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Center of Mass : Introduction 0/4

Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4


Momentum Conservation & Collision 0/4

Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Collision Process
Collision process always starts with different velocity and due to which the colliding object deformed and after some time they acquire common velocity along the line of impact. Thereafter due to reformation of the two objects separate with different velocities.
Depending on reformation collision is of three types.
 Complete reformation also known as elastic collision
 Partial reformation also known as inelastic collision
 No reformation also known as perfectly inelastic collision
Elastic collision
Inelastic collision internal forces are conservative and there is no loss of energy in the formation reformation process.
Impulse of deformation
It is the impulse of impact force during the time object acquires common velocity along the line of impact.
Impulse is always equal and opposite.
Impulse of reformation
It is the impulse of impact force during the time object acquires different velocity from common velocity and separate.
Newton’s experimental law
The ratio of impulse of reformation impulse of reformation remains constant for the two colliding objects. It depends on the pair of colliding objects.
e = 1 for perfectly elastic body, that means which reforms completely.
e = 0 for perfectly plastic body, (inelastic) that means no reformation.
This ratio is also defined in terms of velocities.
If elasticity (e) = 1, velocity of separation is equal to velocity of approach.
If elasticity (e) = 0, impulse of reformation is zero. Which means velocity of separation will be zero.
If 0 < e < 1, , velocity of separation < velocity of approach
Q15.
 Find out the common velocity during the collision.
 Find out impulse of deformation.
 Find out impulse of reformation.
 Find out impulse of impact.
 Find out the maximum energy stored in the deformation.
 Final velocity of the two objects.
 Loss in mechanical energy.
Ans:
Momentum of the colliding objects will be conserved if external force on the system is zero.
Impulse of deformation,
Impulse of reformation,
Impulse of impact,
Maximum energy stored in deformation,
Final velocities of objects,
Apply momentum conservation,
Energy loss in collision,
When one object is much heavier than the other
When one of the colliding object is much much heavier than the other object than the velocity of heavier object can be considered unchanged irrespective of type of collision.
Q16.
Find out the maximum height raised by the smaller ball. Assume that all collisions are elastic.
Ans:
In this case there are two collisions are taking place. First collision will be between bigger ball and the ground; second collision will be between bigger ball and is smaller ball. As we know that all collisions are elastic, velocity of separation will be equals to velocity of approach in both collisions.
First collision,
Velocity of ground can be taken is constant before and after collision is a mass of ground is much heavier than that of bigger ball. (Velocity of bigger ball just before hitting the ground = .
Second collision between big ball and smaller ball,
In case of second collision mass of bigger ball is much heavier than a smaller ball; hence we can assume that velocity of bigger ball remains unchanged before and after second collision.
Velocity of approach = velocity of separation i.e.
Now we can apply Newton’s third equation of motion to find out maximum height of smaller ball with given initial velocity.
Q17. Oblique collision.
Assume elastic collision. Two discs collides obliquely, above diagram shows the top view of two discs. Find out the common velocity along the line of impact, impulse of deformation and final velocity of both objects.
Ans:
Let’s draw collision diagram,
One thing to note is here that velocity remains unchanged in the perpendicular direction of line of impact as there is no impulse force is acting in this direction.
Let’s apply momentum conservation along the line of impact,
Apply condition of elasticity (e=1),
Velocity of approach = velocity of separation
Solving both equation,
v_2 = \frac{2u \cos \theta}{3} \newline v_1 = \frac{u \cos \theta}{3} latex Mu \cos \theta = (3M) v \newline v = \frac{u \cos \theta}{3} latex I_D = m_b v – m_b \cdot 0 \newline = (2m)\frac{u \cos \theta}{3} $
Whenever there is a sudden change in velocity of an object and elasticity (e) is not equal 1, there will be lost in energy.