Course: Physics 1 (6441) Semester: Spring, 2023
Level: B.Ed. (2.5 & 4 Years)
Assignment 2
Q. 1(a): Vector Sum of Momenta and Center of Mass (CoM)
The first question involves proving that the vector sum of momenta of individual particles is equal to the mass of the whole system times the velocity of the Center of Mass (CoM).
Solution:
To begin, let’s consider a system of ‘n’ particles, each with mass ‘m_i’ and velocity ‘v_i.’ The total momentum of the system, ‘P_total,’ is given by the sum of individual momenta:
P_total = m_1 * v_1 + m_2 * v_2 + … + m_n * v_n
Next, let’s define the total mass of the system as ‘M_total,’ which is the sum of the masses of all particles:
M_total = m_1 + m_2 + … + m_n
Now, we know that the velocity of the Center of Mass (CoM) of the system is the weighted average of the velocities of individual particles, with masses as weights:
V_CoM = (m_1 * v_1 + m_2 * v_2 + … + m_n * v_n) / (m_1 + m_2 + … + m_n)
Substituting the expression for ‘P_total’ into the equation for ‘V_CoM,’ we get:
V_CoM = P_total / M_total
This result shows that the vector sum of momenta of individual particles is indeed equal to the mass of the whole system times the velocity of the Center of Mass.
Q. 1(b): Law of Conservation of Linear Momentum
The second part of the question requires stating and proving the law of Conservation of Linear Momentum.
Solution:
The law of Conservation of Linear Momentum states that the total momentum of an isolated system remains constant if no external forces act on it.
Mathematically, this can be expressed as:
ΣP_initial = ΣP_final
Where ΣP_initial is the initial total momentum of the system, and ΣP_final is the final total momentum of the system after any internal interactions.
The proof of this law is based on Newton’s Third Law of Motion, which states that every action has an equal and opposite reaction. In an isolated system, the internal forces between particles are equal and opposite, resulting in no net external force acting on the system. As a consequence, the total momentum remains unchanged.
Q. 2: Two-Dimensional Collision
The second question involves discussing two-dimensional collisions as an extension of one-dimensional analysis.
Solution:
In one-dimensional collisions, we consider the motion of objects along a single axis. However, in real-world scenarios, objects often collide in two dimensions, involving both horizontal and vertical motions.
The conservation of momentum still holds in two-dimensional collisions, but we need to consider both components of momentum independently. By applying principles of vector analysis, we can analyze the motion of objects after the collision accurately.
Q. 3: Derivation of Rotational Kinetic Energy
The third question seeks the mathematical expression for calculating the kinetic energy of a rotational body.
Solution:
To derive the formula for rotational kinetic energy, let’s consider a rigid body rotating about a fixed axis with angular velocity ‘ω.’ The rotational kinetic energy ‘K_rot’ is given by:
K_rot = (1/2) * I * ω^2
Where ‘I’ represents the moment of inertia of the body about the axis of rotation. It depends on the mass distribution and the axis’s location relative to the body. The rotational kinetic energy is essential in various mechanical systems, such as wheels, flywheels, and gyros.
Q. 4: Torque and Angular Momentum Relationship
The fourth question focuses on discussing the effect of torque and its relationship with angular momentum.
Solution:
Torque is the rotational equivalent of force, causing an object to rotate about an axis. The relationship between torque ‘Ï„’ and angular momentum ‘L’ is given by:
Ï„ = dL / dt
This equation illustrates that torque is the rate of change of angular momentum with respect to time. When a torque acts on a rotating body, it changes the body’s angular momentum, causing it to accelerate or decelerate.
Q. 5: Calculation of Escape Velocity
The final question requires deriving the formula for calculating the escape velocity of a body.
Solution:
The escape velocity ‘v_escape’ is the minimum velocity required for an object to break free from a celestial body’s gravitational pull. It is given by the formula:
v_escape = √(2 * G * M / r)
Where ‘G’ is the universal gravitational constant, ‘M’ is the mass of the celestial body, and ‘r’ is the distance between the object and the center of the celestial body.