Given vectors: $\mathbf{F} = 3\hat{\mathbf{i}}+ 4 \hat{\mathbf{j}} -5\hat{\mathbf{k}}$ and $\mathbf{d} = 5\hat{\mathbf{i}}+ 4 \hat{\mathbf{j}}+3\hat{\mathbf{k}}$. The unit is not specified (e.g., Newtons, meters). Let's assume they are in some consistent units where the dot product is meaningful.

The scalar product is $\mathbf{F} \cdot \mathbf{d} = F_x d_x + F_y d_y + F_z d_z = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$ unit.

The magnitude of $\mathbf{F}$ is $F = |\mathbf{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ unit.

The magnitude of $\mathbf{d}$ is $d = |\mathbf{d}| = \sqrt{d_x^2 + d_y^2 + d_z^2} = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}$ unit.

We know $\mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$, where $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{d}$.

$\cos \theta = \frac{\mathbf{F} \cdot \mathbf{d}}{Fd} = \frac{16}{\sqrt{50} \times \sqrt{50}} = \frac{16}{50} = 0.32$.

The angle between the vectors is $\theta = \cos^{-1}(0.32)$.

The projection of $\mathbf{F}$ on $\mathbf{d}$ is the component of $\mathbf{F}$ along the direction of $\mathbf{d}$. This is given by $F \cos \theta$.

From $\mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$, we have $F \cos \theta = \frac{\mathbf{F} \cdot \mathbf{d}}{d}$.

Projection of $\mathbf{F}$ on $\mathbf{d} = \frac{16}{\sqrt{50}} = \frac{16}{5\sqrt{2}} = \frac{16\sqrt{2}}{10} = 1.6\sqrt{2} \approx 1.6 \times 1.414 \approx 2.262$ unit.

The angle between the force and displacement is $\cos⁻¹(0.32)$, and the projection of $\mathbf{F}$ on $\mathbf{d}$ is $1.6\sqrt{2}$ unit (approximately 2.26 unit).

Given: Mass of raindrop $m = 1.00 \text{ g} = 0.001 \text{ kg}$. Initial height $h_i = 1.00 \text{ km} = 1000 \text{ m}$. Initial speed $v_i = 0$ (assuming it starts from rest). Final height $h_f = 0$ (ground level). Final speed $v_f = 50.0 \text{ m s}^{-1}$. Resistive force $F_r$ opposes motion.

Let the downward direction be positive. The gravitational force $F_g = mg$ acts downwards (positive). The displacement is downwards (positive), $d = h_i - h_f = 1000 - 0 = 1000 \text{ m}$. The resistive force $F_r$ acts upwards (negative).

(a) Work done by the gravitational force ($W_g$). The gravitational force is constant ($F_g = mg$). Assuming $g \approx 9.8 \text{ m s}^{-2}$ (standard value unless specified differently). Let's use $g=10$ as in previous chapter if applicable, but typically $g=9.8$ or $g=9.81$ is standard for physics problems. The example solution seems to use $g=10$ implicitly in calculating $W_g$. Let's use $g=10 \text{ m s}^{-2}$.

$W_g = \mathbf{F}_g \cdot \mathbf{d}$. Force and displacement are in the same direction (downwards is positive). $W_g = F_g d \cos 0^\circ = (mg) d = (0.001 \text{ kg})(10 \text{ m s}^{-2})(1000 \text{ m}) = 10.0 \text{ J}$.

The work done by the gravitational force is 10.0 J.

(b) What is the work done by the unknown resistive force? The net force on the raindrop is the vector sum of the gravitational force and the resistive force: $\mathbf{F}_{net} = \mathbf{F}_g + \mathbf{F}_r$. Both forces are vertical. $F_{net, y} = F_{g, y} + F_{r, y}$. $F_{g, y} = +mg$. $F_{r, y} = -F_r$ (since it opposes motion). So $F_{net} = mg - F_r$ (downwards positive).

According to the Work-Energy theorem, the total work done by the net force equals the change in kinetic energy:

$W_{net} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.

$W_{net} = W_g + W_r$ (Total work is the sum of work done by individual forces).

$W_g + W_r = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.

We need to find $W_r$. $W_r = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 - W_g$.

We know $W_g = 10.0 \text{ J}$. Initial speed $v_i = 0$. Final speed $v_f = 50.0 \text{ m s}^{-1}$.

$K_i = \frac{1}{2}m v_i^2 = 0$.

$K_f = \frac{1}{2}m v_f^2 = \frac{1}{2}(0.001 \text{ kg})(50.0 \text{ m s}^{-1})^2 = \frac{1}{2}(0.001)(2500) \text{ J} = 1.25 \text{ J}$.

$\Delta K = K_f - K_i = 1.25 - 0 = 1.25 \text{ J}$.

$W_r = \Delta K - W_g = 1.25 \text{ J} - 10.0 \text{ J} = -8.75 \text{ J}$.

The work done by the resistive force is -8.75 J. The negative sign indicates that the resistive force opposes the motion, causing energy dissipation.

Given: Displacement of the cycle $d = 10 \text{ m}$. Force on the cycle due to the road (frictional force, $F_{road \to cycle}$) = 200 N, opposed to the motion. Let the direction of motion be positive. Then the force is $F_{road \to cycle} = -200 \text{ N}$. The displacement is $d = +10 \text{ m}$. The angle between the force and displacement is $180^\circ$ ($\cos 180^\circ = -1$).

(a) Work done by the road on the cycle ($W_{road \to cycle}$) is the work done by the frictional force on the cycle.

$W_{road \to cycle} = \mathbf{F}_{road \to cycle} \cdot \mathbf{d} = F_{road \to cycle} d \cos 180^\circ = (200 \text{ N})(10 \text{ m})(-1) = -2000 \text{ J}$.

The work done by the road on the cycle is -2000 J.

(b) How much work does the cycle do on the road? By Newton's Third Law, the force on the road due to the cycle ($F_{cycle \to road}$) is equal in magnitude and opposite in direction to the force on the cycle by the road. So, $F_{cycle \to road} = +200 \text{ N}$ (in the direction of the cycle's motion).

However, the road is assumed to be a rigid surface that does not move or displace during this interaction. The displacement of the road is zero.

Work done by the cycle on the road ($W_{cycle \to road}$) = $\mathbf{F}_{cycle \to road} \cdot \mathbf{d}_{road}$.

$W_{cycle \to road} = (200 \text{ N}) \times (0 \text{ m}) \times \cos \theta = 0 \text{ J}$.

The work done by the cycle on the road is zero.

Given: Mass of bullet $m = 50.0 \text{ g} = 0.050 \text{ kg}$. Initial speed $v_i = 200 \text{ m s}^{-1}$. Final kinetic energy $K_f$ is 10% of initial kinetic energy $K_i$.

Initial kinetic energy $K_i = \frac{1}{2}m v_i^2 = \frac{1}{2}(0.050 \text{ kg})(200 \text{ m s}^{-1})^2 = \frac{1}{2}(0.050)(40000) \text{ J} = 0.025 \times 40000 \text{ J} = 1000 \text{ J}$.

Final kinetic energy $K_f = 10\% \text{ of } K_i = 0.10 \times 1000 \text{ J} = 100 \text{ J}$.

Let $v_f$ be the emergent speed of the bullet. The final kinetic energy is also given by $K_f = \frac{1}{2}m v_f^2$.

$100 \text{ J} = \frac{1}{2}(0.050 \text{ kg})v_f^2$.

$100 = 0.025 v_f^2$.

$v_f^2 = \frac{100}{0.025} = \frac{100}{1/40} = 100 \times 40 = 4000$.

$v_f = \sqrt{4000} = \sqrt{400 \times 10} = 20\sqrt{10} \text{ m s}^{-1}$.

$20\sqrt{10} \approx 20 \times 3.162 \approx 63.24 \text{ m s}^{-1}$.

The emergent speed of the bullet is approximately 63.2 m s⁻¹.

Let the displacement be along the x-axis, starting from $x=0$.

Applied force by the woman, $F_{applied}(x)$:

• From $x = 0$ to $x = 10 \text{ m}$: $F_{applied}(x) = 100 \text{ N}$ (constant).
• From $x = 10 \text{ m}$ to $x = 20 \text{ m}$: The force reduces linearly from 100 N to 50 N. This is a straight line with a negative slope. The equation of the line is $F_{applied}(x) = mx + c$. At $x=10$, $F=100$; at $x=20$, $F=50$. Slope $m = \frac{50-100}{20-10} = \frac{-50}{10} = -5 \text{ N/m}$. Using point $(10, 100)$: $100 = -5(10) + c \implies 100 = -50 + c \implies c = 150$. So, $F_{applied}(x) = -5x + 150$ for $10 \le x \le 20$.

Frictional force, $F_{friction}$: The problem states the frictional force is 50 N and opposes motion. Assuming the motion is in the positive x-direction, the frictional force is $F_{friction} = -50 \text{ N}$ (constant and negative since it opposes the positive displacement) over the entire 20 m distance.

Plotting the forces versus displacement:

Calculate the work done by each force over the total displacement of 20 m. Work done is the area under the respective force-displacement curve.

Work done by the applied force ($W_{applied}$): This is the area under the $F_{applied}(x)$ curve from $x=0$ to $x=20$. This area is a rectangle from 0 to 10 m plus a trapezium from 10 m to 20 m.

$W_{applied} = (\text{Area of rectangle } 0-10 \text{ m}) + (\text{Area of trapezium } 10-20 \text{ m})$.

$W_{applied} = (100 \text{ N}) \times (10 \text{ m}) + \frac{1}{2}(\text{height})((\text{base } 1) + (\text{base } 2))$.

The trapezium has height $20 - 10 = 10 \text{ m}$. Base 1 is the force at $x=10$ (100 N). Base 2 is the force at $x=20$ (50 N).

$W_{applied} = 1000 \text{ J} + \frac{1}{2}(10 \text{ m})(100 \text{ N} + 50 \text{ N}) = 1000 \text{ J} + \frac{1}{2}(10)(150) \text{ J} = 1000 \text{ J} + 750 \text{ J} = 1750 \text{ J}$.

The work done by the woman is 1750 J.

Work done by the frictional force ($W_{friction}$): This is the area under the $F_{friction}(x)$ curve from $x=0$ to $x=20$. The force is constant at -50 N.

$W_{friction} = F_{friction} \times d = (-50 \text{ N}) \times (20 \text{ m}) = -1000 \text{ J}$.

The work done by the frictional force is -1000 J.

Given: Mass of block $m = 1 \text{ kg}$. Initial speed $v_i = 2 \text{ m s}^{-1}$ at $x = 0.10 \text{ m}$. The block enters a rough patch from $x_i = 0.10 \text{ m}$ to $x_f = 2.01 \text{ m}$. The retarding force in this patch is $F_r(x) = -k/x$, where $k = 0.5 \text{ J}$. The force is zero outside this range.

The net force on the block in the rough patch is the retarding force, $F_{net}(x) = F_r(x) = -k/x$. We use the Work-Energy theorem: $K_f - K_i = W_{net}$.

Initial kinetic energy $K_i = \frac{1}{2}m v_i^2 = \frac{1}{2}(1 \text{ kg})(2 \text{ m s}^{-1})^2 = \frac{1}{2}(1)(4) \text{ J} = 2 \text{ J}$.

Work done by the net force ($W_{net}$) is the integral of the net force over the displacement from $x_i$ to $x_f$. In the rough patch, the net force is $F_r(x)$.

$W_{net} = \int_{x_i}^{x_f} F_{net}(x) dx = \int_{0.10}^{2.01} (-\frac{k}{x}) dx = -k \int_{0.10}^{2.01} \frac{1}{x} dx$.

$W_{net} = -k [\ln |x|]_{0.10}^{2.01} = -k (\ln(2.01) - \ln(0.10)) = -k \ln(\frac{2.01}{0.10}) = -k \ln(20.1)$.

Given $k = 0.5 \text{ J}$.

$W_{net} = -(0.5 \text{ J}) \ln(20.1)$.

$\ln(20.1) \approx 2.9997$.

$W_{net} \approx -(0.5 \text{ J})(2.9997) \approx -1.500 \text{ J}$.

The Work-Energy theorem states $K_f - K_i = W_{net}$.

$K_f = K_i + W_{net} = 2 \text{ J} + (-1.500 \text{ J}) = 0.500 \text{ J}$.

The final kinetic energy of the block as it crosses the patch is 0.500 J.

Now find the final speed $v_f$. $K_f = \frac{1}{2}m v_f^2$.

$0.500 \text{ J} = \frac{1}{2}(1 \text{ kg})v_f^2$.

$0.500 = 0.5 v_f^2$.

$v_f^2 = \frac{0.500}{0.5} = 1.0$.

$v_f = \sqrt{1.0} = 1.0 \text{ m s}^{-1}$ (speed must be positive).

The final speed of the block is 1.0 m s⁻¹.

The example solution's calculation of $K_f$ matches ours (0.5 J). Its calculation of $v_f$ also matches (1.0 m/s). The logarithm calculation in the example $2 - 0.5 \ln (20.1) = 2 - 1.5 = 0.5$ seems to round $\ln(20.1)$ to 3, which is a large rounding error. Using $\ln(20.1) \approx 3.00$, $0.5 \times 3 = 1.5$. $2 - 1.5 = 0.5$. This is acceptable for a quick estimate, but using the full value of $\ln(20.1)$ is more accurate. The example result is likely based on this approximation, which fortuitously gave a round number.

The forces acting on the bob are gravity (conservative) and tension in the string (non-conservative, but does no work since displacement is always perpendicular to string). Since tension does no work, the total mechanical energy (potential + kinetic) is conserved.

Let the potential energy be zero at the lowest point A ($y=0$). The height at point B is $L$ (at horizontal level through pivot), and at point C is $2L$ (topmost point). Let the speed at A be $v_0$, at B be $v_B$, and at C be $v_C$.

(i) Obtain an expression for $v_0$. We use the condition that the string becomes slack only at the topmost point C. At point C, the forces on the bob are tension ($T_C$) downwards and gravity ($mg$) downwards. The net force downwards provides the centripetal force towards the center of the circle (which is the pivot point). The radius of the circle at C is $L$. Applying Newton's Second Law in the downward direction at C:

$T_C + mg = \frac{mv_C^2}{L}$.

The string becomes slack just at point C, meaning the tension $T_C$ is just becoming zero at this point.

$0 + mg = \frac{mv_C^2}{L}$.

$gL = v_C^2$, so $v_C = \sqrt{gL}$.

Now, use conservation of mechanical energy between point A and point C. Energy at A = Energy at C.

$K_A + V_A = K_C + V_C$.

$V_A = 0$ (chosen at A). $V_C = mg(2L)$ (height $2L$ above A).

$K_A = \frac{1}{2}mv_0^2$. $K_C = \frac{1}{2}mv_C^2$.

$\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv_C^2 + mg(2L)$.

$\frac{1}{2}mv_0^2 = \frac{1}{2}m(gL) + 2mgL = \frac{1}{2}mgL + \frac{4}{2}mgL = \frac{5}{2}mgL$.

Cancel mass $m$:

$\frac{1}{2}v_0^2 = \frac{5}{2}gL$.

$v_0^2 = 5gL$.

$v_0 = \sqrt{5gL}$.

The initial horizontal velocity at A is $v_0 = \sqrt{5gL}$.

(ii) The speeds at points B and C. We already found $v_C = \sqrt{gL}$ from the condition at the top point C. Point B is at a height $L$ above A.

Use conservation of mechanical energy between point A and point B. Energy at A = Energy at B.

$K_A + V_A = K_B + V_B$.

$V_A = 0$. $V_B = mgL$ (height $L$ above A).

$K_A = \frac{1}{2}mv_0^2 = \frac{1}{2}m(5gL) = \frac{5}{2}mgL$.

$K_B = \frac{1}{2}mv_B^2$.

$\frac{5}{2}mgL + 0 = \frac{1}{2}mv_B^2 + mgL$.

$\frac{5}{2}mgL - mgL = \frac{1}{2}mv_B^2$.

$\frac{3}{2}mgL = \frac{1}{2}mv_B^2$.

Cancel mass $m$:

$\frac{3}{2}gL = \frac{1}{2}v_B^2$.

$v_B^2 = 3gL$.

$v_B = \sqrt{3gL}$.

The speed at point B is $v_B = \sqrt{3gL}$, and the speed at point C is $v_C = \sqrt{gL}$.

(iii) The ratio of the kinetic energies (KB/KC) at B and C.

$K_B = \frac{1}{2}mv_B^2 = \frac{1}{2}m(3gL) = \frac{3}{2}mgL$.

$K_C = \frac{1}{2}mv_C^2 = \frac{1}{2}m(gL) = \frac{1}{2}mgL$.

Ratio $K_B/K_C = \frac{(3/2)mgL}{(1/2)mgL} = \frac{3/2}{1/2} = 3$.

The ratio of kinetic energies $K_B/K_C$ is 3:1.

Comment on the nature of the trajectory of the bob after it reaches point C. At point C, the string becomes slack, meaning the tension becomes zero. The velocity at point C is horizontal (towards the left, $v_C = \sqrt{gL}$). After the string becomes slack, the bob is only under the influence of gravity. This is exactly the condition for projectile motion. Since the initial velocity at C is horizontal, the bob will follow a parabolic trajectory, like a ball thrown horizontally from a cliff edge.

Given: Mass of car $m = 1000 \text{ kg}$. Initial speed of car $v_i = 18.0 \text{ km/h}$. Convert speed to m/s: $v_i = 18.0 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 18.0 \times \frac{5}{18} \text{ m/s} = 5.0 \text{ m s}^{-1}$.

The road is smooth, so friction is negligible. The collision is with a spring of spring constant $k = 5.25 \times 10^3 \text{ N m}^{-1}$. The spring is horizontally mounted, so gravity and normal forces do no work in the horizontal direction. We can assume the only significant forces doing work are the spring force and the car's initial pushing force (which establishes its initial kinetic energy).

We can use the principle of conservation of mechanical energy. Initially, the car has kinetic energy and the spring is unstretched (potential energy = 0). At the point of maximum compression ($x_m$), the car momentarily comes to rest ($v_f = 0$), so its kinetic energy is zero, and all the initial energy is stored as potential energy in the compressed spring.

Initial energy (at $x=0$, $v=v_i$) = $K_i + V_i = \frac{1}{2}mv_i^2 + \frac{1}{2}k(0)^2 = \frac{1}{2}mv_i^2$.

Final energy (at $x=x_m$, $v=0$) = $K_f + V_f = \frac{1}{2}m(0)^2 + \frac{1}{2}kx_m^2 = \frac{1}{2}kx_m^2$.

By conservation of mechanical energy, Initial Energy = Final Energy:

$\frac{1}{2}mv_i^2 = \frac{1}{2}kx_m^2$.

Cancel the factor of 1/2:

$mv_i^2 = kx_m^2$.

We want to find $x_m$. Rearrange the equation:

$x_m^2 = \frac{mv_i^2}{k}$.

$x_m = \sqrt{\frac{mv_i^2}{k}} = v_i \sqrt{\frac{m}{k}}$.

Substitute the numerical values:

$x_m = (5.0 \text{ m s}^{-1}) \sqrt{\frac{1000 \text{ kg}}{5.25 \times 10^3 \text{ N m}^{-1}}}$.

$x_m = 5.0 \sqrt{\frac{1000}{5250}} = 5.0 \sqrt{\frac{100}{525}} = 5.0 \sqrt{\frac{4}{21}} \text{ m}$.

$x_m \approx 5.0 \sqrt{0.190476} \approx 5.0 \times 0.4364 \text{ m} \approx 2.182 \text{ m}$.

The maximum compression of the spring is approximately 2.18 m.

(The example solution result of 2.00 m suggests a possible value difference for g or other constants, or different initial data were used in that calculation. Let's recalculate using $v_i = 5$, $m=1000$, $k=5250$: $x_m = \sqrt{1000 \times 5^2 / 5250} = \sqrt{1000 \times 25 / 5250} = \sqrt{25000 / 5250} = \sqrt{2500 / 525} = \sqrt{100 / 21} \approx \sqrt{4.76} \approx 2.18$. My calculation is consistent. The example result 2.00 m implies $x_m^2 = 4$. $4 = 25000 / k$, so $k = 25000/4 = 6250$. Perhaps the spring constant in the example's calculation was $6.25 \times 10^3$ N/m instead of 5.25? If $k=6.25 \times 10^3$: $x_m = \sqrt{1000 \times 5^2 / 6250} = \sqrt{25000 / 6250} = \sqrt{4} = 2$. Yes, the example result matches if $k = 6.25 \times 10^3$ N/m. Given the discrepancy, I will provide the answer based on the provided number $k = 5.25 \times 10^3 \text{ N m}^{-1}$).

Given: Total mass $m = 1800 \text{ kg}$. Constant upward speed $v = 2 \text{ m s}^{-1}$. Frictional force $F_f = 4000 \text{ N}$ opposing the motion (downwards).

Since the elevator is moving at a constant velocity, its acceleration is zero. By Newton's Second Law, the net force on the elevator must be zero. The forces acting on the elevator are its weight ($W$) downwards, the frictional force ($F_f$) downwards, and the force exerted by the motor ($F_m$) upwards.

$W = mg = (1800 \text{ kg})(9.8 \text{ m s}^{-2}) = 17640 \text{ N}$. (Using $g=9.8$ unless specified otherwise). Let's use $g=10$ as per example instruction.

$W = (1800 \text{ kg})(10 \text{ m s}^{-2}) = 18000 \text{ N}$.

For zero net force (constant velocity), the upward force must balance the total downward forces:

$F_m - W - F_f = 0$.

$F_m = W + F_f = 18000 \text{ N} + 4000 \text{ N} = 22000 \text{ N}$.

The motor must exert an upward force of 22000 N to lift the elevator at a constant speed against gravity and friction. The power delivered by the motor is $P = \mathbf{F}_m \cdot \mathbf{v}$. Since the force and velocity are in the same direction (upwards), $\theta = 0^\circ$, $\cos \theta = 1$.

$P = F_m v \cos 0^\circ = (22000 \text{ N})(2 \text{ m s}^{-1})(1) = 44000 \text{ N m/s} = 44000 \text{ J/s} = 44000 \text{ W}$.

The minimum power delivered by the motor in watts is 44000 W.

Convert power from watts to horsepower (1 hp = 746 W):

$P (\text{hp}) = \frac{P (\text{W})}{746 \text{ W/hp}} = \frac{44000}{746} \text{ hp} \approx 58.98 \text{ hp}$.

The minimum power delivered by the motor in horsepower is approximately 59 hp.

Consider an elastic collision in one dimension between a neutron (mass $m_1$, initial velocity $v_{1i}$) and a stationary moderating nucleus (mass $m_2$, initial velocity $v_{2i}=0$). We want to see how much kinetic energy the neutron loses (or the moderating nucleus gains) after the collision.

For a 1D elastic collision where $v_{2i}=0$, the final velocities are given by:

$v_{1f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1i}$

$v_{2f} = \left(\frac{2m_1}{m_1 + m_2}\right) v_{1i}$

The initial kinetic energy of the neutron is $K_{1i} = \frac{1}{2}m_1 v_{1i}^2$.

The final kinetic energy of the neutron is $K_{1f} = \frac{1}{2}m_1 v_{1f}^2 = \frac{1}{2}m_1 \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 v_{1i}^2 = K_{1i} \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2$.

The fractional kinetic energy lost by the neutron is $\frac{K_{1i} - K_{1f}}{K_{1i}} = 1 - \frac{K_{1f}}{K_{1i}} = 1 - \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2$.

Let the mass of the neutron be $m_n$.

Case 1: Collision with deuterium ($^{2}\text{H}$). The mass of a deuterium nucleus is approximately 2 atomic mass units (u). The mass of a neutron is approximately 1 u. So, $m_2 \approx 2m_1$.

Fractional energy lost = $1 - \left(\frac{m_1 - 2m_1}{m_1 + 2m_1}\right)^2 = 1 - \left(\frac{-m_1}{3m_1}\right)^2 = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$.

This means the neutron loses 8/9 (about 89%) of its kinetic energy in a head-on elastic collision with a deuterium nucleus. The remaining energy is $1/9$ of the initial energy.

Case 2: Collision with carbon ($^{12}\text{C}$). The mass of a carbon nucleus is approximately 12 u. So, $m_2 \approx 12m_1$.

Fractional energy lost = $1 - \left(\frac{m_1 - 12m_1}{m_1 + 12m_1}\right)^2 = 1 - \left(\frac{-11m_1}{13m_1}\right)^2 = 1 - \left(-\frac{11}{13}\right)^2 = 1 - \frac{121}{169} = \frac{169 - 121}{169} = \frac{48}{169}$.

$\frac{48}{169} \approx 0.284$.

This means the neutron loses about 28.4% of its kinetic energy in a head-on elastic collision with a carbon nucleus.

Comparing the two cases, the neutron loses a much larger fraction of its energy when colliding with deuterium (8/9 $\approx$ 0.89) compared to carbon (48/169 $\approx$ 0.284). However, even with carbon, a significant portion of energy is lost in a single head-on collision.

Yes, a neutron can lose most of its kinetic energy in an elastic collision with light nuclei like deuterium or carbon. Deuterium is more effective per head-on collision than carbon. In a real reactor, there are many collisions, and the process works even if collisions are not perfectly head-on.

Given: Elastic collision between two billiard balls with equal masses ($m_1 = m_2 = m$). The target ball is initially at rest ($v_{2i} = 0$). The initial velocity of the cue ball is $v_{1i}$. The target ball moves off at an angle $\theta_2 = 37^\circ$ relative to the initial direction of the cue ball ($v_{1i}$). We need to find the angle $\theta_1$ at which the cue ball moves off after the collision, relative to its initial direction.

Let the initial direction of the cue ball be along the x-axis. So $\mathbf{v}_{1i} = v_{1i} \hat{\mathbf{i}}$ and $\mathbf{v}_{2i} = 0$. After the collision, the final velocities are $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$ at angles $\theta_1$ and $\theta_2$ with the initial direction.

By conservation of linear momentum ($\mathbf{p}_{total, i} = \mathbf{p}_{total, f}$):

$m_1 \mathbf{v}_{1i} + m_2 \mathbf{v}_{2i} = m_1 \mathbf{v}_{1f} + m_2 \mathbf{v}_{2f}$

$m \mathbf{v}_{1i} + m(0) = m \mathbf{v}_{1f} + m \mathbf{v}_{2f}$ (since $m_1=m_2=m$)

$\mathbf{v}_{1i} = \mathbf{v}_{1f} + \mathbf{v}_{2f}$ (Divide by m). This is a vector equation.

Since the collision is elastic, kinetic energy is conserved:

$K_{total, i} = K_{total, f}$

$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

$\frac{1}{2}m v_{1i}^2 + 0 = \frac{1}{2}m v_{1f}^2 + \frac{1}{2}m v_{2f}^2$

$v_{1i}^2 = v_{1f}^2 + v_{2f}^2$ (Divide by $\frac{1}{2}m$). This is a scalar equation involving the magnitudes of the velocities.

Now, consider the vector equation $\mathbf{v}_{1i} = \mathbf{v}_{1f} + \mathbf{v}_{2f}$. Take the dot product of this equation with itself:

$\mathbf{v}_{1i} \cdot \mathbf{v}_{1i} = (\mathbf{v}_{1f} + \mathbf{v}_{2f}) \cdot (\mathbf{v}_{1f} + \mathbf{v}_{2f})$

$v_{1i}^2 = \mathbf{v}_{1f} \cdot \mathbf{v}_{1f} + \mathbf{v}_{1f} \cdot \mathbf{v}_{2f} + \mathbf{v}_{2f} \cdot \mathbf{v}_{1f} + \mathbf{v}_{2f} \cdot \mathbf{v}_{2f}$

$v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2(\mathbf{v}_{1f} \cdot \mathbf{v}_{2f})$ (since dot product is commutative).

We also have the scalar equation from kinetic energy conservation: $v_{1i}^2 = v_{1f}^2 + v_{2f}^2$.

Substitute this into the equation from the dot product:

$v_{1f}^2 + v_{2f}^2 = v_{1f}^2 + v_{2f}^2 + 2(\mathbf{v}_{1f} \cdot \mathbf{v}_{2f})$.

This implies $2(\mathbf{v}_{1f} \cdot \mathbf{v}_{2f}) = 0$. Since we are considering a collision where the balls move off, $v_{1f} \ne 0$ and $v_{2f} \ne 0$. Therefore, the dot product of the final velocities must be zero: $\mathbf{v}_{1f} \cdot \mathbf{v}_{2f} = 0$.

The dot product of two non-zero vectors is zero if and only if the vectors are perpendicular. Thus, $\mathbf{v}_{1f}$ is perpendicular to $\mathbf{v}_{2f}$.

The final velocity vector of the cue ball ($\mathbf{v}_{1f}$) makes an angle $\theta_1$ with the initial direction. The final velocity vector of the target ball ($\mathbf{v}_{2f}$) makes an angle $\theta_2 = 37^\circ$ with the initial direction. The angle between $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$ is the angle between their directions. If one is at angle $\theta_1$ and the other at angle $\theta_2$, the angle between them is $|\theta_1 - \theta_2|$ or $360^\circ - |\theta_1 - \theta_2|$. However, if they are perpendicular, the angle between them is $90^\circ$.

Since $\mathbf{v}_{1f}$ is perpendicular to $\mathbf{v}_{2f}$, the angle between their directions is $90^\circ$. The direction of $\mathbf{v}_{1f}$ is at angle $\theta_1$ from the initial direction, and the direction of $\mathbf{v}_{2f}$ is at angle $\theta_2 = 37^\circ$ from the initial direction.

The angle between $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$ is also equal to $|\theta_1| + |\theta_2|$ if they are in different half-planes relative to the initial direction, or $|\theta_1 - \theta_2|$ if they are in the same half-plane. Looking at the diagram Fig. 5.10, $\mathbf{v}_{1f}$ is above the initial line and $\mathbf{v}_{2f}$ is below. So the angle between them is $\theta_1 + \theta_2$.

Therefore, $\theta_1 + \theta_2 = 90^\circ$.

We are given $\theta_2 = 37^\circ$.

$\theta_1 + 37^\circ = 90^\circ$.

$\theta_1 = 90^\circ - 37^\circ = 53^\circ$.

The angle $\theta_1$ is 53°.

This result applies specifically to elastic collisions between objects of equal mass where one is initially at rest: after the collision, the two objects move off at right angles to each other.