A 70 g ball moves 45m/s. If its speed is measured to an accuracy of 0.10% what is the minimum uncertainty in its position?

To find the minimum uncertainty in the position, we need to determine the range of positions that are consistent with the given accuracy in speed measurement.

The accuracy of speed measurement is given as 0.10%. This means that the measured speed can differ from the actual speed by a maximum of 0.10%.

Let's calculate the maximum and minimum possible speeds within this range:

Maximum speed = Actual speed + (0.10% of actual speed)
= 45 m/s + 0.10% * 45 m/s
= 45 m/s + 0.10 * 0.01 * 45 m/s
= 45 m/s + 0.045 m/s
= 45.045 m/s

Minimum speed = Actual speed - (0.10% of actual speed)
= 45 m/s - 0.10% * 45 m/s
= 45 m/s - 0.10 * 0.01 * 45 m/s
= 45 m/s - 0.045 m/s
= 44.955 m/s

To calculate the minimum uncertainty in position, assuming linear motion, we use the formula:

Δx = v * Δt

where Δx is the uncertainty in position, v is the speed, and Δt is the uncertainty in time.

Given that the ball has a velocity of 45 m/s, we need to find the uncertainty in time that corresponds to the maximum and minimum speeds.

Δt_max = (v_max - v) / v
= (45.045 m/s - 45 m/s) / 45 m/s
≈ 0.00099 s

Δt_min = (v - v_min) / v
= (45 m/s - 44.955 m/s) / 45 m/s
≈ 0.001 s

Now, we can calculate the minimum uncertainty in position for the maximum and minimum speeds:

Δx_max = v * Δt_max
= 45 m/s * 0.00099 s
≈ 0.04455 m

Δx_min = v * Δt_min
= 45 m/s * 0.001 s
= 0.045 m

Therefore, the minimum uncertainty in the position is approximately 0.04455 m to 0.045 m.

Heisenberg uncertainty principle

Δx•Δp ≥h/2•2•π,
Note! the right part of uncertainty principle in different sources may be: h, h/2 π, h/2, etc. But all these magnitudes are ~10^-34 .
m = 0.07 kg.
v = 45 m/s.
Δv = 0.1 •v/100 = 0.045 m/s.
Δ p =Δ(m•v) = m• Δv = 0.07•0.045=3.15•10^-3.
Δx ≥h/4•π•Δp = 6.63•10^-34/4• π•3.15•10^-3 =1.67•10^-32
Δx ≥1.67•10^-32 m.