After reading through the numerous comments, solutions and observations it's obvious a lot of you have forgotten Engineering 101. First a definition to refresh your memory and then a few assumptions:
1. Elastic collision means NO LOSS of ENERGY. There is, absolutely, no deformation allowed in an elastic collision. All momentum and energy is transferred.
2. For such short distances we can safely neglect air resistance.
3. To use real numbers let's assume that "neglible" means the mass of the larger ball m1 is 100 times larger than the mass of the very small ball m2. This is very important as we'll see later.
4. We will assume a proper mechanism will release the balls in contact with each other.
5. The impacted surface is sufficiently rigid and massive to provide for the elastic collision given.
Now we have a nice simple free fall momemtum problem. But we need to define our variables and givens:
m1 = mass of 0.1m diameter ball.
m2 = mass of very small ball, the diameter not important but the ratio of masses is, remember we previously assumed m2 = m1/100 to achieve neglible status.
r1 = radius of larger ball, m1, Given r1 = 0.05 meters.
d1 = diameter of larger ball, m1, 2*r1 = 0.10 meters.
s0 = Initial height from surface to base of larger ball (set at datum - this will become clear later when we will need to introduce vector directions). Given s0 = 0.5 meters.
s1 = Surface.
s2 = Bottom of very small ball at impact above surface.
s3 = Height of roof.
s4 = Height of face looking at very small ball.
v0 = Initial velocity at s0 = 0.5 meter height to bottom of larger ball.
v1 = Instantaneous velocity at impact.
v2 = Instantaneous velocity of very small ball at elastic collision transfer.
v3 = Instantaneous velocity at face.
v4 = Average velocity of very small ball from collision to face.
t0 = Time at s0 and v0.
t1 = Time at s1 and v1.
t2 = Time at elastic collision.
t3 = Time to face.
Ao = Constant acceleration, for Earth's gravity this is 9.81 m/sec^2 (nominal value).
Initial values:
s0 = 0
t0 = 0
v0 = 0
The general distance equation for a free falling body is:
s(t) = (Ao*(t(t) - t0)^2)/2 + v0*(t(t)-t0) + s0
Find the time to fall from 50 centimeters (0.5 meters).
For (t)=1 and (s)=1 we have:
s1 = Ao*(t1)^2 + 0 + 0
0.5 = (Ao/2)*t1^2, solving for t1;
t1 = (2*0.5/9.81)^.5,
t1 = 0.319275 seconds from release to impact.
The general velocity equation for a free falling body is:
v(t) = Ao*(t1-t0) + v0
Find the velocity at impact.
For t1 = 0.319275, t0=0, and v0 = 0 we have:
v1 = 9.81*.319275 + 0 = 3.13209 meters/second
Momentum:
Remember F = ma? It is also written F = m * dv/dt and since we're doing elastic collisions the force of impact must equal the force of reaction.
F1=F2
F1=m1v1/(t(t)-t(t-1))
F2=m2v2/(t(t)-t(t-1))
m1v1/(t(t)-t(t-1)) = m2v2/(t(t)-t(t-1))
Zero deformation equates to equal time so time cancels out and we then have:
m1v1=m2v2
Recall that we assumed that m2 = m1/100 as being neglible? We now have:
m1*v1 = m1*100*v2, NOTE the masses cancel out! Solving for v2.
v2 = 100*v1, Very important! No matter what material you use it's the ratio of masses which becomes the multiplier for the initial velocity of the very small ball.
v2= 100*3.13209 = 313.209 meters/second. WOW! And I don't mean World of Warcraft.
Because the problem statement used elastic collisions all the kinetic energy was transferred to the very small ball. Thus the larger ball remains motionless and in contact with the surface. And yes, the very small ball will hit you in the face if directly over the balls.
Of course in real life that is dependent on the actual mass and hardness of the ball materials and considering the coefficient of restitution and air resistance. But for hard materials it won't be much of a decrease.
Check the velocity at your face in case you want to see how much effect gravity has on slowing the very small ball's vertical velocity.
Distance from very small ball to face is:
s4 = s3 - d1
s4 = 4.5 - 0.1 = 4.4 meters.
Solve for time (using general equation again), and remembering that v2 is in v0 position AND the direction vector is now upward so Ao is now negative as we are working against gravity.:
First root:
(-313.092 + ((313.209^2-4*(-4.905)*(-4.4))^.5)/2*(-4.905) = 0.01405 seconds.
Second root:
(-313.092 - ((313.209^2-4*(-4.905)*(-4.4))^.5)/2*(-4.905) = 63.841 seconds.
Since second root is entirely unreasonable and the first root is reasonable.
t3 = 0.01405 seconds
Velocity at face (remembering again that Ao is negative, working against gravity)
v3 = -Ao*t3 + v2
v3 = -9.81*0.01405 + 313.209
v3 = 312.954 meters/second instantaneous velocity at your face (if you're really looking down at it).
If you choose to do attempt this experiment for real at home or in a lab, please be careful and take needed precautions to prevent injury and possible death.
For example a steel ball 10 cm in diameter is about 4.11 kg, 1/100 of that mass equates to a steel ball approximately 2.1 cm in diameter.
At 300 meters/second it's going to hurt real bad.
You can more easily do this with a basketball and a tennis ball.
Approximate mass of a basketball is 600 gm and is 24 cm in diameter.
Approximate mass of a tennis ball is 57 gm and is 6.5 cm in diameter.
Or roughly a ten and a half to one ratio of masses.
Changing the diameters and heights, the exit velocity of the tennis ball is about 30 meters per second, that's impressive. Include the coefficients of restitution and a little air drag and the velocity should drop into a range that you can experiment with in your driveway and confirm the equations.
