Consider two elastic bodies of mases `m_(1) and m_(2)` moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure given below.
In order to have collision, we assume that the mass `m_(1)` move faster than mass `m_(2)` i.e, `u_(1) gt u_(2)`. For elastic collison, the total linear momentum and kinetic energy of the two bodies before and after collision must remain the same.
From the law of conservation of lienear momentum,
Total momentum before collision `(p_(i))`= total momentum after collision `(p_(f)0`
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)` ...(1)
or `m_(1)(u_(1)-v_(1))=m_(2)(v_(2)-u_(2))` ...(2)
Further,
Total kinetic energy before collisioon `KE_(i)`=Total kinetic energy after collision `KE_(f)`
`1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)` ....(3)
After simplifying and rearranging the terms.
`m_(1)(u_(1)^(2)-v_(1)^(2))=m_(2)(v_(2)^(2)-u_(2)^(2))`
Using the formula `a^(2)-b^(2)=(a+b)(a-b)` , we can rewrite the above equation as
`m_(1)(u_(1)+v_(1))(u_(1)-v_(1))=m_(2)(v_(2)+u_(2))(v_(2)-u_(2))` ...(4)
Dividing equation (4) by (2) gives,
`(m_(1)(u_(1)+v_(1))(u_(1)-v_(1)))/(m_(1)(u_(1)-v_(1)))=(m_(2)(v_(2)+u_(2))(v_(2)-u_(2)))/(m_(2)(v_(2)-u_(2))`
`u_(10+v_(1)=v_(2)+u_(2)`
`u_(1)-U_(2)=u_(2)=v_(2)-v_(1)`
Rearranging, (5)
Equation (5) can be rewritten as `u_(1)-u_(2)=-(v_(1)-v_(2))`
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
`v_(1)=v_(2)+u_(2)-u_(2)` ...(5)
or `v_(2)=u_(1)+v_(1)-u_(2)` ...(7)
To find the final velocities `v_(1) and v_(2)` :
Substituting equation (7) in equation (2) gives the velocity of `m_(1)` as
`m_(u_(1)-v_(1)=m_(2)(u_(1)+v_(1)-u_(2)-u_(2))`
`m_(1)(u_91)-v_(1))=m_(2)(u_(1)+v_(1)-2u_(2))`
`m_(1)u_(1)-m_(1)v_(1)=m_(2)u_(1)+m_(2)v_(1)+2m_(2)u_(2)`
`m_(1)u_(1)-m_(2)u_(1)+2m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(1)`
`(m_(1)-m_(2))u_(1)+2m_(2)u_(2)=(m_(1)+m_(2))v_(1)`
or `v_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)((2m_(2))/(m_(1)+m_(2)))u_(2)` ...(8)
Similarly by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of `m_(2)` as
`v_(2)=((2m_(1))/(m_(1)+m_(2)))u_(1)+((m_(2)-m_(1))/(m_(1)+m_(2)))u_(2)` ... (9)
Case 1 : When bodies has the same mass i.e.,`m_(1)=m_(2)`
equation (8) `rArr v_(1)=(0)u_(1)+((2m_(2))/(2m_(2)))u_(2)`
`v_(1)=u_(2)` ... (10)
equation (9) `rArr v_(2)=((2m_(1))/(2m_(1)))u_(1)+(0)u_(2)`
`v_(2)=u_(1)` ...(11)
The equations (10) and (11) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e, `m_(1)=m_(2)` and second body (usually called target) is at rest `(u_(2)=0)`,
By substituting `m_(1)=m_(2)=andu_(2)` , in equations (8) and equations (9) we get, from equation (8) `rArr v_(1)=0)`...(12)
from equation (9) `rArr v_(2)=u_(1)` ...(13)
Equations (12) and (13) show that when the first body comes to rest the second body move with the initial velocity of the first body
Case3 : The first body is very much lighter than the second body
`(m_(1)ltltm_(2),m_(1)/m_(2)ltlt1)` than the ratio `m_(1)/m_(2)=0` and also if the traget is at rest `(u_(2)=0)`
Dividing numerator and denomintor of equation (8) by `m_(2)`, we get
`v-(1)=(((m_(1)/m_(2))-1)/(m_(1)/(m_(2))+1))u_(1)+((2)/(m_(1)/(m_(2))+1))(0)`
`v_(1)=((0-1)/(0+1))u_(1)`
`v_(1)=-u_(1)` ...(14)
Similarly , dividing numerator and denominator of equation (9) by `m_(2)` ,we get
`v_(2)=((2(m_(1))/(m_(2)))/(m_(1)/m_(2)+1))u_(1)+((1-m_(1)/m_(2))/(m_(1)/m_(2)+1))(0)`
`v_(2)=(0) mu_(i)+ ((1-m_(1)/m_(2))/(m_(1)/m_(2)+1)) (0)`
`v^(2)=0` ...(15)
The equation (14) implies that the first body which is higher is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (15) implies that the second body whifch is heavier in mass continuous to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was throuw but in opposite direction.
Case 4: The second body is very much lighter than the first body `(m_(2) lt lt m_(1), m_(2)/m_(1) lt lt 1)` then the ratio `m_(2)/m_(1)=0` and also if the target is at rest `(u_(2)=0)`
Dividing numerator and denominator of equation (8) by `m_(1)`, we get
`v_(1)=((1-m_(2)/m_(1))/(1+m_(2)/m_(1)))u_(1)+((2 m_(2)/m_(1))/(1+m_(2)/m_(1)))(0)`
`v_(1)=((1-0)/(1+0))u_(1)+((0)/(1+0)) (0)`
`v_(1)=u_(1)`
Similary,
Dividing numerator and denominator of equation (xiii) by `m_(1)", we get "v_(2)=((2)/(1+0)) u_(1)`
`v_(2)=2u_(1)`
The equation (16) implies that the first body which is heavier continuous to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will mass with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.
