A missile launcher, mass M, on a horizontal track fires a missile, mass m, horiz

Question

A missile launcher, mass M, on a horizontal track fires a missile, mass m, horizontally. To stop the recoil of the launcher, it rolls smoothly (and without friction) up an inclined plane and comes to a stop h meters above ground level. (a) What is the launch speed, vo, of the missile in terms of M, m, h, and g? (The launcher goes from ground level to the incline with no loss in momentum) (b) If the missile launcher’s mass is 4400 kg, the missile’s mass is 110 kg, and the final height of launcher is 4.0 m. Find missile launch speed.

Explanation / Answer

here,

mass of missile launcher is M

mass of missile is m

height on incline above the ground is h

(a)

let the speed of launcher after collison be u

using conservation of momentum

M * u + m * v0 = 0

u = - m * v0 /M

u si in the opposite direction

using conservation of energy

change in kinetic energy = potential energy gained

0.5 * M * u ^2 = M * g * h

0.5 * (m*v0/M)^2 = g * h

v0 = (M/m) * sqrt(2*g*h)

the launch speed v0 is (M/m) * sqrt(2*g*h)

(b)

M = 4400 kg

m = 110 kg

h = 4 m

the launch speed , v0 = (M/m) * sqrt(2*g*h)

v0 = (4400/110)* sqrt(2 * 9.8 * 4)

v0 = 354.18 m/s

missile launch speed is 354.18 m/s

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