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Physics

Centre of Mass, Linear Momentum, Collision

Centre of Mass of a Uniform Straight Rod

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⚡ Quick Summary
The center of mass of a uniform rod is located at its midpoint. For calculations involving multiple objects, each object can be treated as a point particle located at its center of mass.
X = (1/M) ∫ x dm = (1/M) ∫ x (M/L) dx (integrated from 0 to L) = L/2
To find the center of mass of a uniform rod of mass M and length L, consider an element of length dx at position x. The mass of this element is dm = (M/L)dx. The x-coordinate of the center of mass is then calculated by integrating x dm over the length of the rod and dividing by the total mass M.

Centre of Mass of Continuous Bodies

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For continuous bodies, the center of mass is found by replacing summation with integration. Instead of summing over individual particles, we integrate over small elements of mass 'dm'.
X = (1/M) ∫ x dm, Y = (1/M) ∫ y dm, Z = (1/M) ∫ z dm
When dealing with continuous mass distributions, the summation used for discrete particles is replaced by integration. We consider a small element of the body with mass dm and coordinates (x, y, z). The coordinates of the center of mass are found by integrating over the entire body.

Centre of Mass of a Uniform Semicircular Wire

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To locate the centre of mass of a uniform semi circular wire
no formula present in the text
To find the centre of mass of a uniform semi circular wire of mass M and radius R, take the centre as origin, the line joining the ends as the X-axis and Y axis in the middle. Using integration, it can be derived.

Centre of Mass

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The centre of mass of a system of particles is a point that represents the average position of all the particles in the system, weighted by their masses. The motion of the centre of mass is determined only by the external forces acting on the system.
<ul><li>Velocity of Centre of Mass: Vcm = (m1v1 + m2v2) / (m1 + m2)</li><li>Acceleration of Centre of Mass: Acm = (m1a1 + m2a2) / (m1 + m2)</li><li>M * Acm = Fext</li></ul>
  • Position of Centre of Mass: For a two-particle system, the position of the centre of mass changes as the positions of the individual particles change.
  • Velocity of Centre of Mass: The velocity of the centre of mass at time t is given by Vcm = (m1v1 + m2v2) / (m1 + m2).
  • Acceleration of Centre of Mass: The acceleration of the centre of mass is given by Acm = (m1a1 + m2a2) / (m1 + m2).
  • Effect of Internal Forces: Internal forces between particles do not affect the motion of the centre of mass. If no external force acts on the system and the centre of mass is initially at rest, it remains at rest. If the centre of mass is moving with respect to an inertial frame, it continues its motion along the same straight line with the same speed.
  • System of N Particles: For a system of N particles, the acceleration of the ith particle is given by ai = (1/mi) * (sum of Fij + Fi_ext), where Fij is the force on the ith particle due to the jth particle, and Fi_ext is the external force on the ith particle. Summing over all particles, the sum of (mi * ai) = F_ext, where F_ext is the sum of all external forces.
  • External Forces and Centre of Mass: If the external forces acting on the system add up to zero, the acceleration of the centre of mass is zero, and its velocity is constant.

Rocket Propulsion

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The acceleration of a rocket increases as time passes due to the consumption of fuel. The velocity of the rocket is derived considering the exhaust velocity of the gas and the decreasing mass of the rocket.
['v_gas,ground = v_gas,rocket + v_rocket,ground = -u + v', 'Δv = (ΔM)u / (M - ΔM)', 'dv/dt = (ru) / (M - rt)', 'v = u * ln(M0 / (M0 - rt))']
The velocity of the gas with respect to the ground is given by v_gas,ground = v_gas,rocket + v_rocket,ground = -u + v. The linear momentum of the mass M at t + Δt is (M - ΔM)(v + Δv) + ΔM(v - u). Assuming no external force, (M - ΔM)(v + Δv) + ΔM(v - u) = Mv, which simplifies to Δv = (ΔM)u / (M - ΔM). This leads to dv/dt = (ru) / (M - rt), where r = dM/dt. Integrating this gives v = u * ln(M0 / (M0 - rt)).

Collision with a Spring

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⚡ Quick Summary
When two blocks with different velocities collide with a spring, the spring compresses until the blocks reach equal velocities. After that, the spring expands until it reaches its natural length, and the blocks separate with constant velocities. Throughout the process, the system's momentum is conserved, and the total energy (kinetic plus elastic potential) remains constant.
['m1v1 + m2v2 = P (constant momentum)', 'm1V + m2V = P (at maximum compression)', "m1v1' + m2v2' = P (after separation)"]
Two blocks m1 and m2 moving with velocities v1 and v2 (v1 > v2) collide with a spring. The spring compresses, exerting forces kx on the blocks. The compression is maximum when the blocks have equal velocities. After that, the spring expands until it reaches its natural length, and the blocks separate. Momentum of the two-block system is conserved throughout the process: m1v1 + m2v2 = m1V + m2V = m1v1' + m2v2'. The total energy (kinetic plus elastic potential) remains constant as there is no friction.

Elastic, Inelastic, and Partially Elastic Collisions

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Collisions can be elastic (kinetic energy conserved), inelastic (kinetic energy lost), or partially elastic (some kinetic energy lost).
Elastic Collision:<br>m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub> = m<sub>1</sub>v<sub>1</sub>' + m<sub>2</sub>v<sub>2</sub>'<br>1/2 m<sub>1</sub>v<sub>1</sub><sup>2</sup> + 1/2 m<sub>2</sub>v<sub>2</sub><sup>2</sup> = 1/2 m<sub>1</sub>v<sub>1</sub>'<sup>2</sup> + 1/2 m<sub>2</sub>v<sub>2</sub>'<sup>2</sup><br><br>Inelastic Collision:<br>m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub> = m<sub>1</sub>V + m<sub>2</sub>V<br><br>Velocity of separation (after collision) = Velocity of approach (before collision):<br>v<sub>1</sub> - v<sub>2</sub> = v<sub>2</sub>' - v<sub>1</sub>'<br><br>Final velocities in one-dimensional elastic collision:<br>v<sub>1</sub>' = [(m<sub>1</sub> - m<sub>2</sub>) / (m<sub>1</sub> + m<sub>2</sub>)]v<sub>1</sub> + [2m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>)]v<sub>2</sub><br>v<sub>2</sub>' = [2m<sub>1</sub> / (m<sub>1</sub> + m<sub>2</sub>)]v<sub>1</sub> + [(m<sub>2</sub> - m<sub>1</sub>) / (m<sub>1</sub> + m<sub>2</sub>)]v<sub>2</sub>
Elastic Collision: Kinetic energy is conserved (Kf = Ki).
Inelastic Collision: Kinetic energy is not conserved (Kf < Ki), and in a perfectly inelastic collision, the two objects stick together (v1' = v2' = V).
Partially Elastic Collision: Kinetic energy is not conserved (Kf < Ki), but the loss of kinetic energy (ΔK = Ki - Kf) is less than in a perfectly inelastic collision. The objects try to regain their original shapes after maximum deformation, pushing each other. The final kinetic energy is less than the initial kinetic energy. The shapes are not completely recovered, and some energy remains inside the deformed ball.

Elastic Collision in One Dimension - Special Cases (Heavy Body vs. Light Body)

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Analyzing elastic collisions where one body is much heavier than the other, both when the heavy body hits the light one, and vice versa.
Heavy hits Light (m1 >> m2):<br>v<sub>1</sub>' ≈ v<sub>1</sub><br>v<sub>2</sub>' ≈ 2v<sub>1</sub> - v<sub>2</sub><br>If v<sub>2</sub>=0, v<sub>2</sub>' ≈ 2v<sub>1</sub>
Case 1: Heavy body (m1) hits a light body (m2) from behind (m1 >> m2):
Approximations: (m1 - m2) / (m1 + m2) ≈ 1, 2m2 / (m1 + m2) ≈ 0, 2m1 / (m1 + m2) ≈ 2.
The heavier body continues to move with almost the same velocity (v1' ≈ v1). The lighter body's velocity after the collision is approximately v2' ≈ 2v1 - v2. If the lighter body is initially at rest (v2 = 0), then v2' = 2v1, meaning the lighter body flies away with twice the velocity of the heavier body.

Case 2: Light body (m1) hits a heavy body (m2) from behind (m1 << m2):

Impulse and Impulsive Force

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When bodies collide, they exert forces on each other for a short time, causing a change in momentum. Large forces acting for short durations are impulsive forces. Impulse is the change in momentum caused by an impulsive force and is the area under the Force-time curve.
Impulse = ∫(t_i to t_f) F dt = P_f - P_i
When two bodies collide, they exert forces on each other while in contact. The momentum of each body is changed due to the force on it exerted by the other. On an ordinary scale, the time duration of this contact is very small and yet the change in momentum is sizeable. This means that the magnitude of the force must be large on an ordinary scale. Such large forces acting for a very short duration are called impulsive forces. The force may not be uniform while the contact lasts. The change in momentum produced by such an implusive force is P - P = ∫dP = ∫F dt This quantity ∫F dt is known as the impulse of the force F during the time interval t_i to t_f and is equal to the change in the momentum of the body on which it acts. Obviously, it is the area under the F - t curve for one-dimensional motion

Collision of Two Bodies (Non Head-on)

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When two bodies collide and the collision is not head-on, they move along different lines. The conservation of momentum can be applied in X and Y directions. If the collision is elastic, kinetic energy is also conserved. However, the final motion cannot be uniquely determined with only this information; it depends on the angle between the line of force and the initial velocity. Momentum is conserved perpendicular to the force.
m1u1 = m1v1cosθ + m2v2cosΦ ; 0 = m1v1sinθ - m2v2sinΦ; (1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
If the collision is not head-on (the force during the collision is not along the initial velocity), the objects move along different lines. Suppose the object A moves with a velocity v1 making an angle θ with the X-axis and the object B moves with a velocity v2 making an angle Φ with the same axis. Also, suppose v1 and v2 lie in X-Y plane. Using conservation of momentum in X and Y directions, we get: m1u1 = m1v1cosθ + m2v2cosΦ 0 = m1v1sinθ - m2v2sinΦ If the collision is elastic, the final kinetic energy is equal to the initial kinetic energy. Thus: (1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2 We have four unknowns v1, v2, θ and Φ to describe the final motion whereas there are only three relations. Thus, the final motion cannot be uniquely determined with this information. In fact, the final motion depends on the angle between the line of force during the collision and the direction of initial velocity. The momentum of each object must be individually conserved in the direction perpendicular to the force. The motion along the line of force may be treated as a one-dimensional collision.

Centre of Mass

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Deals with the motion of system of particles under external forces.
a_CM = F / M, where M is the total mass and F is the external force. x_CM = (m1x1 + m2x2) / (m1 + m2)
The position of the centre of mass at time t is given by x = a t^2 = (F / 4m) * t^2, where a is the acceleration of the centre of mass.

Conservation of Linear Momentum

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In a closed system with no external forces, the total linear momentum remains constant.
M(4000 km/h) = 5MV/6 + M * 3900 km/h 0 = MV - mw or, MV = m (v - V)
If the final velocity of the shuttle is V then the total final linear momentum is considered. The linear momentum of the system remains constant. Initially, both the man and the platform were at rest.

Relative Velocity

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The velocity of an object with respect to the Earth can be calculated by adding its velocity relative to another object and the velocity of that object relative to the Earth.
V_earth = V_shuttle + V_module
The velocity of the ejected module with respect to the earth = its velocity with respect to the shuttle + the velocity of the shuttle with respect to the earth

Recoil Velocity

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When a person moves on a platform, the platform recoils in the opposite direction to conserve momentum. The recoil velocity can be calculated using conservation of momentum.
0 = (25 kg) * (5 m/s) - (10 kg)v
Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant.

Elastic Collision of Equal Masses

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In an elastic collision between two objects of equal mass, the objects exchange velocities.
Not applicable.
When two balls of equal mass undergo an elastic collision, their velocities are interchanged after the collision.

Collision with a Rigid Wall

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When a ball collides with a rigid wall, it rebounds with the same speed if the collision is perfectly elastic.
Not applicable. Velocity of separation = Velocity of approach
In a collision with a rigid wall (considered to have infinite mass), momentum conservation doesn't give a useful result. The velocity of separation equals the velocity of approach. If the ball approaches the wall with speed *v*, it rebounds with speed *v* in the opposite direction.

Head-on Collision with Coefficient of Restitution

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After a head-on collision, the velocities of the objects can be found using conservation of momentum and the coefficient of restitution.
m1v1 + m2v2 = m1u1 + m2u2; e = (u2 - u1) / (v1 - v2)
For a head-on collision between two masses *m1* and *m2* with initial velocities *v1* and *v2*, and final velocities *u1* and *u2*, conservation of momentum gives: *m1v1 + m2v2 = m1u1 + m2u2*. The coefficient of restitution *e* is defined as: *e = (u2 - u1) / (v1 - v2)*.

Collision with a Floor at an Angle

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When a ball collides with a floor at an angle, the parallel component of velocity remains unchanged, and the normal component is affected by the coefficient of restitution.
v'sinθ' = vsinθ; v'cosθ' = evcosθ
When a ball hits the floor at an angle θ, the component of velocity parallel to the floor remains constant (v'sinθ' = vsinθ). The normal component is affected by the coefficient of restitution (v'cosθ' = evcosθ).

Centre of Mass

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The center of mass of a system is the unique point at which the weighted relative position of the distributed mass sums to zero. It's the average position of all the parts of the system, weighted by their masses.
None explicitly mentioned in the provided text.
  • If the external forces acting on a system have zero resultant, the center of mass
    • may move
    • may accelerate.
  • A nonzero external force acts on a system of particles. The velocity and the acceleration of the center of mass are found to be v and a at an instant t. It is possible that
    • v = 0, a ≠ 0
    • v ≠ 0, a = 0
    • v ≠ 0, a ≠ 0.

Elastic Collision

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An elastic collision is one where the total kinetic energy and linear momentum of the system are conserved (remain constant).
Conservation of Kinetic Energy and Linear Momentum
  • In an elastic collision
    • the kinetic energy remains constant
    • the linear momentum remains constant
    • the final kinetic energy is equal to the initial kinetic energy
    • the final linear momentum is equal to the initial linear momentum.
  • In a head-on elastic collision of two bodies of equal masses
    • the velocities are interchanged
    • the speeds are interchanged
    • the momenta are interchanged
    • the faster body slows down and the slower body speeds up.

Inelastic Collision

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An inelastic collision is one where kinetic energy is not conserved. Momentum is still conserved if there are no external forces.
Conservation of Momentum
  • A ball hits a floor and rebounds after an inelastic collision. In this case
    • the total momentum of the ball and the earth is conserved
    • the total energy of the ball and the earth remains the same.

Linear Momentum

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Linear momentum is a measure of mass in motion. It is conserved in a closed system.
P = mv (where P is momentum, m is mass, and v is velocity)
  • For a system of particles where a block moving in air breaks in two parts and the parts separate, the total momentum must be conserved.

Linear Momentum

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Linear momentum is the product of an object's mass and velocity. It's a measure of how difficult it is to stop a moving object. The total momentum of a closed system remains constant if no external forces act on it (Law of Conservation of Linear Momentum).
p = mv (momentum = mass x velocity)
The problems deal with concepts of momentum conservation, recoil, and impulse.

Conservation of Momentum

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In a closed system (no external forces), the total momentum before an event (like a collision or explosion) is equal to the total momentum after the event.
m1v1 + m2v2 = m1v1' + m2v2' (for a two-object system)
The law of conservation of linear momentum states that if there is no net external force on a system, the momentum of the system remains constant.

Recoil

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When a gun fires a bullet, the gun moves backward. This backward motion is called recoil and is due to conservation of momentum.
0 = m1v1 + m2v2 (where one velocity will be negative, indicating opposite direction)
Recoil velocity occurs due to the conservation of momentum. When a projectile is launched, the launching object recoils in the opposite direction to conserve the total momentum of the system.

Impulse

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Impulse is the change in momentum of an object. It's also equal to the force applied to the object multiplied by the time the force acts.
Impulse = Δp = FΔt
Impulse is defined as the change in momentum of an object. It is also equal to the average force applied on the object multiplied by the time interval over which it acts.

Centre of Mass

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⚡ Quick Summary
Deals with the concepts of center of mass, linear momentum, and collisions.
Formulas related to center of mass calculation, linear momentum (p=mv), impulse (J = Δp), and coefficients of restitution for collisions are expected.
This section likely covers the definition of the center of mass, how to calculate it for systems of particles and continuous bodies, the concept of linear momentum and its conservation, different types of collisions (elastic, inelastic, perfectly inelastic), and the impulse-momentum theorem.

Linear Momentum

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Quantity of motion of a moving body, equal to the product of its mass and velocity.
p = mv, where p is linear momentum, m is mass, and v is velocity.
Linear Momentum is a vector quantity defined as the product of an object's mass and its velocity. The principle of conservation of linear momentum states that in a closed system (no external forces), the total momentum remains constant.

Collision

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An event in which two or more bodies exert forces on each other for a relatively short time.
Coefficient of restitution (e) = (relative velocity of separation) / (relative velocity of approach). For elastic collisions, e = 1; for inelastic collisions, 0 < e < 1; for perfectly inelastic collisions, e = 0.
Collisions can be elastic (kinetic energy is conserved), inelastic (kinetic energy is not conserved), or perfectly inelastic (objects stick together after the collision). The coefficient of restitution (e) quantifies the 'bounciness' of a collision.

Impulse and Momentum

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Deals with the concepts related to centre of mass, linear momentum, and collisions. It involves understanding how momentum changes during collisions and how external forces affect the motion of the center of mass of a system.
['No specific formulas are explicitly mentioned in the provided text excerpt. The formulas related to this chapter includes:\n', 'P = mv (Linear Momentum)', 'Impulse = Change in Momentum', 'Conservation of Linear Momentum']
This section focuses on understanding the principles governing the behavior of systems of particles, particularly during collisions. Key concepts include: * **Centre of Mass:** The point representing the average position of all the mass in a system. * **Linear Momentum:** A measure of mass in motion, given by the product of mass and velocity. * **Collisions:** Interactions between objects that result in an exchange of momentum and energy. Collisions can be elastic (kinetic energy conserved) or inelastic (kinetic energy not conserved).