Vardaan Institute - Detailed Solutions: Master Sheet (Work, Energy & Power)

Section A: Work-Energy Theorem Solutions

A bullet of mass $20\text{ g}$ strikes a target at $100\text{ m/s}$ and penetrates $10\text{ cm}$. Find average retarding force.

Sol: $W = \Delta K \implies -F \cdot s = 0 - \frac{1}{2} m u^2 \implies F(0.1) = \frac{1}{2}(0.02)(100^2) = 100\text{ J} \implies F = 1000\text{ N}$.

A spring of spring constant $k = 500\text{ N/m}$ is compressed by $x = 0.2\text{ m}$. Calculate work done and potential energy.

Sol: $U = \frac{1}{2} k x^2 = \frac{1}{2} (500) (0.2)^2 = 250 \times 0.04 = 10\text{ J}$. Work done against spring $= 10\text{ J}$.

Section B: Conservation of Energy Solutions

A body of mass $m$ is tied to a string of length $L$ and whirled in a vertical circle. Find minimum velocity at bottom.

Sol: At top $v_{top} = \sqrt{gL}$. By conservation of mechanical energy: $\frac{1}{2} m v_{bottom}^2 = \frac{1}{2} m v_{top}^2 + m g (2L) \implies v_{bottom} = \sqrt{5gL}$.

If momentum of a body increases by $50\%$, find percentage increase in its kinetic energy.

Sol: $p' = 1.5 p$. Since $K = \frac{p^2}{2m}$, $K' = \frac{(1.5 p)^2}{2m} = 2.25 K$. Percentage increase $= (2.25 - 1) \times 100\% = 125\%$.

Section C: Collisions & Power Solutions

An engine pumps $1000\text{ kg}$ of water to a height of $10\text{ m}$ in $20\text{ s}$. Calculate power delivered.

Sol: $P = \frac{m g h}{t} = \frac{1000 \times 9.8 \times 10}{20} = 4900\text{ W} = 4.9\text{ kW}$.

Derive expression for final velocities in 1D head-on elastic collision of two unequal masses $m_1$ and $m_2$.

Sol: $v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2 m_2}{m_1 + m_2}\right)u_2$ and $v_2 = \left(\frac{2 m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2$.

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