3. Collision

Vertical circle

Let an object of mass ‘m’ move in a circle that lines in the vertical plane. This is not a uniform circular motion as the velocity of the object in the vertical circle is not constant.

Let T₁ be the tension in the string at the lowest point of the vertical circle and T₂ be the tension in the string at the highest point of the vertical circle

There is a requirement of centripetal force mv2/r  to make the object move in a circular path.

Let v1 and v2 be the velocities of the object at the lowest and highest points of the vertical circle.

At lowest point    mg-T1= -mv1^2/r

T1= mg +m v1^2/r

At highest point  -T2-mg=-mv2^2/r

T2= m v2^2/r - mg

• Work done by the Tension force is zero and tension is perpendicular to the displacement direction always.
• We can use the conservation of energy in the case of a vertical circle.

Important results for Vertical circle

•  The difference in the tension in the string at the lowest and highest point is 6mg. It is independent of the velocity.   T1-T2=6mg
• The difference in the square of the velocities at the lowest and highest points is 4gr.    v1^2-v2^2= 4gr
• The minimum velocity at the highest point is √rg for the object to move in a complete vertical circle.
• The minimum velocity at the lowest point is √5rg for the object to move in a complete vertical circle.

Collision

Collision, also called impact, in physics, is the sudden, forceful coming together in direct contact of two bodies, such as, for example, two billiard balls, a golf club and a ball, a hammer and a nail head, two railroad cars when being coupled together, or a falling object and a floor.

• In all of the examples of colliding bodies here referred to, the time of contact is extremely short and the force of contact extremely large.
• there is an instantaneous change in the velocity of a body but no change in its position during the period of contact.
• Forces of this nature are known as impulsive forces and, being difficult to measure or estimate, their effects are measured by the change in the momentum  (mass times velocity) of the body.

Two types of collision:

Elastic collision: Collision between two bodies is said to be elastic when there is no energy loss during the collision

Inelastic collision: Collision between two bodies is said to be inelastic when there is some loss of kinetic energy during the collision. In the case of perfectly inelastic collisions bodies stick together after collision and move with a common velocity.

• Conservation of momentum holds for both case elastic collision as well as inelastic collision.
• Conservation of energy holds only in the case of elastic collision.

Coefficient of restitution   e= (velocity of separation )/(velocity of approach )= (v2-v1)/(u2-u1)

For perfectly elastic collision e=1 and for perfectly inelastic collision  e=0

Elastic collision in one Dimension.

Consider two bodies of masses m1 and m2 moving with initial velocities u1 and u2 along the same direction. After the collision, their final velocities are v1 and v2 respectively.

Since it is an elastic collision so there will be no energy loss. So we can apply conservation of energy and momentum both

By conservation of momentum we have Pi= Pf

m1 u1+m2 u2= m1 v1+ m2 v2

Rearranging we have  m1 (u1-v1)=m2 (v2-u2) .. (1)

By conservation of energy we have  K.Ei= K.Ef

1/2 m1 u1^(2 )+1/2 m2 u2^(2 )= 1/2 m1 v1^2+1/2 m2 v2^2

So we have  m1( u1^2-v1^2)=m2 ( v2^2-u2^2)   ..(2)

Divide equation 2 by 1 we will get 

u1+v1= v2+u2

rearranging we get ,   u1-u2= v2-v1

Thus v1=v2+u2-u1    , v2=v1+u1-u2    ,

When we put the value of v1 and v2 separately in the equation of momentum conservation we get the values of final velocities of the object  v1 and v2  in terms of their masses m1 and m2 and initial velocities  u1  and u2.

v1= ((m1-m2))/((m1+m2)) u1 +  2m2/((m1+m2)) u2

v2= 2m1/((m1+m2)) u1 +  ((m1-m2))/((m1+m2)) u2

Special case 1 :  when the target is at rest  u2=0

v1= ((m1-m2))/((m1+m2)) u1

v2= 2m1/((m1+m2)) u1

Special case 2: when the masses of the objects are equal and both objects are moving.

v1=u2  , v2= u1

When the masses of the objects are the same, the velocities of objects get interchanged after the collision.

Special case 3: when the target is at rest and the masses of both the objects are equal.

v1=0 , v2=u1

When the target is at rest and the masses of objects are the same, the first object becomes stationary and the second object starts moving with the initial velocity of the first object.

Elastic Collision in two dimensions.

Suppose we have two objects, one is at rest and the other is moving with initial velocity v1i, after collision object m1 moves with final velocity v1f making an angle θ with the horizontal direction and mass m2 moves with final velocity v2f making an angle Φ with the horizontal direction.

• First, we will resolve the components of v1f and v2f along a horizontal and vertical direction which is the x and y direction respectively. 

(v1f)x= v1f cosθ ,  (v1f) y= v1f sinθ.

(v2f)x= v2f cosΦ  , (v2f)y= v2f sinΦ

• Then we will apply conservation of momentum separately along x and y-direction.px i= px f  ;  py i= py f.

Putting the values we get

• From the conservation of energy, we have K.E i= K.Ff

1/2 m1 v1i^(2 )+0 =1/2 m1 v1f^2  +  1/2 m2 v2f^2

Analyzing the above equations reveals that finding values for four unknown quantities v₁, v₂, θ₁, and θ₂ using the above three equations is not possible. As a result, it cannot predict the variable because there are four of them. However, if we measure any one variable, we can uniquely determine the other variable using the above equation.

Inelastic collision

Suppose we have two objects of mass m1 and m2  moving with initial speeds v1i and v2i respectively. After collision, they stick together and start moving together with final velocity vf.  This is the case of a perfectly inelastic collision.

We cannot use conservation of energy in inelastic collisions. But we can use conservation of momentum to find the final velocity vf

Initial momentum = final momentum

 m1 v1i + m2 v2i =(m1+m2) vf

vf= (m1 v1i+m2 v2i) /(m1+m2)

Loss of kinetic energy in case of inelastic collision