanical energy per unit mass is also conserved (again, because of the symmetry of the emission), the final energy is the sum of the initial kinetic and potential energy when the ball is thrown times that fraction of remaining mass when the ball is caught.

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