Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fRead more
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fatigue is very important in engineering and in material selection, especially when it comes to components in bridges, vehicles, and machinery that are subject to fluctuating loads.
Poisson's ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as folloRead more
Poisson’s ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as follows:
u = – (transverse strain)/(axial strain) = – (Δd/d)/(ΔL/L)
Where:
– Δd is the change in diameter (transverse deformation),
– d is the original diameter,
– ΔL is the change in length (axial deformation),
– L is the original length.
Important Points
Poisson’s ratio is a dimensionless quantity.
It generally lies between 0 and 0.5 for most materials with values near to 0.5 showing that the material is almost incompressible.
A value of 0 means that there is no transverse deformation when the material is stretched or compressed.
To find the isothermal bulk modulus K of an ideal gas we can use the formula: K = - V (∂P/∂V)_T For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle's Law: PV = nRT Differentiating this equation while keeping the temperature constant giRead more
To find the isothermal bulk modulus K of an ideal gas we can use the formula:
K = – V (∂P/∂V)_T
For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle’s Law:
PV = nRT
Differentiating this equation while keeping the temperature constant gives us:
∂P/∂V = -nRT/V²
Thus the isothermal bulk modulus becomes:
K = -V (-nRT/V²) = nRT/V
Since nRT = PV we can substitute that in the equation too;
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young's modulus: The stress in the wire, σ, can be found by the relation as follows: σ = F/A where F represents the appliedRead more
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young’s modulus:
The stress in the wire, σ, can be found by the relation as follows:
σ = F/A
where F represents the applied force. The strain ε is described by:
ε = x/L
As related by Young’s modulus Y,
Y = σ/ε = (F/A)/(x/L)
From this we can write the force F:
F = (YAx)/L
The work done W when the wire is stretched by an amount x is given by the area under the stress-strain curve which is the integral of force over displacement:
W = ∫ F dx = ∫ (YAx/L) dx
Evaluating the integral we get:
W = (Y A/L) ∫ x dx = (Y A/L) * [x²/2] from 0 to x = (Y A/L) * (x²/2)
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution. Data given are as follows: - Mass of ball 1 is m₁ = m Velocity of ball 1 is u₁ = 2 m/s Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)Read more
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution.
Data given are as follows:
– Mass of ball 1 is m₁ = m
Velocity of ball 1 is u₁ = 2 m/s
Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)
Velocity of ball 2 is u₂ = 0 m/s (ball is at rest)
Coefficient of restitution is e = 0.5
Step 1: Apply the principle of conservation of momentum
The total momentum before the collision is equal to the total momentum after the collision:
m₁ * u₁ + m₂ * u₂ = m₁ * v₁ + m₂ * v₂
Substitute the given values:
m * 2 + 2m * 0 = m * v₁ + 2m * v₂
2m = m * v₁ + 2m * v₂
Divide both sides by m:
2 = v₁ + 2v₂ (Equation 1)
Step 2: Utilization of the coefficient of restitution
Coefficient of restitution is defined by the following equation:
e = (relative speed after collision) / (relative speed before collision)
In this case, it can be derived as
e = (v₂ – v₁) / (u₁ – u₂)
Using the values:
0.5 = (v₂ – v₁) / (2 – 0)
0.5 = (v₂ – v₁) / 2
Multiply both sides by 2
1 = v₂ – v₁ Equation 2
Step 3: Solution of equations
We have two equations
1. v₁ + 2v₂ = 2
2. v₂ – v₁ = 1
From the last equation we may write
v₂ = v₁ + 1
This must be put in the first:
v₁ + 2v₁ + 2 = 2
3v₁ + 2 = 2
3v₁ = 0
v₁ = 0
Inserting the latter into the last of our initial equations:
v₂ = 0 + 1
v₂ = 1
Final velocities after collision:
Velocity of ball 1 (v₁) = 0 m/s
Velocity of ball 2 (v₂) = 1 m/s
Final Answer:
The final velocities after the collision will be 0 m/s, 1 m/s.
What is elastic fatigue? What is its importance?
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fRead more
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fatigue is very important in engineering and in material selection, especially when it comes to components in bridges, vehicles, and machinery that are subject to fluctuating loads.
Show more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
Define Poisson’s ratio.
Poisson's ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as folloRead more
Poisson’s ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as follows:
u = – (transverse strain)/(axial strain) = – (Δd/d)/(ΔL/L)
Where:
– Δd is the change in diameter (transverse deformation),
– d is the original diameter,
– ΔL is the change in length (axial deformation),
– L is the original length.
Important Points
Poisson’s ratio is a dimensionless quantity.
It generally lies between 0 and 0.5 for most materials with values near to 0.5 showing that the material is almost incompressible.
A value of 0 means that there is no transverse deformation when the material is stretched or compressed.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is
To find the isothermal bulk modulus K of an ideal gas we can use the formula: K = - V (∂P/∂V)_T For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle's Law: PV = nRT Differentiating this equation while keeping the temperature constant giRead more
To find the isothermal bulk modulus K of an ideal gas we can use the formula:
K = – V (∂P/∂V)_T
For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle’s Law:
PV = nRT
Differentiating this equation while keeping the temperature constant gives us:
∂P/∂V = -nRT/V²
Thus the isothermal bulk modulus becomes:
K = -V (-nRT/V²) = nRT/V
Since nRT = PV we can substitute that in the equation too;
K = PV/V = P
Thus the isothermal bulk modulus of the gas is:
K = P
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
A wire of length L, and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by an amount x, the work done is
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young's modulus: The stress in the wire, σ, can be found by the relation as follows: σ = F/A where F represents the appliedRead more
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young’s modulus:
The stress in the wire, σ, can be found by the relation as follows:
σ = F/A
where F represents the applied force. The strain ε is described by:
ε = x/L
As related by Young’s modulus Y,
Y = σ/ε = (F/A)/(x/L)
From this we can write the force F:
F = (YAx)/L
The work done W when the wire is stretched by an amount x is given by the area under the stress-strain curve which is the integral of force over displacement:
W = ∫ F dx = ∫ (YAx/L) dx
Evaluating the integral we get:
W = (Y A/L) ∫ x dx = (Y A/L) * [x²/2] from 0 to x = (Y A/L) * (x²/2)
Work done is hence,
W = (Y A x²)/(2 L)
Click for more solutions:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution. Data given are as follows: - Mass of ball 1 is m₁ = m Velocity of ball 1 is u₁ = 2 m/s Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)Read more
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution.
Data given are as follows:
– Mass of ball 1 is m₁ = m
Velocity of ball 1 is u₁ = 2 m/s
Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)
Velocity of ball 2 is u₂ = 0 m/s (ball is at rest)
Coefficient of restitution is e = 0.5
Step 1: Apply the principle of conservation of momentum
The total momentum before the collision is equal to the total momentum after the collision:
m₁ * u₁ + m₂ * u₂ = m₁ * v₁ + m₂ * v₂
Substitute the given values:
m * 2 + 2m * 0 = m * v₁ + 2m * v₂
2m = m * v₁ + 2m * v₂
Divide both sides by m:
2 = v₁ + 2v₂ (Equation 1)
Step 2: Utilization of the coefficient of restitution
Coefficient of restitution is defined by the following equation:
e = (relative speed after collision) / (relative speed before collision)
In this case, it can be derived as
e = (v₂ – v₁) / (u₁ – u₂)
Using the values:
0.5 = (v₂ – v₁) / (2 – 0)
0.5 = (v₂ – v₁) / 2
Multiply both sides by 2
1 = v₂ – v₁ Equation 2
Step 3: Solution of equations
We have two equations
1. v₁ + 2v₂ = 2
2. v₂ – v₁ = 1
From the last equation we may write
v₂ = v₁ + 1
This must be put in the first:
v₁ + 2v₁ + 2 = 2
3v₁ + 2 = 2
3v₁ = 0
v₁ = 0
Inserting the latter into the last of our initial equations:
v₂ = 0 + 1
v₂ = 1
Final velocities after collision:
Velocity of ball 1 (v₁) = 0 m/s
Velocity of ball 2 (v₂) = 1 m/s
Final Answer:
The final velocities after the collision will be 0 m/s, 1 m/s.
Click this for more solutions:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/