Applied Maths - Mechanics 33

(a) Let the origin be assigned at P.
Let the position of particle A be denote by Ax and Ay
Let the position of particle B be denote by Bx and By
Let the particle B be projected at an angle φ and velocity v
Ax = ucosθt … (1)
Ay = usinθt – gt2/2 … (2)
Bx = l – vcosφt … (3)
By = h + usinφt – gt2/2 … (4)
For collision to happen, Ax = Bx and Ay = By at certain time t
(1) = (3) => ucosθt = l – vcosφt
(ucosθ + vcosφ)t = l … (5)
(2) = (4) => usinθt – gt2/2 = h + usinφt – gt2/2
(usinθ – vsinφ)t = h … (6)
(6)/(5) (usinθ – vsinφ)/(ucosθ + vcosφ) = h/l … (7)
if h/l >= tanθ, then (7) becomes
(usinθ – vsinφ)/(ucosθ + vcosφ) >= tanθ
usinθ – vsinφ >= (ucosθ + vcosφ)tanθ = usinθ + vcosφtanθ
0 >= vcosφtanθ + vsinφ which is not possible
So if h/l >= tanθ, no collision will occur


(b) (usinθ – vsinφ)/(ucosθ + vcosφ) = h/l
l(usinθ – vsinφ) = h(ucosθ + vcosφ)
u(lsinθ – hcosθ) = v(hcosφ + lsinφ) … (8)
Let tanβ = l/h
sinβ = l / √(l2 + h2)
cosβ = h / √(l2 + h2)
(8) becomes
u(lsinθ – hcosθ) / √(l2 + h2) = v(hcosφ + lsinφ) / √(l2 + h2)
u(lsinθ – hcosθ) / √(l2 + h2) = v(cosβcosφ + sinβlsinφ)
u(lsinθ – hcosθ) / √(l2 + h2) = vcos(φ – β)
v is minimum when cos(φ – β) is maximum = 1
Min v = u(lsinθ – hcosθ) / √(l2 + h2)


(c) At this speed, cos(φ – β) = 1
φ – β = 0
φ = β
Time of collision is given by (6)
(usinθ – vsinφ)t = h
{usinθ – [u(lsinθ – hcosθ) / √(l2 + h2)] [l / √(l2 + h2)]}t = h
ut{sinθ – [l(lsinθ – hcosθ) / (l2 + h2)} = h
ut(l2 sinθ + h2 sinθ – l2sinθ + lhcosθ) / (l2 + h2) = h
ut(hsinθ + lcosθ) = (l2 + h2)
t = (l2 + h2)/[u(hsinθ + lcosθ)]