Applied Maths - Mechanics 32

See the figure:
http://img441.imageshack.us/img441/8373/sphmsp.jpg
Let the speed of the hemisphere be Vh after the impact which is taken as the positive direction. There is no vertical component of the velocity.
Let the horizontal velocity of the sphere be Vx and vertical speed be Vy, and the downward direction taken as positive.
To simplify presentation, I use c in place of cos and s in place of sin.

Momentum is conserved only in the horizontal direction:
mVx + MVh = 0 (zero initial horizontal moment)
Vh = -mVx/M … (1)
Resolve velocities into the radial and tangential direction.
For radial direction, apply coefficient of restitution
Vhs - Vxs - Vyc = euc … (2)
For tangential direction, there is no change for the velocity of the sphere since the objects are smooth.
Vys – Vxc = us … (3)

Sub (1) into (2),
-msVx/M – Vxs – Vyc = euc
-Vy = eu + mVxs/Mc + Vxs/c … (4)
Sub into (3),
-eus – mVxs2/Mc – Vxs2/c – Vxc = us
– mVxs2/Mc – Vx/c = us(1 + e)
– mVxs2 – MVx = Musc(1 + e)
Vx = -Musc(1 + e)/(M + ms2) … (A)

From (1) Vh = musc(1 + e)/(M + ms2) … (B)

(4) => -Vy = eu + (ms/Mc + s/c)[-Musc(1 + e)/(M + ms2)]
-Vy = eu + (ms + Ms)[-us(1 + e)/(M + ms2)]
-Vy = [eu(M + ms2) – (ms + Ms)us(1 + e)]/(M + ms2)
-Vy = u[eM + ems2 – ms2 – Ms2 – ems2 – eMs2]/(M + ms2)
-Vy = u[eM – ms2 – Ms2 – eMs2]/(M + ms2)
Vy = -u[eMc2 – (m + M)s2]/(M + ms2) … (C)

By (1), if |Vx| = |v|, then m = M
Put into (C)
Vy is simplified to -u[ec2 – 2s2]/(1 + s2)
The sphere will fly above P if Vy is negative (not relevant for Vx in horizontal direction), i.e.
-u[ec2 – 2s2]/(1 + s2) < 0
ec2 – 2s2 > 0
ec2 > 2s2
e/2 > s2/c2
0.5e > tan2