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A ball A suffers an oblique elastic collision with a ball B that is at rest initially if there masses are same then after collision what will happen

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An oblique collision
of two equal masses with one at rest initially and the other with u1 initially. After collision they move in directions which
are perpendicular to each other. Possibly, the ball B moves with u1 CosФ along
the line L joining the centers of masses at the collision time. Ball
A moves perpendicular to that line L with velocity = u1 * Sin Ф.

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See diagram.

Let ball A have mass M and B have mass M. Let the Ball B be at origin as shown and at rest (u2 =0). Let A move in x direction and collide with B. The velocity u1 of A makes an angle Ф with the line joining the two centers of masses.

Let us assume that there are no external forces acting on them. Let us assume that they are on a frictionless surface or in a frictionless medium. Let us assume that the collision is elastic. There is no loss of energy.

Let the velocities of A and B after collision be:

__V1__ = V1 Cos Ф1__ i__ + V1 Sin Ф1 __ ____j__ , Ф1 = angle of vector __V1__ with x axis.

__V2__ = V2 CosФ2 __i__ - V2 Sin Ф2__
j __ , (-Ф2) =
angle of vector __V2 __with x axis

Applying conservation of momentum :

M u1__ ____i __= M V1
Cos Ф1__ ____i __+ M V1 Sin Ф1__ ____j__ + M V2 Cos Ф2__ i -__ M
V2 Sin Ф2 __j__

V1 Cos Ф1 + V2 Cos Ф2 = u1 -- eq 1

V1 Sin Ф1 = V2 Sin Ф2 -- eq 2

From the conservation of energy we have : 1/2 M u1² = 1/2 M V1² + 1/2 M V2²

V1² + V2² = u1² -- eq 3

From eq 2, V2 = V1 Sin Ф1 / sin Ф2 -- eq 4.

Substitute in eq 1, V1 [ Cos Ф1 + Sin Ф1 Cos Ф2 / Sin Ф2 ] = u1

V1 Sin (Ф1+Ф2) = u1 Sin Ф2

V1 = u1 Sin Ф2 / Sin (Ф1+Ф2) --- eq 5

V2 = u1 Sin Ф1 / Sin (Ф1+Ф2) -- eq 6

Substitute V1 and V2 in eq 3 and then cancelling u1² on both sides:

Sin² Ф2 + Sin² Ф2 = Sin² (Ф1+Ф2) -- eq 7

2 Sin² Ф2 + 2 Sin² Ф2 = 2 Sin² (Ф1+Ф2)

1 - Cos 2Ф1 + 1 - Cos 2Ф2 = 1 - Cos 2(Ф1+Ф2)

Cos 2Ф1 + Cos 2Ф2 = Cos 2(Ф1+Ф2) + 1 -- eq 8

2 Cos (Ф1+ Ф2) Cos (Ф1-Ф2) = 2 Cos² (Ф1+Ф2)

It means that either, Cos (Ф1 + Ф2) = 0, or Cos (Ф1-Ф2) = Cos (Ф1 + Ф2)

Cos (Ф1 + Ф2) = 0 =>

Solution (1) Ф1 + Ф2 = π/2 -- eq 9

Solution (2) Ф1 + Ф2 = 3π/2 --- eq 10

Cos (Ф1-Ф2) = Cos (Ф1+Ф2) =>

Ф1-Ф2 = Ф1+Ф2 or Ф1-Ф2 = -(Ф1+Ф2) or Ф1-Ф2 = 2π - (Ф1+Ф2)

So, Ф2 = 0 or Ф1 = 0 or both Ф1 = Ф2 = 0 or Ф1 = 180.

Solution 3) Ф2 = 0 -- eq 11

Solution 4) Ф1 = 0 -- eq 12

solution 5) both Ф1 = Ф2 = 0 -- eq 13

Solution 6) Ф1 = 180 = π -- eq 14

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Solution 1 means after collision they move in directions perpendicular to each other. Ball Moves in 1st quadrant and ball B moves in quadrant 4.

V1 = u1 Cos Ф and V2 = u1 Sin Ф where Ф = angle with x-axis

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Solution 2 is NOT POSSIBLE as momentum will not be conserved.

Solution 3 is V1 = 0 and V2 = u1. This is the HEAD-ON collision case. So Ball A stops and Ball B moves with the same initial velocity of A.

Solution 4 is NOT POSSIBLE. Ball B is at rest and Ball A moves with initial velocity. There is no collision at all.

Solution 5 is Balls A and B move along the same X-direction after collision. So it is HEAD-ON collision and not oblique collision. Same solution as solution 3. Ball A stops. Ball B moves with initial velocity of A.

Solution 6: Ball A rebounds back in the same direction. The solution is same as solution 3. Ball A stops and Ball B moves with initial velocity of A.

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Finally, An oblique collision of two equal masses with one at rest initially and the other with u1 initially. After collision they move in directions which are perpendicular to each other. Possibly, the ball B moves with u1 CosФ along the line L joining the centers of masses at the collision time. Ball A moves perpendicular to that line L with velocity = u1 * Sin Ф.

===============================================

See diagram.

Let ball A have mass M and B have mass M. Let the Ball B be at origin as shown and at rest (u2 =0). Let A move in x direction and collide with B. The velocity u1 of A makes an angle Ф with the line joining the two centers of masses.

Let us assume that there are no external forces acting on them. Let us assume that they are on a frictionless surface or in a frictionless medium. Let us assume that the collision is elastic. There is no loss of energy.

Let the velocities of A and B after collision be:

Applying conservation of momentum :

M u1

V1 Cos Ф1 + V2 Cos Ф2 = u1 -- eq 1

V1 Sin Ф1 = V2 Sin Ф2 -- eq 2

From the conservation of energy we have : 1/2 M u1² = 1/2 M V1² + 1/2 M V2²

V1² + V2² = u1² -- eq 3

From eq 2, V2 = V1 Sin Ф1 / sin Ф2 -- eq 4.

Substitute in eq 1, V1 [ Cos Ф1 + Sin Ф1 Cos Ф2 / Sin Ф2 ] = u1

V1 Sin (Ф1+Ф2) = u1 Sin Ф2

V1 = u1 Sin Ф2 / Sin (Ф1+Ф2) --- eq 5

V2 = u1 Sin Ф1 / Sin (Ф1+Ф2) -- eq 6

Substitute V1 and V2 in eq 3 and then cancelling u1² on both sides:

Sin² Ф2 + Sin² Ф2 = Sin² (Ф1+Ф2) -- eq 7

2 Sin² Ф2 + 2 Sin² Ф2 = 2 Sin² (Ф1+Ф2)

1 - Cos 2Ф1 + 1 - Cos 2Ф2 = 1 - Cos 2(Ф1+Ф2)

Cos 2Ф1 + Cos 2Ф2 = Cos 2(Ф1+Ф2) + 1 -- eq 8

2 Cos (Ф1+ Ф2) Cos (Ф1-Ф2) = 2 Cos² (Ф1+Ф2)

It means that either, Cos (Ф1 + Ф2) = 0, or Cos (Ф1-Ф2) = Cos (Ф1 + Ф2)

Cos (Ф1 + Ф2) = 0 =>

Solution (1) Ф1 + Ф2 = π/2 -- eq 9

Solution (2) Ф1 + Ф2 = 3π/2 --- eq 10

Cos (Ф1-Ф2) = Cos (Ф1+Ф2) =>

Ф1-Ф2 = Ф1+Ф2 or Ф1-Ф2 = -(Ф1+Ф2) or Ф1-Ф2 = 2π - (Ф1+Ф2)

So, Ф2 = 0 or Ф1 = 0 or both Ф1 = Ф2 = 0 or Ф1 = 180.

Solution 3) Ф2 = 0 -- eq 11

Solution 4) Ф1 = 0 -- eq 12

solution 5) both Ф1 = Ф2 = 0 -- eq 13

Solution 6) Ф1 = 180 = π -- eq 14

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Solution 1 means after collision they move in directions perpendicular to each other. Ball Moves in 1st quadrant and ball B moves in quadrant 4.

V1 = u1 Cos Ф and V2 = u1 Sin Ф where Ф = angle with x-axis

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Solution 2 is NOT POSSIBLE as momentum will not be conserved.

Solution 3 is V1 = 0 and V2 = u1. This is the HEAD-ON collision case. So Ball A stops and Ball B moves with the same initial velocity of A.

Solution 4 is NOT POSSIBLE. Ball B is at rest and Ball A moves with initial velocity. There is no collision at all.

Solution 5 is Balls A and B move along the same X-direction after collision. So it is HEAD-ON collision and not oblique collision. Same solution as solution 3. Ball A stops. Ball B moves with initial velocity of A.

Solution 6: Ball A rebounds back in the same direction. The solution is same as solution 3. Ball A stops and Ball B moves with initial velocity of A.

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Finally, An oblique collision of two equal masses with one at rest initially and the other with u1 initially. After collision they move in directions which are perpendicular to each other. Possibly, the ball B moves with u1 CosФ along the line L joining the centers of masses at the collision time. Ball A moves perpendicular to that line L with velocity = u1 * Sin Ф.