**COLLISION OF ELASTIC BODIES**

We can classify bodies into two types

- An elastic body is one that rebound after impact.
- An inelastic Body is one that does not rebound after Impact.

**COLLISION **

The bodies after collision come momentarily to rest. The two bodies tend to compress each other, so long as they are compressed to the maximum value. The time taken is called the time of compression. The two bodies attempt to regain their original shape due to their elasticity. This process is known as restitution and the time taken during restitution is called the time of restitution. The line of impact is the common normal to both bodies. The plane of impact is the common tangent plane of bodies.

**Time of collision = time of compression + time of restitution**

**A direct collision** is a collision in which the two bodies, before impact, are moving along the line of impact. **Indirect collision** if the two bodies, before impact are not moving along the line of impact.

**DIRECT COLLISION**

**LAW OF CONSERVATION OF MOMENTUM:**

“The total momentum of two bodies remains constant after their collision.”

m_{1} = mass of the first body

u_{1} = Initial velocity of the first body

v_{1} = Final velocity of the first body

m_{2}, u_{2}, v_{2} = corresponding values for the second body.

m_{1}u_{1}+m_{2} u_{2} = m_{1}v_{1}+ m_{2 }v_{2}

_{}

**Fig:1**

**NEWTON’S LAW OF COLLISION OF ELASTIC BODIES:**

“When two bodies collide with each other, their velocity of separation bears a constant ratio to their velocity of approach.”

m_{1} = mass of the first body

u_{1} = velocity of the first body before impact

v_{1} =velocity of the first body after impact

m_{2}, u_{2}, v_{2} = Corresponding values for the second body.

Let us assume that u_{1} > u_{2} and v_{2} > v_{1}

v_{2} – v_{1} = e(u_{1} – u_{2}), Where e = coefficient of restitution

- e varies between 0 and 1.
- Perfectly elastic collision: When e = 1, the velocity of approach = velocity of separation
- Inelastic collision: when e = 0, v
_{2}= v_{1}

**LOSS OF KINETIC ENERGY:**

- Only in Inelastic collision.
- Difference between kinetic Energies of the system before and after the collision.

**DIRECT IMPACT OF A BODY WITH A FIXED PLANE:**

Let u = Initial velocity of first body

v = Final velocity of the first body

The fixed plane will not move after impact. Hence, Acc. to Newton’s lass of collision v = eu.

We do not apply the principle of momentum as the plane has infinite mass.

**INDIRECT COLLISION**

**INDIRECT IMPACT OF TWO BODIES:**

m_{1} = mass of the first body

u_{1} = Initial velocity of the first body

v_{1} = Final velocity of the first body

a_{1} = Angle which is the initial velocity of the first body makes with the line of impact

q_{1} = Angle which is the initial velocity of the first body makes with the line of impact

m_{2}, u_{2}, v_{2}, a_{2}, q_{2} = Corresponding values for the second body.

**Fig:2**

According to the Law of Conservation of Momentum along the line of impact,

m_{1}u_{1} Cos a_{1 }+ m_{2} u_{2}Cos a_{2} = m_{1}v_{1 }Cos q_{1 }+ m_{2}v_{2} Cos q_{2}

According to Newton’s Law of elastic collision along the line of impact,

v_{2} cos q_{2} – v_{1} cos q_{1} = e(u_{1} cos a-u_{2} cos a_{2})

**INDIRECT IMPACT OF A BODY WITH A FIXED PLAN:**

Consider a body having an indirect impact on a fixed plane as shown in Fig

Let u = Initial velocity of the body,

v = Final velocity of the body,

α = Angle, which the initial velocity of the body makes with the line of impact,

θ = Angle which the final velocity of the body makes with the line of impact, and

e = Coefficient of restitution.

A little consideration will show, that the component of u, along the line of impact, will cause the direct ‘impact’ of the body with the fixed plane. The other component of u (i.e. along the perpendicular to the line of impact) will not affect the phenomenon of impact and will be equal to the other component of v (i.e., along the perpendicular to the line of impact).

We know that velocity of approach = u cos α

and velocity of separation = v cos θ

The Newton’s Law of Collision of Elastic Bodies also holds good for this impact i.e.,

v cos θ = eu cos α

**Notes:**

- In this impact also, we do not apply the principle of momentum (i.e. equating the initial momentum and the final momentum) since the fixed plane has infinite mass.
- The components of initial and final velocities at right angles to the line of impact are same u sin α = v sin θ

**Fig:3**

**INTRODUCTION TO VIRTUAL WORK**

The virtual work is the work done by a force (or a couple) on a body due to a small virtual (i.e. imaginary) displacement of the body. A virtual displacement is any arbitrary very small change in the position of a body. The displacement is only imagined, as the body is not actually

displaced. for which the displacement is called virtual displacement. The product of the force and the virtual displacement in the direction of the force is called virtual work.

**VIRTUAL WORK:**

When a system or a body is in equilibrium under the action of a number of forces, there is no displacement and accordingly, no work is done. However, if the system is imagined to undergo any arbitrary but small displacement, some work can be imagined to have been done.

The imaginary small-displacement given to the system is called virtual displacement. The product of the net resultant force and the visual displacement in the direction of the force is pro virtual work." The concept of virtual work is used in solving problems related to static equilibrium.

**PRINCIPLE OF VIRTUAL WORK**

Consider a particle at A, subjected to several forces F_{1}, F_{2}, F_{3}, ..., F_{n} as shown in Fig. We assume that the particle undergoes a small virtual displacement δr from A to A'. The work of each of the forces F_{1}, F_{2}, F_{3} F_{n} during the virtual displacement δr is called virtual work. The virtual work of all the forces acting on the particle is

**Fig.4:Principle of virtual work**

dU = F_{1} .δr + F_{2} .δr + F_{3} .δr + ….. + Fn.δr

= (F_{1}+ F_{2} + F_{3} + ….. + Fn).δr

dU = (ΣF).δr, .......(i)

where ΣF is the resultant of the given forces.

If we express ΣF and δr in terms of their rectangular components, we have

dU= (iΣFx+ jΣFy + kΣFz).(iδx + jδy + kδz)

dU = ΣFx δx + ΣFy δy +ΣFz z.

The principle of virtual work for a particle states that "if a particle is in an equilibrium, the total virtual work of the forces acting on the particle is zero for any virtual displacement of the particle". This condition is necessary; if the particle is in equilibrium, the resultant ΣF of the forces is zero, and it follows from Equation (1) that the total virtual work dU is zero. The condition is also sufficient: if the total virtual work dU is zero for any virtual displacement, the scalar product (ΣF) . δr is zero for any δr, and the resultant ΣF must be zero. The principle of virtual work for a rigid body states that "if a rigid body is in equilibrium the total virtual work of the external forces acting on the rigid body is zero for any virtual displacement of the body.

**SIGN CONVENTIONS:**

There are different sign conventions for finding out the virtual work done in different

books; we shall the following sign conventions.

• Forces acting towards the right are considered as positive, whereas those towards left as

negative

• Upward forces are considered as positive, whereas the downward forces as negative.

• Moments or couples acting in the clockwise direction are considered as positive, whereas

the anticlockwise couples as negative.

• Tensile forces are considered as positive whereas the compressive forces as negative.

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## Comments

write a commentMonisasishOct 31, 2017

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Ranu SachanJul 31, 2020

Wq=1.δ

And

By energy conservation (WQ = UQ), we have the expression to find truss deformation due to applied loads using Virtual Work:

Why

Load is always unit

And U is always unit load strain energy