# Oblique Collision In 2D

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On a horizontal smooth surface a disc is placed at rest. Another disc of same mass is coming with impact parameter equal to its own radius. First disc is of radius r. What should be the radius of the coming disc so that after collision first disc moves at an angle 45° to the direction of motion of incoming disc?
(A)2r. (B)r(√2 -1). (C) r/(√2 - 1) (D)r√2

asked Jun 24, 2018
edited Jun 25, 2018
What difficulty are you having with this question? What have you tried? Have you drawn a diagram?

Draw a diagram, as above.

The incoming disk (blue), with centre C and radius $R$, enters from the left moving horizontally. The outgoing disk (grey), with centre A and radius $r$, is initially stationary.

The impact parameter is the distance between the centres A and C measured perpendicular to the direction of motion BA - ie it is distance CB. Here the impact parameter equals the radius $R$ of the incoming disk and BA is horizontal.

The stationary disk moves off at $45^{\circ}$ to the direction AB (red arrow). Assuming that there is no friction during impact, the impulse forces between the disks are purely normal to the contact surface. The change in momentum of disk A is along the line of their centres CA at the moment of impact. Since disk A was initially stationary, CA is also the direction in which A moves after the collision.

Note that the relative mass of disk A has no effect on its final direction but it does affect the final speed of disk A. This is because A had no initial momentum, so its final momentum is in the direction of the impulse. Whereas for disk C there was some initial momentum so the final momentum is not in the direction of the impulse.

From the geometry of the triangle ABC you can work out the relation between the two radii $R$ and $r$.

answered Jun 25, 2018 by (27,948 points)
edited Jun 25, 2018