Q.A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Take the common direction of trains A and B as positive. Then A has velocity $+V_A$, B has $+V_B$, and C (opposite direction) has $-V_C$, where $V_C>0$ is its speed.

Speed of C with respect to B (they move oppositely, so speeds add):

Speed of A with respect to B:

Given $V_{CB}=2\,V_{AB}$: …

The core concept here is the conservation of the center of mass. For a system of particles, if there are no external forces acting on the system in a particular direction, then the center of mass of the system will remain stationary or continue to move with a constant velocity in that direction.

In this problem, the system consists of the boat and the boy. The forces acting on this system are:

1. Gravity (acting vertically downwards on both the boat and the boy).
2. Normal force from the water (acting vertically upwards on the boat).
3. The force exerted by the boy on the boat, and by the boat on the boy (these are internal forces within the system).

Crucially, we assume there are no external horizontal forces, such as water resistance or wind, acting on the boat-boy system. Therefore, the center of mass of the combined boat-boy system will not shift horizontally. Any movement of the boy relative to the boat will cause the boat to move in the opposite direction, ensuring the system's center of mass stays in its original position.

Let's solve the problem step-by-step.

1. Define the Coordinate System and Initial Positions:

   Let the shore be at the origin, $x=0$.

   The boat has a mass $M_B = 200$ kg and length $L = 10$ m. Its center of mass is at its geometric center.

   The boy has a mass $M_P = 50$ kg.

   The boy is initially standing at one end of the boat, at a distance of 20 m from the shore. This means the end of the boat closer to the shore is at $x=20$ m.

   • Initial position of the boy ($x_{P,i}$): $x_{P,i} = 20$ m.
   • The boat extends from $x=20$ m to $x=20+10 = 30$ m.
   • Initial position of the boat's center of mass ($x_{B,i}$): $x_{B,i} = 20 + \frac{L}{2} = 20 + \frac{10}{2} = 20 + 5 = 25$ m.

2. Calculate the Initial Center of Mass of the System:

   The formula for the center of mass of a two-particle system is:

   $X_{CM} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$

   Using this, the initial center of mass of the boat-boy system ($X_{CM,i}$) is:

3. Determine the Final Positions:

   The boy moves to the center of the boat. Let the new position of the boat's center of mass be $x_{B,f}$.

   Since the boy is now at the center of the boat, his final position ($x_{P,f}$) will be the same as the boat's center of mass:

   • Final position of the boy ($x_{P,f}$): $x_{P,f} = x_{B,f}$.

4. Calculate the Final Center of Mass of the System:

   The final center of mass of the boat-boy system ($X_{CM,f}$) is:

   $X_{CM,f} = \frac{M_B x_{B,f} + M_P x_{P,f}}{M_B + M_P}$ …

The center of mass of a system of particles is a unique point that represents the average position of the total mass of the system. It's a crucial concept in mechanics because the motion of the center of mass often simplifies the analysis of a complex system.

The position of the center of mass depends on two factors: the masses of the individual particles and their respective positions. If either of these changes, the center of mass will generally shift. In this problem, we are changing the mass of one particle, which will cause the center of mass to move closer to the particle whose mass has increased, as it now contributes more to the "average" position.

For a system of two particles, say with masses $m_1$ and $m_2$ located at positions $x_1$ and $x_2$ along an axis, the position of their center of mass ($X_{CM}$) is given by the weighted average:

Let's apply this formula to find the initial and final positions of the center of mass and then determine the shift.

1. Define the initial setup and calculate the initial center of mass: Let's place one of the particles at the origin of our coordinate system for simplicity.
   • Particle 1: mass $m_1 = m$, position $x_1 = 0$.
   • Particle 2: mass $m_2 = m$, position $x_2 = d$. The total mass of the system is $M_{initial} = m_1 + m_2 = m + m = 2m$. The initial position of the center of mass, $X_{CM, initial}$, is:

So, initially, the center of mass is exactly halfway between the two particles, which is expected for two equal masses.

2. Define the final setup and calculate the new center of mass:

The problem states that the mass of one of the particles is doubled. Let's assume the mass of Particle 1 is doubled. The distance between them remains $d$.

* Particle 1: new mass $m'_1 = 2m$, position $x'_1 = 0$.

* Particle 2: mass $m'_2 = m$, position $x'_2 = d$. …

The concept first. For a body made of parts,

Each uniform plate can be replaced by a point mass at its own geometric centre. Because both plates here have the same dimensions ($L$ long, $a$ wide) and the same material and thickness, they have equal masses $m$, so the composite centre of mass is simply the average of the two plate-centres. All we then need is the height of each plate's centre above the ground in the two orientations.

Step 1 — Set up the upright T.

Measure $y$ upward from the horizontal surface.

• Stem (vertical plate, length $L$, width $a$): it stands on the ground, so it spans $y = 0$ to $y = L$. Its centre: $y = L/2$.
• Cross-bar (horizontal plate, length $L$, thickness $a$): it rests on top of the stem, so it spans $y = L$ to $y = L+a$. Its centre: $y = L + a/2$.

Step 2 — Invert the T (now it looks like $\perp$).

• Cross-bar now lies flat on the ground: it spans $y = 0$ to $y = a$. Its centre: $y = a/2$.
• Stem stands on the bar: it spans $y = a$ to $y = a + L$. Its centre: $y = a + L/2$. …

Concept: The Center of Mass Must Remain Fixed

The center of mass of a system is the weighted average position of all masses. When we require that the center of mass stays at the same location, any displacement of one mass must be balanced by an appropriate displacement of the other mass. The key insight is that heavier masses need to move less to produce the same effect on the center of mass position.

Let's set up a coordinate system with the center of mass at the origin initially.

Step-by-Step Solution

1. Initial positions relative to center of mass

Let the initial positions of $m_1$ and $m_2$ relative to the center of mass be $x_1$ and $x_2$ respectively. Since the center of mass is at the origin:

This tells us that $x_1$ and $x_2$ have opposite signs (the masses are on opposite sides of the center of mass).

2. After the displacements

Particle $m_1$ is moved toward the center of mass by distance $d$. Since it's moving toward the center of mass, its distance from the center decreases, so its new position is $x_1 - d$ (assuming $x_1 > 0$; if $x_1 < 0$, it would be $x_1 + d$, but the algebra works out the same).

Let particle $m_2$ move by distance $\Delta x_2$ (positive means away from center of mass in its direction). Its new position is $x_2 + \Delta x_2$.

3. Condition for center of mass to remain fixed

For the center of mass to stay at the origin:

4. Solve for the required displacement

Expanding:

Since $m_1 x_1 + m_2 x_2 = 0$ from step 1:

5. Interpret the sign …

The key idea here is that momentum is a vector quantity. When two bodies collide and stick together, the total momentum before the collision equals the total momentum after the collision — but you must add the momenta as vectors, not just as numbers. The masses are equal, so the problem reduces to finding the vector sum of the two initial velocities and then dividing by the total mass.

Since the initial velocities are perpendicular, the vector sum is the hypotenuse of a right triangle. That’s the entire conceptual backbone: momentum conservation in two dimensions, with perpendicular components.

Let’s work through it step by step.

1. Set up the coordinate system. Let body A move along the x-axis with speed $V_0$, so its velocity vector is $\vec{v}_A = V_0 \hat{i}$. Let body B move along the y-axis with speed $2V_0$, so its velocity vector is $\vec{v}_B = 2V_0 \hat{j}$. Both have the same mass $m$.

2. Write the total momentum before collision.

   Momentum is mass times velocity. So:

3. After collision, the bodies merge. The combined mass is $2m$, and it moves with some final velocity $\vec{v}_f$. By conservation of momentum:

Cancel $m$ (it’s nonzero):

4. Solve for the final velocity vector:

5. Find the speed (magnitude of $\vec{v}_f$). The components are perpendicular, so: $$$$

We are told a car covers three equal thirds of a total distance $D$ at speeds $60$ kmph, $10$ kmph, and $v$ kmph respectively. The mean (average) speed over the whole trip is given as $5$ m/s. We need $v$ in kmph.

Concept and intuition

The most common mistake here is to think mean speed is the arithmetic mean of the three speeds. It is not. Mean speed is defined as

Since each third has a different speed, the time taken for each third is different. We must find the total time in terms of $v$, set the mean speed equal to the given value (after converting units), and solve for $v$.

1. Set up the distance and times Let the total distance be $3d$ km, so each third is $d$ km. Time for first third: $t_1 = \dfrac{d}{60}$ hours. Time for second third: $t_2 = \dfrac{d}{10}$ hours. Time for last third: $t_3 = \dfrac{d}{v}$ hours. Total time:

2. Write the mean speed in kmph Mean speed in kmph is

This is given as $5$ m/s. We must convert to kmph:

(Recall: $1 \text{ m/s} = 3.6$ kmph, or multiply by $18/5$.)

3. Set up the equation

Invert both sides:

Multiply by 3:

Concept & Intuition

The center of mass is the “balance point” — the average position weighted by how much mass is at each location. When density varies, you can’t just take the midpoint; you must integrate. Here, density increases linearly from the left end (density $0$ at $x=0$) to the right end (density $2L$ at $x=L$). That means more mass is concentrated toward the right, so the center of mass should be to the right of the midpoint $L/2$. Among the options, only $\frac{2}{3}L$ is greater than $L/2$ and less than $L$, so it’s the plausible choice — but we’ll verify by calculation.

Step-by-step solution

1. Set up the integrals For a one-dimensional rod along the $x$-axis from $x=0$ to $x=L$, the center of mass coordinate is

where $dm = \lambda(x)\,dx$ and $\lambda(x)=2x$.

2. Compute the total mass

3. Compute the first moment (numerator) $\int_0^L x \, dm = \int_0^L x \cdot 2x \, dx = 2\int_0^L x^2 \, dx = 2\left[ \frac{x^3}{3} \right]_0^L = \frac{2L^3}{3}$ …

Concept & Intuition

The centre of mass is the “balance point” — the average position weighted by mass. When density varies, you can’t just take the geometric centre; you must account for where the mass is concentrated. Here, density increases linearly from the left end ($x=0$) to the right end ($x=L$), so more mass lies toward the right. That pulls the centre of mass to the right of the midpoint — but how far? We’ll compute it exactly.

Step-by-step solution

1. Set up the integrals For a one-dimensional rod, the centre of mass coordinate $\bar{x}$ (from the left end) is

where $dm = \lambda(x)\,dx$ and $\lambda(x)=2x$.

2. Compute the total mass

3. Compute the first moment of mass $\int_0^L x\,dm = \int_0^L x \cdot 2x\,dx = 2\int_0^L x^2\,dx = 2\left[ \frac{x^3}{3} \right]_0^L = \frac{2L^3}{3}$ …

The key is to see that rolling motion combines translation and rotation, and the "without slipping" condition locks the two together. When the disc rolls down, gravitational potential energy gets split into two forms: translational kinetic energy of the centre of mass and rotational kinetic energy about the centre. The moment of inertia of a disc is $\frac{1}{2}MR^2$, and the rolling condition gives $\omega = v/R$. That ratio is what makes mass and radius disappear from the final speed.

1. Energy conservation — The disc starts from rest at height $h$. Its initial energy is purely gravitational potential: $Mgh$. At the bottom, the energy is all kinetic:

2. Moment of inertia — For a solid disc about its centre,

3. Rolling condition — No slipping means the point of contact is instantaneously at rest, so

4. Substitute into energy equation —

Simplify the rotational term:

5. Combine terms —

Cancel $M$ (mass does not matter):

$gh = \frac{3}{4}v^2 \quad\Rightarrow\quad v = \sqrt{\frac{4}{3}gh}.$ …

The key idea is the conservation of centre of mass in a system with no external horizontal force. The man and boat are initially at rest on still water — no friction, no engine, no wind. So the horizontal position of the system’s centre of mass cannot change, no matter how the man moves inside the boat.

When the man runs from one end to the other, he pushes backward on the boat to move forward. The boat, in turn, moves in the opposite direction. Once he stops, both are again at rest relative to the water, but shifted relative to their original positions. The centre of mass, however, is exactly where it started.

Let’s set this up.

1. Define the system and coordinates.

   Let the boat’s mass be $M = 750\ \text{kg}$ and its length $L = 10\ \text{m}$. The man’s mass is $m = 80\ \text{kg}$.

   Place the origin at the initial position of the left end of the boat. The boat’s own centre of mass is at its geometric centre, initially at $x = 5\ \text{m}$. The man starts at the left end, so his initial position is $x = 0$.

2. Find the initial centre of mass of the system.

   The centre of mass of the boat alone is at $5\ \text{m}$. The man is at $0$. So:

Simplify:

3. Describe the final configuration.

   After the man runs to the other end and stops, he is at the right end of the boat. But the boat itself has moved. Let the displacement of the boat relative to the water be $d$ (positive if to the right).

   The boat’s centre is now at $5 + d$. The man is at the right end, which is at $10 + d$ (since the boat’s length is 10 m and its left end is now at $d$).

4. Apply centre-of-mass conservation.

   The final centre of mass must equal the initial one:

Multiply through by $830$:

Expand: