Thursday
November 27, 2014

Homework Help: Physics - Kinematics

Posted by Rod on Thursday, November 25, 2010 at 1:31pm.

Qn: Ball 1 is launched up an inclined plane from the bottom of the plane with an initial speed that is the minimum speed for it just to reach the top of the plane. At the same moment as ball 1 is launched up the plane, Ball 2 is released from rest at the top of the plane. They make their first contact somewhere along that plane. What is the value of (distance from the bottom of the plane to the point of contact)/(distance from top of plane to the point of contact)?

My views: What I managed to deduce from this question is that since 1/2mv^2 = mgh, initial speed of ball 1 = sqrt(2gh) and final speed of ball 1 = 0. But after that I'm abit lost on what to do. I'm thinking that since the energy changes for both balls are similar as they travel on the same plane, they should meet each other at the middle of the plane, hence the value should be 1. But I'm really not sure, anyone can help?

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