Question

Heisenberg's uncertainty principle can be expressed mathematically as Δx*Δp = h/4π, where Δx and Δp denote the uncertainty in position and momentum respectively and h is Planck's constant. What would be the uncertaintry in the position of a pitched baseball (mass = 0.2574 kg) that was traveling at a velocity of 84.50 m/s if the velocity has an uncertainty of 2.275%?

please explain step by step

Answer #1

According to Heisenberg uncertainty principle,

Δx*Δp = h/4π

where, Δp - the uncertainty in momentum;

Δx - the uncertainty in position

and Planck's Constant, h=6.626 X 10^{-34}
m^{2}kgs^{-1}

Now, The uncertainty in momentum can be written as

Δp=m⋅Δv , where

Δv - the uncertainty in velocity;

=2.275% of 84.50 m/s i.e. 1.922 m/s

m - the mass of the particle.

In the given case, we are dealing with 0.2574 kg base ball
having an uncertainty in velocity is 1.922 ms^{-1}

so, the uncertainty in momentum will be

Δp=0.2574 kg X 1.922 ms^{-1}

=0.4947 kgms^{-1}

Now, From the given equation we can calculate the uncertainty in position as

Δx = h/4π.(1/Δp)

=(6.626 X 10^{-34} m^{2}kg s^{-1}/4 x
3.14 ).(1/0.4947 kg^{-1}m^{-1}s^{1})

=(6.626 X 10^{-34} m^{2}kg
s^{-1}/12.56). (2.02
kg^{-1}m^{-1}s^{1})

=(0.527 X 10^{-34}m^{2}kg s^{-1}).(2.02
kg^{-1}m^{-1}s^{1})

=1.064.X 10^{-34} m

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do all five questions
Question 1
20 pts
Ignoring the effects of air resistance, if a ball falls freely
toward the ground, its total mechanical energy
Group of answer choices
increases
remains the same
not enough information
decreases
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Question 2
20 pts
A child jumps off a wall from an initial height of 16.4 m and lands
on a trampoline. Before the child springs back up into the air the
trampoline compresses 1.8 meters. The spring constant...

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