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down the sides of a hemispherical bowl. Both start with zero velocity at

the lip, which is 12 cm above the bottom. The objects collide at the bottom of the bowl. Just

before the collision, they are moving at right angles to each other. To what maximum height

above the bowl will the objects move if the collision is completely inelastic? You may treat the

objects as small particles.

Solution 1:

Using mgy and 1/2mv^2 find out the velocity of both masses just before they collide at the bottom of the bowl. Then using conservation of momentum, determine the final velocity of the 8kg mass(completely inelastic) and then by using mgy and 1/2mv^2 again find out the distance the 8k mass travels up the bowl. (answer is 7.5cm, this 7.5 cm is the vertical distance from the bottom of the bowl, not the arc length over which the mass moves)

I was wondering if it would be possible to do this problem by using the idea of center of mass of the 2 masses just before they collide and consider it as a 8kg mass with a resultant velocity and calculate the answer ignoring the collision altogether since the collision represents internal forces and there is no external force on the cm besides mg.