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NCERT Exemplar · Q39

Q.Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured within an accuracy of 4% what will be the uncertainty in speed and position?

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Heisenberg's uncertainty principle links the precision of momentum and position measurements. For a 10 g table-tennis ball moving at 90 m/s with 4% speed uncertainty, the uncertainty in speed is 3.6 m/s and the minimum uncertainty in position is 1.46×10−341.46 \times 10^{-34} m — utterly negligible for a macroscopic object.

Why Heisenberg's principle matters here

When we measure a particle's momentum (or speed) with some precision, quantum mechanics places a fundamental limit on how precisely we can simultaneously know its position. The uncertainty principle states:

Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

where Δx\Delta x is the uncertainty in position, Δp\Delta p is the uncertainty in momentum, and h=6.626×10−34h = 6.626 \times 10^{-34} J·s is Planck's constant. For macroscopic objects like a table-tennis ball, this quantum limit is so tiny that it has no practical consequence — but the calculation reveals exactly why classical physics works so well at our scale.

Step-by-step calculation

  1. Find the uncertainty in speed The speed is measured within 4% accuracy, so:

Δv=4100×90=3.6 m/s\Delta v = \frac{4}{100} \times 90 = 3.6 \text{ m/s}

  1. Convert mass to SI units The mass is given as 10 g:

m=10 g=10×10−3 kg=0.01 kgm = 10 \text{ g} = 10 \times 10^{-3} \text{ kg} = 0.01 \text{ kg}

  1. Calculate the uncertainty in momentum Momentum p=mvp = mv, so the uncertainty in momentum is:

Δp=m⋅Δv=0.01×3.6=0.036 kg⋅m/s\Delta p = m \cdot \Delta v = 0.01 \times 3.6 = 0.036 \text{ kg·m/s}

  1. Apply Heisenberg's uncertainty principle The minimum uncertainty in position is:

Δx≥h4πΔp\Delta x \geq \frac{h}{4\pi \Delta p}

Substituting the values:

Δx≥6.626×10−344×3.14159×0.036\Delta x \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14159 \times 0.036}

Δx≥6.626×10−340.4524≈1.46×10−34 m\Delta x \geq \frac{6.626 \times 10^{-34}}{0.4524} \approx 1.46 \times 10^{-34} \text{ m} …

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