Collision
Collision: The term collision or impact represents an event during which two bodies are approaching each other, a force comes into play between them for a finite time and brings about a measurable change in their velocities, momenta and energy according to the respective laws of conservation.
During collisions, two particles may or may not touch physically. For example, when an alpha particle collides with the nucleus of gold, due to the repulsive force between them, it may not touch the nucleus.
Let us consider, the colliding objects as spheres with elasticity properties. During contact the time is composed of a compression time during which slide deformation may take place and restitution time during which the shape is restored.
A collision can be direct or oblique. In a direct collision, the direction of motion of both sphere is along the common normal [the line passing through the centres of two colliding bodies] at the point of contact both before and after collision.
A collision which is not direct is called oblique collision.
Newton’s collision rule: If U12 and V12 be the relative velocities of the spheres along the common normal before and after impact respectively, then V12 = – eU12
The quantity e is called the coefficient of restitution = 
.
It is a constant value lying between 0 and 1 depending on the materials of which the objects are made.
For perfectly elastic collision V12 = U12. So, e = 1
For perfectly inelastic collision V12 = 0. So, e = 0
For partly elastic collision V12 < U12. So, e < 1.
Collisions of particles: Two particles of mass m1,m2 travelling in a straight line collide each other.Calculate the velocities of the particles after collision.
Let us consider, two particles are moving along a straight-line(x axis) and the velocities of the particles before and after collision are u1, u2 and v1, v2 respectively.
By Newton’s collisions rule, v12 = – eu12 [u12 = relative velocity of m1 with respect to m2 before collision and v12 = relative velocity of m1 with respect to m2 after collision].
Or, v1 – v2 = e(u2 – u1) ——- (I)
By the principle of conservation of linear momentum, m1u1 + m2u2 = m1v1 + m2v2 ——- (II)
Solving equation (i) and (ii) we get, v1 = 
—— (III)
And v2 = 
——- (IV)
For perfectly elastic collision we put e = 1in equation (III) and (IV) to obtain v1 = 
——(V) and v2 = 
——— (VI).
For perfectly inelastic collision we put e = 0 in equation (III) and (IV) to obtain v1 = v2 = 
—— (VII).
Perfectly elastic collision: In perfectly elastic collision the relative velocity between two bodies before and after collision remain same and the linear momentum and kinetic energy of the bodies remain conserved.
Properties:1. In perfectly elastic collision two bodies of same mass interchange their velocities: Let us consider two bodies of same mass m are moving along the straight line with same direction with velocities u1 and u2 respectively. The relative velocity between the bodies before collision is (u1 – u2). After collision two bodies are moving with velocities v1 and v2 respectively. The relative velocity between the bodies after collision is (v2 – v1).
Using conservation of linear momentum between the bodies we get, mu1 + mu2 = mv1 + mv2
Or, u1 + u2 = v1 + v2 ——-(VIII)
From the condition of perfectly elastic collision we get, u1 – u2 = v2 – v1 —–(IX)
From equations (VIII) and (IX) we get, v1 = u2 and v2 = u1.
- If m1>> m2 and u2 = 0 ,then,
≈ 0 and from equation (V) v1 = 2u2 +
u1= u1 + 0 =u1.From equation (VI) v2 = 2 u1 + 0 = 2u1.
- If m2>> m1 and u2 = 0 then,
≈ 0 and from equation (V) v1 = – u1.From equation (VI) v2 = 0 .
Kinetic energy is conserved in perfectly elastic collision: Let us consider two bodies of masses m1 and m2 are moving along the straight line with same direction with velocities u1 and u2 respectively. The relative velocity between the bodies before collision is (u1 – u2). After collision two bodies are moving with velocities v1 and v2 respectively. The relative velocity between the bodies after collision is (v2 – v1).
Using conservation of linear momentum between the bodies we get, m1u1 + m2u2 = m1v1 + m2v2 ——-(I)
From the condition of perfectly elastic collision we get, u1 – u2 = v2 – v1 —–(II)
The kinetic energy of two bodies before collision is Ei = m1u12 + 
m2u22 —–(III)
Or, Ei =
Or, Ei = 
Or, Ei = 
Or, Ei = 
——-(IV)
From equation (I) and (II) we get, Ei = 
——-(V)
As equation (IV) is obtained from equation (III), then comparing equation (III), (IV) and (V) we get, Ei = m1v12 + m2v22 = Ef = the kinetic energy of two bodies after collision.
So, kinetic energy is conserved in perfectly elastic collision.
Perfectly inelastic collision: In perfectly inelastic collision, there is a relative velocity between the bodies before collision but the relative velocity between the bodies after collision is zero. So, after collision two bodies stick each other.
Let us consider two bodies of masses m1 and m2 are moving along the straight line with same direction with velocities u1 and u2respectively. After collision two bodies are moving with velocity v.
Using conservation of linear momentum between the bodies we get, m1u1 + m2u2 = (m1 + m2)v
v =
.
Loss of kinetic energy in perfectly inelastic collision: Let us consider two bodies of masses m1 and m2 are moving along the straight line with same direction with velocities u1 and u2 respectively. After collision two bodies are moving with velocity v.
Using conservation of linear momentum between the bodies we get, m1u1 + m2u2 = (m1 + m2)v
Or, v ——(I)
The kinetic energy of the bodies before collision is Ei = [m1u12+ m2u22]
The kinetic energy of the bodies after collision is Ef = [m1 + m2]v2
The lost kinetic energy of the bodies in perfectly inelastic collision is E = Ei – Ef = [m1u12+ m2u22] –[m1 + m2]v2
Or, E = [(m1u12+ m2u22) – (m1 + m2)
[using equation (I)]
Or, E = [(m1u12+ m2u22) – 
]
Or, E =
Or, E = 
Or, E = .
Example: 1. Block A of mass m is connected with the spring of force constant K is rest on a frictionless surface.Another identical block B moves with velocity v0 on that surface and collides with the spring.Calculate compression of the string during collision.
As the surface is frictionless, therefore the nature of force is conservative. Hence, we can use conservation of mechanical energy.
Before collision,total energy of the system = kinetic energy of block B= 
.
After collision, the total energy of the system = kinetic energy of block A and B+ potential energy of spring due to corporation.
The potential energy of spring due to x compression of spring is
.
As there is no frictional force and spring is considered as massless, hence during collision two blocks will move together with speed v.
Therefore, = 
+ ——- (I)
Using conservation of linear momentum we get, mv0 = 2mv
Or, v = 
——- (II)
From equation (I) and (II) we get, = 
+
Or, =
x = 
.
- A body of mass 3 kg collides elastically with another body at rest and then continues to move in the original direction with one-half of its original speed. What is the mass of the struck body?
Let us consider, m and M are the masses of the two bodies and u is the initial velocity of m.Let v and w be the velocities of m and M after collision respectively.
By the conservation of momentum we have, mu = mv + Mw
Or, 3u = 3(
) + Mw [ v = ]
Or,
= Mw
Or, w = 
——- (I)
By the conservation of mechanical energy we have, 
= 
+ 
Or, mu2 = 
+ Mw2
Or,
= Mw2
Or, = 
M = 
= 1 kg.
Partly elastic collision: In partly elastic collision the relative velocity between two bodies after collision is less than the relative velocity between two bodies before collision. Some amount of mechanical energy is converted into potential energy to maintain the shape of the deformed body after collision.
Oblique collision:The collision between two bodies is considered as oblique collision if just before collision at least one of the colliding objects was moving in the directiondifferent from the line of common normal. At the point of contact between two bodies the perpendicular direction with respect to common normal direction is called as common tangent direction
Properties along common normal and common tangent direction:
- A pair of equal and opposite impulse act on the colliding bodies during collision along common normal direction (XX/).
- No component of impulse acts along common tangent direction (YY/) during collision.If any linear momentum presents along common tangent direction, then it remains unchanged after collision.
- The net impulse on both the bodies is zero during collision.The conservation of linear momentum can be used along both the direction (XX/, YY/)
- Along common normal direction, the coefficient of restitution [relative speed of separation = e relative speed of approach] can be used.
Example: 1. A ball of mass mcollides elastically with another identical ball at rest (shown in figure). Show that if the collision is oblique, the bodies go at right angle to each other after collision.
Before collision the component of velocity of ball 1 alone normal direction vcos 
a nd along the tangent direction is vsin respectively.
As the collision is perfectly elastic and the masses are identical,therefore, the velocity is interchanged between two bodies along common normal direction.
Hence, after collision alongcommon normal direction, the velocity of body 1 and 2are respectively zero and vcos.
The velocity component along common tangent direction remains unchanged. Therefore, after collision along common tangent direction, velocity of body1 and 2 are respectively vsin and 0.
Hence after collision, body 1 moves along common tangent direction and body 2 moves along common normal direction. Therefore, two bodies after collision moves at right angle to each other.
- A ball of mass m hits the floor with a speed v making an angle with normal direction. The coefficient of restitution is e.Find velocity and angle of reflection of the ball after collision.
Before collision with the floor the velocity components of the ball alone common normal direction and common tangent direction are respectively vcos and vsin.The velocity component along common tangent direction of the ball remains unchanged after collision. The coefficient of restitution is applicable along common normal direction only.
Therefore, after collision the velocity of the ball along common normal direction is evcos.
Therefore, the resultant velocity of the ball after collision is v/ = 
. If φ is the angle between v/with normal direction then, tanφ = 
= 
.
- A small ball strikes a frictionless horizontal floor which speed u at an angle with normal to the surface. If the coefficient of restitution for the impact is e, then calculate the time of flight and range of the ball during second time collision with floor.
Here AB and CD are the common tangent and common normal axes respectively.
Before collision the velocity component of the ball alone AB and CD are respectively usin and ucos
After collision, v is the velocity of the ball creates angle Ø with AB. The component of v along AB and CD are respectively vx = vcos Ø= usin (no change of along common tangent direction) and vy = vsinØ= eucos (coefficient of restitution acts along common normal axis).
Time of flight of the ball is T =
= 
.
Range of the ball is R = 
= 
= 
.
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