Class 11 Physics | Work, Energy, and Power MCQs with Answers – Part 5

Explanation: \( \textbf{Initial mechanical energy:} \) \[ E=K_1+U_1 \] \[ E=18+42=60\,\text{J} \] Since no non-conservative work is done, mechanical energy remains constant: \[ E=K_2+U_2 \] At the second point, \[ U_2=36\,\text{J} \] So, \[ K_2=60-36=24\,\text{J} \] Use kinetic energy: \[ K_2=\frac{1}{2}mv^2 \] \[ 24=\frac{1}{2}(3)v^2 \] \[ 24=1.5v^2 \] \[ v^2=16 \] \[ v=4\,\text{m s}^{-1} \] The decrease in potential energy appears as an increase in kinetic energy because no non-conservative work is done. \( \textbf{Final answer:} \) The speed at the second point is \(4\,\text{m s}^{-1}\).

Explanation: Row P is correct because without air resistance, gravitational potential energy and kinetic energy can transform while mechanical energy remains constant. Row Q is correct because friction is non-conservative and its work must be included in energy accounting. Row R is correct because in a short isolated collision, total momentum is conserved even when kinetic energy is not. Row S is incorrect because when power varies with time, the total energy transferred is the area under the \(P-t\) graph, not just the final power value. The method must match the condition of the process being studied.

Explanation: For a one-dimensional elastic collision, both momentum and kinetic energy are conserved. A useful result for a stationary target is that the incident body's final velocity is \(v_1=\frac{m_1-m_2}{m_1+m_2}u\). The sign of this result is important. If \(m_1\gt m_2\), the first body continues forward after collision. If \(m_1\lt m_2\), the first body rebounds because the expression becomes negative. The formula is not just a speed formula; it gives velocity along the chosen line.

Explanation: \( \textbf{Given:} \) \(m_1=1\,\text{kg}\), \(m_2=3\,\text{kg}\), \(u_1=6\,\text{m s}^{-1}\), and \(u_2=0\). For an elastic collision with the second body initially at rest, \[ v_1=\frac{m_1-m_2}{m_1+m_2}u_1 \] Substitute: \[ v_1=\frac{1-3}{1+3}(6) \] \[ v_1=\frac{-2}{4}(6) \] \[ v_1=-3\,\text{m s}^{-1} \] The negative sign means the lighter incoming ball rebounds after hitting the heavier stationary ball. \( \textbf{Final answer:} \) The velocity of the \(1\,\text{kg}\) ball is \(-3\,\text{m s}^{-1}\).

Explanation: In a head-on elastic collision with the second body initially at rest, the final velocities come from momentum conservation and kinetic energy conservation together. The initially stationary body's final velocity is \(v_2=\frac{2m_1}{m_1+m_2}u\). This expression shows how the transferred speed depends on the incident mass compared with the target mass. If the two masses are equal, \(v_2=u\), which matches the familiar velocity-exchange result. The factor \(2m_1\) appears because the elastic condition includes rebound or separation-speed information, not momentum conservation alone.

Explanation: \( \textbf{Given:} \) \(m_1=4\,\text{kg}\), \(u_1=5\,\text{m s}^{-1}\), \(m_2=1\,\text{kg}\), and \(u_2=0\). For an elastic collision with the second body initially at rest, \[ v_2=\frac{2m_1}{m_1+m_2}u_1 \] Substitute: \[ v_2=\frac{2(4)}{4+1}(5) \] \[ v_2=\frac{8}{5}(5) \] \[ v_2=8\,\text{m s}^{-1} \] The lighter target can move faster than the original speed of the heavier cart because momentum and kinetic energy are both shared after collision. \( \textbf{Final answer:} \) The \(1\,\text{kg}\) cart moves at \(8\,\text{m s}^{-1}\).

Explanation: In a collision with a fixed smooth wall, the impulsive contact force acts along the normal to the wall. Therefore, it affects only the normal component of velocity. The tangential component remains unchanged because the wall is smooth and gives no tangential impulse. The coefficient of restitution applies to the normal component, so the rebound normal speed is \(eu_n\) in the opposite normal direction. The full velocity direction changes according to this component-wise change.

Explanation: \( \textbf{Normal component before collision:} \) \[ u_n=10\,\text{m s}^{-1} \] \( \textbf{Tangential component before collision:} \) \[ u_t=24\,\text{m s}^{-1} \] For a smooth fixed wall, the tangential component remains unchanged: \[ v_t=24\,\text{m s}^{-1} \] The normal component reverses direction and its magnitude becomes \[ v_n=eu_n \] \[ v_n=(0.6)(10)=6\,\text{m s}^{-1} \] The speed after collision is found from perpendicular components: \[ v=\sqrt{v_n^2+v_t^2} \] \[ v=\sqrt{6^2+24^2} \] \[ v=\sqrt{36+576} \] \[ v=\sqrt{612}\approx24.7\,\text{m s}^{-1} \] Only the normal component is reduced by the coefficient of restitution; the tangential component remains unchanged for a smooth wall. \( \textbf{Final answer:} \) The speed just after collision is approximately \(24.7\,\text{m s}^{-1}\).

Explanation: After each collision with a fixed floor, the speed just after rebound is multiplied by \(e\). Since height reached is proportional to the square of upward speed, the rebound height is multiplied by \(e^2\) after each bounce. After the first rebound, the height is \(e^2h\). After the second rebound, it is \(e^4h\). After the third rebound, it is \(e^6h\). The exponent doubles the rebound count because energy and height depend on speed squared.

Explanation: \( \textbf{Initial gravitational potential energy before fall:} \) \[ mgh=(0.10)(10)(4)=4.0\,\text{J} \] Just before impact, this becomes kinetic energy if air resistance is neglected: \[ K_{\text{before}}=4.0\,\text{J} \] For collision with a fixed floor, rebound speed is multiplied by \(e\). Kinetic energy depends on speed squared, so after rebound: \[ K_{\text{after}}=e^2K_{\text{before}} \] Substitute: \[ K_{\text{after}}=(0.5)^2(4.0) \] \[ K_{\text{after}}=0.25(4.0)=1.0\,\text{J} \] The kinetic energy is reduced by the factor \(e^2\), not by \(e\). \( \textbf{Final answer:} \) The kinetic energy just after rebound is \(1.0\,\text{J}\).

Explanation: Energy transfer from a \(P-t\) graph is found from the signed area under the graph. The positive area adds energy to the body: \[ E_+=+300\,\text{J} \] The negative area removes energy: \[ E_-=-80\,\text{J} \] The net energy transfer is \[ E_{\text{net}}=E_++E_- \] \[ E_{\text{net}}=300-80 \] \[ E_{\text{net}}=220\,\text{J} \] The negative power interval must reduce the total rather than be added as an ordinary area magnitude. \( \textbf{Final answer:} \) The net energy transferred is \(+220\,\text{J}\).

Explanation: Power is the slope of a \(W-t\) graph. For the first interval: \[ P_1=\frac{\Delta W_1}{\Delta t_1} \] \[ P_1=\frac{100-0}{5-0} \] \[ P_1=\frac{100}{5}=20\,\text{W} \] For the second interval: \[ P_2=\frac{\Delta W_2}{\Delta t_2} \] \[ P_2=\frac{180-100}{9-5} \] \[ P_2=\frac{80}{4}=20\,\text{W} \] Both segments have the same slope, so the power is the same in both intervals. \( \textbf{Final answer:} \) The powers are \(20\,\text{W}\) and \(20\,\text{W}\).

Explanation: From the work-energy theorem in one dimension, \(\frac{dK}{dx}=F_{\text{net}}\). A horizontal \(K-x\) graph has zero slope. Therefore, \(\frac{dK}{dx}=0\), so the net force along the \(x\)-direction is zero. This does not mean that no forces are present; it means their net work-producing component along the motion is zero. The kinetic energy remains constant over that interval.

Explanation: In one-dimensional motion, the work-energy theorem gives \[ dK=F_{\text{net}}\,dx \] So, \[ F_{\text{net}}=\frac{dK}{dx} \] The slope of the \(K-x\) graph is given as \[ \frac{dK}{dx}=-5\,\text{J m}^{-1} \] Since \[ 1\,\text{J m}^{-1}=1\,\text{N} \] the net force is \[ F_{\text{net}}=-5\,\text{N} \] The negative sign means the net force is opposite to the positive \(x\)-direction. \( \textbf{Final answer:} \) The net force is \(-5\,\text{N}\).

Explanation: At a local maximum of \(U(x)\), the slope of the potential energy curve is zero. Since \(F=-\frac{dU}{dx}\), the force is zero exactly at that point, so equilibrium is possible. However, a small displacement to either side lowers the potential energy and the force tends to move the particle away from the maximum. This makes the equilibrium unstable. The value of \(U\) need not be zero for equilibrium; the shape of the curve near the point matters.

Explanation: A flat potential energy curve has zero slope, so \(F=-\frac{dU}{dx}=0\) throughout the flat region. Since the potential energy does not change with position inside the interval, a small displacement within it does not produce a restoring or repelling force. This corresponds to neutral equilibrium inside the flat portion. If the particle has kinetic energy, it can move freely within the flat region until it reaches a boundary set by its total energy. Neutral equilibrium is about the absence of a preferred nearby position.

Explanation: \( \textbf{Given potential energy:} \) \[ U=2x^3-6x \] For a one-dimensional conservative force, \[ F=-\frac{dU}{dx} \] Differentiate: \[ \frac{dU}{dx}=6x^2-6 \] Therefore, \[ F=-(6x^2-6) \] At \(x=1\,\text{m}\): \[ F=-(6(1)^2-6) \] \[ F=-(6-6) \] \[ F=0\,\text{N} \] The force is zero because the potential energy curve has zero slope at that point. \( \textbf{Final answer:} \) The force is \(0\,\text{N}\).

Explanation: For motion along a straight line, useful power is \(P=Fv\). If \(P\) is constant, the driving force is \(F=\frac{P}{v}\). As the speed \(v\) increases, the available driving force decreases. Since acceleration is \(a=\frac{F}{m}\), the acceleration also decreases. Constant power does not mean constant force, so the motion is not uniformly accelerated in this situation.

Explanation: \( \textbf{Useful power:} \) \[ P=10\,\text{kW}=10000\,\text{W} \] \( \textbf{Speed:} \) \[ v=10\,\text{m s}^{-1} \] For force along motion, \[ P=Fv \] So, \[ F=\frac{P}{v} \] \[ F=\frac{10000}{10}=1000\,\text{N} \] Using \(F=ma\): \[ a=\frac{F}{m} \] \[ a=\frac{1000}{500}=2\,\text{m s}^{-2} \] The acceleration is the instantaneous value at \(10\,\text{m s}^{-1}\); at a higher speed with the same power, it would be smaller. \( \textbf{Final answer:} \) The acceleration is \(2\,\text{m s}^{-2}\).

Explanation: \( \textbf{Efficiency:} \) \[ \eta=40\%=0.40 \] \( \textbf{Useful output energy:} \) \[ E_{\text{out}}=2000\,\text{J} \] Efficiency is \[ \eta=\frac{E_{\text{out}}}{E_{\text{in}}} \] Rearrange: \[ E_{\text{in}}=\frac{E_{\text{out}}}{\eta} \] Substitute: \[ E_{\text{in}}=\frac{2000}{0.40} \] \[ E_{\text{in}}=5000\,\text{J} \] The input energy is greater than the useful output because \(60\%\) of the supplied energy is not delivered as useful work. \( \textbf{Final answer:} \) The input energy required is \(5000\,\text{J}\).