A ball of m1 is held slightly above ball of mass m2 . The two balls are droped s

Question

A ball of m1 is held slightly above ball of mass m2. The two balls are droped simultaneously, falling distance h vertically from rest. Air resistance is negligible. The second ball collides with the floor, then with the first ball. All collisions have coefficient of restitution r. After the second collision, the second ball momentarily stops dead.

a. Calculate the mass ratio m1/m2.

b. Calculate the height to which the first ball rebounds. Treat the diameters of the balls themselves as neglibible.

Explanation / Answer

For m2:

u2 = (2gh)^1/2 before first collision

v2 = r*u2 after first collision

Since the delay between 1st and 2nd collisions is ~0,

v2 is also the speed at which m2 collides with m1.

So, u1 = (2gh)^1/2

and after collision,

(v1 + 0) = r*(u1 + v2)

So, v1 = r*(u1 + r*u2)

But, in falling a height h, u1 = u2.

So, v1 = u1*r*(1+r) = (r^2 + r)*(2gh)^1/2.

So, using conservation of momentum,

m1u1 - m2v2 = m1v1 + 0

So, m1u1 - m2*r*u1 = m1*u1*r*(1 + r)

So, m1/m2 = r/[1 - r - r^2]

--------------------

Let the height be x.

So, m1*g*x = m1*v1^2 / 2

So,

g*x = (r^2)*(1+r)^2 * gh

So, x = r^2(r^2 + 2r + 1)*h or (r^2)*(1+r)^2 * h

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