Question

2a.(i) State Hooke's Law's
(ii) An experiment was carried out in the laboratory on an elastic material ,if the Fn was plotted against Extension ΔL,with suitable diagram show the following :(a) Yield Point (b) Elastic limit (c) Permanent deformation point (d) Fracture point
Hence ,highlights sections on the graph where's Hooke's law is obeyed and where there is no proportionality between stress and strain
(b) (i)From the extensive work of Leonardo da Vinci (1452-1519) and that of Charles coloumb (1736-1806),list the Laws of Friction as regard to forced between two surfaces.
(ii)A proton travelling at 10⁷m/s collides with stationery particles and bounces back at 2 x 10⁶m/s .if the particles moves forwards at 3 x 10⁸m/s ,find it's massive [Mass of proton =1.67 x 10^-27kg]

Answer 1

i.

Hooke's laws are:

1.The force needed to extend or compress a spring by some distance is proportional to that distance.

Mathematically, F=-kX where F is the force applied, k is the spring constant and X is the displacement of the spring.

2.Hooke's law states the strain on the material is directly proportional to the stress applied to the material

ii.

Till the proportional limit point ,Hooke's law is followed.Afterwards there is no proportionality.

iii.LAWS OF FRICTION:

1. When a body is moving, the friction is directly proportional to normal force and frictional force direction is perpendicular to the normal force.

2.Friction doesn't depend on area of contact till there is an area of contact.

3.The coefficient of static friction is higher than the coefficient of kinetic friction.

4.Kinetic friction is independent of the velocity of the body

5.Friction depends on the type of the surfaces in contact.

iv.

Here we have to conserve the momentum.

m1u1+m2u2=m1v1+m2v2

Here ,

m1 = mass of proton , m2 is the mass of unknown matter, u1,u2 are the initial velocity and v1,v2 are the final velocity of proton and the unknown mass respectively.

Putting the values we get,

[u2 =0 as given]

(1.67 x 10^-27kG)(10⁷m/s) + m2(0) =(1.67 x 10^-27kG)(-2 x 10⁶m/s) + m2(3 x 10⁸m/s)

Solving this , we get

m2 =6.68 x10^-29kG