A body of mass \( (5 \pm 0.5) \, \text{kg} \) is moving with a velocity of \( (20 \pm 0.4) \, \text{m/s} \). Its kinetic energy will be:

Question

Consider an equilateral prism (refractive index \( \sqrt{2} \)). A ray of light is incident on its one surface at a certain angle \( i \). If the emergent ray is found to graze along the other surface, then the angle of refraction at the incident surface is close to

Answer 1

The problem involves calculating the kinetic energy of a body and its associated uncertainty. The kinetic energy \( KE \) of a body with mass \( m \) and velocity \( v \) is given by the formula:

KE = \frac{1}{2} m v^2

Given:

Mass \( m = (5 \pm 0.5) \text{kg} \)
Velocity \( v = (20 \pm 0.4) \text{m/s} \)

First, calculate the kinetic energy using the central values:

KE = \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 \, \text{J}

Next, calculate the uncertainty in kinetic energy. The formula for the propagation of uncertainty for a product or quotient is:

\left(\frac{\Delta KE}{KE}\right)^2 = \left(\frac{\Delta m}{m}\right)^2 + \left(2 \times \frac{\Delta v}{v}\right)^2

Let's plug in the values:

\frac{\Delta m}{m} = \frac{0.5}{5} = 0.1
\frac{\Delta v}{v} = \frac{0.4}{20} = 0.02

Substituting these into the formula gives:

\left(\frac{\Delta KE}{1000}\right)^2 = (0.1)^2 + (2 \times 0.02)^2 = 0.01 + 0.0016 = 0.0116

Taking the square root of both sides to solve for \frac{\Delta KE}{1000}:

\frac{\Delta KE}{1000} = \sqrt{0.0116} \approx 0.1077

So, \Delta KE = 0.1077 \times 1000 \approx 107.7 \, \text{J}

Rounding this to match the precision of other given uncertainties, we can state:

The kinetic energy is (1000 \pm 140) \, \text{J}.

Thus, the correct option is: (1000 ± 140) J.

Answer 2