Newtonian Mechanics and Conservation Laws Questions and Answers

Answer: C

The correct option is \(R \lt v_{0}^{2}\sin 2\theta_{0}/g\) and \(\theta \gt \theta_{0}\).

Without drag the range is \(v_{0}^{2}\sin 2\theta_{0}/g\) and the path is symmetric, landing at the launch angle. A drag force opposite to the velocity takes energy out throughout the flight, so the projectile cannot go as far: \(R\) is less than the drag-free range.

Drag also keeps reducing the horizontal velocity, \(dv_{x}/dt = -(\gamma/m)v_{x}\), so the horizontal speed at impact is less than \(v_{0}\cos\theta_{0}\). The descent is therefore steeper than the ascent, and the impact angle exceeds the launch angle. The options with '=' apply a drag-free formula, and the options with \(\theta \lt \theta_{0}\) get the asymmetry the wrong way round.

Answer: C

The answer is the speed approaches a constant value when \(\lambda t \gg 1\).

The captured molecules start at rest, so Newton's law for variable mass reads

\[ \begin{aligned} \frac{d(mv)}{dt} &= mg \\ m\frac{dv}{dt} + \lambda m v &= mg \\ v(t) &= \frac{g}{\lambda}\left(1 - e^{-\lambda t}\right) \end{aligned} \]

For \(\lambda t \gg 1\) the exponential dies away and the drop approaches the terminal speed \(g/\lambda\). For \(\lambda t \ll 1\) the speed is about \(gt\), growing linearly, not constant. The speed never grows exponentially; the exponential term decays.

Answer: B

The correct statement is that angular momentum of the system is not conserved.

The potential \(V = -k\,y/r^2\) depends on direction as well as on \(r = \sqrt{x^2 + y^2 + z^2}\). A central force needs a potential that depends on \(r\) alone; this one does not, so the force has components across the radius and exerts a torque \(\vec{r} \times \vec{F} \neq 0\). Angular momentum therefore changes.

The force is not central, so that statement is false. \(V\) depends on \(y\), so \(F_y = -\partial V/\partial y \neq 0\) and linear momentum along \(y\) is not conserved. \(V\) does not depend on time, so the force is conservative and the total energy is conserved, which rules out the last statement.

Answer: C

The magnitude of the change in velocity is \(2v\sin\frac{\theta}{2}\).

The speed stays \(v\), but the velocity, which is tangent to the circle, turns through the same angle \(\theta\) as the radius. Placed tail to tail, the initial and final velocities form an isosceles triangle with two sides \(v\) and angle \(\theta\) between them, and \(\Delta\vec{v}\) is its third side:

\[ \begin{aligned} |\Delta\vec{v}|^{2} &= v^{2} + v^{2} - 2v^{2}\cos\theta \\ &= 4v^{2}\sin^{2}\tfrac{\theta}{2} \end{aligned} \]

So \(|\Delta\vec{v}| = 2v\sin\frac{\theta}{2}\). Zero is the change in speed, not velocity. \(v\cos\theta\) and \(v\cos\frac{\theta}{2}\) are projections, not the length of the difference vector; for \(\theta = 180^\circ\) the correct form gives \(2v\), a full reversal.

Answer: D

The correct plot is (D), a hyperbola branch that starts at a positive \(x\) and never reaches the \(v\) axis.

Energy conservation gives \(\frac{1}{2}mv^{2} = E - U = E + \frac{1}{2}kx^{2}\), so

\[ v^{2} = \frac{k}{m}\left(x^{2} - x_{0}^{2}\right), \]

where \(x_{0}^{2} = -\frac{2E}{k}\).

With \(E \lt 0\), \(x_{0}^{2} \gt 0\). Real velocities exist only for \(|x| \geq x_{0}\), where the particle is on the hill below its energy level; for \(|x| \lt x_{0}\) the potential energy \(U\) is higher than \(E\), which is forbidden. The curve \(v\) against \(x\) is a hyperbola whose branch on the right (the side the ball climbs on) crosses the \(x\) axis at \(x_{0}\), at the turning point where the ball stops.

(A) and (B) are closed arcs of the kind \(v^{2} = c - \frac{k}{m}x^{2}\), the trace of an oscillation in a well. (C) is a constant speed, which happens only with no force. Only (D) matches \(v^{2} \propto x^{2} - x_{0}^{2}\). The official key lists only D.

Answer: A

The ball rises to h/9.

Just before impact the ball's speed is \(u = \sqrt{2gh}\). In a one-dimensional elastic collision of mass \(m_{1} = m\) with a stationary \(m_{2} = 2m\), the first body's velocity after impact is

\[ \begin{aligned} v_{1} &= \frac{m_{1} - m_{2}}{m_{1} + m_{2}}\,u \\ &= \frac{m - 2m}{3m}\,u = -\frac{u}{3} \end{aligned} \]

The ball rebounds with a third of its speed. The height reached scales as the square of the speed:

\[ \begin{aligned} h' &= \frac{v_{1}^{2}}{2g} \\ &= \frac{1}{9}\cdot\frac{u^{2}}{2g} = \frac{h}{9} \end{aligned} \]

The option h/3 forgets to square the speed ratio, and 2h/3 and h/8 do not follow from the elastic collision formulas.

Answer: A

The terminal velocity is \( mg/k \).

The man accelerates until the resisting force balances his weight; then the net force is zero and the speed stays constant:

\[ kV_{t} = mg \quad \Rightarrow \quad V_{t} = \frac{mg}{k} \]

Solving \( m\,dV/dt = mg - kV \) gives \( V = \frac{mg}{k}(1 - e^{-kt/m}) \), which approaches the same value. (Buoyancy is ignored, as the question does.)

The multiples \( 2mg/k \), \( \frac{1}{2}mg/k \) and \( \frac{3}{2}mg/k \) do not satisfy the force balance for a resistance that is simply \( kV \).

Answer: D

The man does work \( \frac{1}{2}\left(M + \frac{M^{2}}{m}\right)v^{2} \).

Momentum is conserved horizontally, so the man's speed \( u \) is fixed by \( mu = Mv \). The work he does appears as the kinetic energy of both bodies:

\[ \begin{aligned} u &= \frac{Mv}{m} \\ W &= \frac{1}{2}Mv^{2} + \frac{1}{2}m\left(\frac{Mv}{m}\right)^{2} \\ &= \frac{1}{2}\left(M + \frac{M^{2}}{m}\right)v^{2} \end{aligned} \]

\( \frac{1}{2}Mv^{2} \) counts only the boat's energy. \( \frac{1}{2}mv^{2} \) gives the man the boat's speed, and \( \frac{1}{2}(M + m)v^{2} \) gives both bodies the speed \( v \), which momentum conservation forbids unless \( m = M \).

Answer: D

The range is proportional to \(v_{0}^{2}\) and sin 2\(\alpha\).

The time of flight is \(T = 2v_{0}\sin\alpha/g\), and the horizontal velocity is \(v_{0}\cos\alpha\), so:

\[ \begin{aligned} R &= v_{0}\cos\alpha \cdot \frac{2v_{0}\sin\alpha}{g} \\ &= \frac{v_{0}^{2}\sin 2\alpha}{g} \end{aligned} \]

The range therefore grows as the square of the launch speed and as \(\sin 2\alpha\), which is why it is greatest at 45°. Options with \(v_{0}\) to the first power, or with \(\sin\alpha\), each miss one of the two factors.

Answer: C

The ratio is I/(I + mR\(^{2}\)), printed as 'I/I + mR\(^{2}\)'.

No external torque acts about the axis, so angular momentum is conserved. The mouse lands at the rim, adding a moment of inertia \(mR^{2}\):

\[ \begin{aligned} I\omega_{0} &= (I + mR^{2})\omega \\ \frac{\omega}{\omega_{0}} &= \frac{I}{I + mR^{2}} \end{aligned} \]

The disc slows down, so the ratio is less than 1. \(I/mR^{2}\) and \((I - mR^{2})/I\) do not follow from conservation, and \(I/(I - mR^{2})\) would mean the disc speeds up when mass is added.