The situation for off-center (not a head-on) collisions between two balls of une

Question

The situation for off-center (not a head-on) collisions between two balls of unequal mass is significantly more difficult to tackle. In this case, even if one of the balls is initially at rest, there are still four unknown quantities, the two final velocities of the two balls and the angle each ball makes with the initial line of travel. However, if the momentum is known for one of the balls in the direction perpendicular to the line of travel, then the momentum for the other ball in the same direction is also known since they are equal and opposite. For the plot below, the point where the steel ball lands (P1) is plotted. Where along the y-axis does the plastic ball land? Click on the y-line where the plastic ball lands. Then click on the "Submit Answer" button. Use the fact that the mass of the steel ball is 23.5 g and that of the plastic ball is 14.6 g.

Explanation / Answer

As there will be no external force on the system during collision, hence momentum will be conserved for the system.

m1u1 + m2u2 = m1v1 + m2v2.

In any direction, u is initial velocity in that direction and v is final velocity in that direction.

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