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?
To solve this problem, we can analyze the motion of both balls along the inclined plane using principles of physics.
Let's start by calculating the time it takes for ball 1 to reach the top of the plane. We can use the equation for motion along an inclined plane:
y = y0 + x*tan(theta) + (1/2)*g*t^2
Since ball 1 just reaches the top of the plane, y = 0 at that point. The initial position of ball 1 is at the bottom of the plane, so y0 = 0. The angle of inclination of the plane is given by theta. The acceleration due to gravity is g. We can ignore the term (1/2)*g*t^2 since the initial and final positions of ball 1 are the same.
0 = 0 + x*tan(theta)
x = 0
This means that ball 1 reaches the top of the plane instantaneously. It implies that ball 2 must start moving at that same moment because they make their first contact somewhere along the plane.
Now, let's consider the motion of ball 1 in the opposite direction (down the plane) and the motion of ball 2 in the same direction (down the plane). Both balls will be subject to the same acceleration due to gravity, g.
We can write the equation for the final velocity of ball 1 as:
v^2 = u^2 + 2*a*s
where v is the final velocity (which is 0), u is the initial velocity (sqrt(2gh)), a is the acceleration (-g), and s is the distance traveled.
0 = (sqrt(2gh))^2 + 2*(-g)*s
0 = 2gh - 2gs
2gs = 2gh
s = h
So, the distance from the top of the plane to the point of contact is equal to the height (h) of the inclined plane.
Now, let's calculate the distance from the bottom of the plane to the point of contact. Since ball 1 reaches the top of the plane instantaneously, and both balls experience the same acceleration due to gravity, they must have traveled the same distance from their starting points when they make contact. Thus, the distance from the bottom of the plane to the point of contact is also h.
Therefore, the value of (distance from the bottom of the plane to the point of contact) / (distance from the top of the plane to the point of contact) is:
h/h = 1
So the value is indeed 1.
To summarize, both balls meet at the half-way point along the inclined plane. The value of (distance from the bottom of the plane to the point of contact) / (distance from the top of the plane to the point of contact) is 1.