If the oscillation is simple harmonic motion then we have,

Inserting given data, we get the spring constant approximately equal to 79N/m.

When they fly apart symmetrically relative to the initial motion direction with the angle of divergenceθ=60,
From the conservation of momentum, along the horizontal and vertical direction

and m1v1sin(θ/2)=m2v2sin(θ/2)

or, m1v1=m2v2 ...(2)

Now, from conservation of kinetic energy,

1/2 m1 v1²+ 0 =1/2 m1 v1²+1/2 m2 v2² ...(3)

From (1) and (2),

m1 u1=cos(θ/2)(m1v1+m1v1m2m2)=2m1v1cos(θ/2)

So, u1=2v1cos(θ/2)...(4)

From (2), (3) and (4)

4m1cos²(θ/2) v1²=m1v1²+m2m1²v1²/m2²

or, 4cos²(θ/2)=1+m1/m2

or,...

Breaking stress may occur when the material is excessively stretched or compressed. To withstand the heavy weight of an elephant, it has thicker legs so that the ratio of the weight to the cross-section area of the legs does not exceed the breaking stress of the bone material. To withstand the weight of a man, he can have thin legs and the ratio of his weight to the cross-section area of the leg will not exceed the breaking stress for the bone material. This is why an elephant has thicker...

Suppose two wires of steel and rubber of equal length L and cross-section A are stretched by the same force F. The Young modulus for steel is steel steel and similarly that for rubber iswhere l_steelis the extension of the steel wire and l_rubberis that of the rubber wire. Thus taking the ratio, we have



As l_rubber> l_steel, Y_steel> Y_rubber. The more the value of the elastic constant, the more the object is elastic because a very large force or stress is needed to make it break. This means...

Solution:
The Young modulus of a wire, Y = 12 1010 N/m2 ;
the diameter of the wire, d = 0.56 mm = 0.56 10^–3 mm
The cross-section area of the wire,

The strain = one-tenth of 1 percent

Solution:
The radius of a sphere, r = 10 cm = 0.10 m;
the increase in pressure, p = 5 10⁸N/m²;
the bulk modulus, B = 3.14 10¹¹N/m²;
the change in volume, V = ?

Solution:
The side of a cube, L = 4 cm = 0.04 m;
the shear length, x = 0.012 cm = 1.2 10^–4m

Solution:
The extension, l = ?;
the force applied, F = 30 N;
the length of a copper wire, L = 2 m;
the diameter of the wire, d = 3 mm = 3 10^–3m
The cross-section area of the wire,

Solution:
The length of a copper wire, LC = 1.5 m;
the length of a steel wire, LS = 1.5 m;
the diameter of the wires, d = 2 mm = 2 10^–3m;
the composite length, L = 3 m;
the extended composite length, L' = 3.003 m;
the total extension, l = L' – L = 3.003 – 3 = 0.003 m;
the extension of the copper wire, l_C= ?;
the extension of the steel wire, l_S= ?

l_C+ l_S= 0.003 … (1)

For the copper wire, Young modulus,

Solution:
The breaking stress = 7.982 10⁸N/m²;
the density of steel,ρ= 8.1 10³kg/m³
Suppose A is the area of cross-section of the steel wire and L is the longest length of the steel wire.
We have,
the weight of the wire = mg = (volume density)g
or F = LAρg

Solution:

Let m be the mass of the copper rod, g be the acceleration due to gravity and A be the cross-sectional area. It is given that L is the length of the rod then assume e to be the elongation produced. If Y be the Young's modulus of elasticity of copper, then

Y = FL/Ae

e = FL/AY

Further, F = mg and centre of mass of the rod is at the centre, so L' = L/2

Therefore, e = mgL/2AY

The strain produced in rubber is much larger compared to that in steel. This means that steel has a larger value of Young's modulus of elasticity and hence, steel has more elasticity than rubber.

Rubber is used as vibration absorbers, because rubber has a relatively high shear modulus compared to other materials. That means when a rubber material is shear stressed, i.e. stressed parallel to its cross-section, rubber can be stressed more before it becomes permanently deformed.

When a wire is bent back and forth, heat is generated due to the area of the elastic hysteresis and frictional force. Hence it becomes hot.

As the temperature increases, the interatomic forces of attraction become weaker. For a given stress, a large strain is produced when temperature increased.

Hence, the modulus of elasticity(stress/strain) decreases with increasing temperature.

So YES, temperature does affect the elasticity of the body, i.e body becomes more plastic in nature at higher temperature.

(a)
Young’s modulus is given by the equation,Y= stress/ strainAt a particular strain, Young’s modulus is directly proportional to the stress.
The slope of stress strain curve for material A has greater than the material B which gives higher value of Young's modulus.

(b)Material B is more brittle since the fracture point (end of solid curve) is at lower corresponding value of stress.

(c) Material A is more ductile since it has a greater plastic region; from elastic limit to breaking point.

(d)The...