Obtain the expression for the velocities of the two bodies after collision in the case of one dimensional elastic collision ond discuss the specical cases.

Explanation

Consider two elastic bodies of massesm `m_(1) and m_(2)` moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass `m_(1)` moves faster than mass `m_(2) i.e, u_(1) gt u_(2)`. For elastic collision, the total linear momentum and kinetic energy of two bodies before and after collsiion must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision `(p_(i))`= Total momentum after collision `(p_(f))`
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2) ........(1)`
`m_(1) (u_(1)-v_(1))=m_(2) (v_2)-u_(2)).......(2)`
Further,

For elastic collision
Total kinetic energy before collision `KE_i` - Total kinetic energy after collision `KE_f`
`1/2 m_(q)u_(1)^(2)=1/2 m_(1)v_(1)^(2)+1/2 m_(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 equations `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_(1)+v_(1)=v_(2)+u_(2)`
`u_(1)-u_(2)=v_(2)-v_(1) ......(5)`
Rearranging. Equation (5) and it can be rewritten as
`u_(1)-u_(2)=-(v_(1)-v_(2))`
This means that for any elastie 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.
Rewriting the above equation for `v_(1) and v_(2)`,
`v_(1)=v_(2)+u_(2)-u_(1) ......(6)`
`(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_(1) (u_(1)-v_(1))=m_(2) (u_(1)+v_(1)-u_(2)-u_(2))`
`m_(1) (u_(1)-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))u_(1)`
`or v_(1) ((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(2))/(m_(1)+m_(2))) u_(2)`
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7) 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)`
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)= and u_(2)= 0` in equations (8) and equations (9) we get,
from equation (8) `rArr v_(1)=0` .........(12)
from equation (9) `rArr v_(2)=0` .........(13)
Equations (12) and (13) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body `(m_(1) lt lt m_(2), m_(1)/m_(2) lt lt 1)` then the ratio `m_(1)/m_(2)=0` and also if the target is at the rest `(u_(2)=0)`
Dividing numerator and denominator 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)`
`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) u_(1)+((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 lighter returns tank (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (15) implies that the second body which is heavier in mass continues to remain at test event collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from wall with the same velocity with which it was thrown 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)`
Similarly,
Dividing numerator and denominator of equation (13) by `m_(1)` we get
`v_(2)=((2)/(1+m_(2)/m_(1))) u_(1)+((m_(2)/m_(1)-1)/(1+m_(2)/m_(1))) (0)`
`v_(2)=((2)/(1+0))u_(1)`
`v_(2)=2u_(1)`
The equation (16) implies that the first body which is heavier continues to move with the initial velocity. The equation (17) suggests that the second body which is lighter will with twice the initial velocity of the first body. It means that the lighter body is throw from the point of collision.