An 80lb boy sitting in a stationary 20lb wagon wants to simu

An 80-lb boy sitting in a stationary 20-lb wagon wants to simulate rocket propulsion by throwing bricks out of the wagon. Neglect horizontal forces on the wagon\'s wheels. If the boy has three bricks weighing 10-lb, each and throws them with a horizontal velocity of 10 ft/sec relative to the wagon, determine the velocity he attains (a) if he throws the bricks one at a time, and (b) if he throws them all at once. if he throws the bricks one at a time, and (b) if he throws them all at once.

Solution

a)

Mass of the boy = 80 lb= 80x0.0311 slug = 2.488 slug

Mass od wagon = 20X0.0311 = 0.622 slug

Mass of each brick = 10x0.0311 = 0.311 slug

Total initial mass of boy+wagon+brickws = 2.488+ 0.622+ 3x0.311 = 4.043 slug

Let V₁ be the velocity of wagon after throwing first brick (backwards).

Absolute velocity of the brick after throwing first brick = 10 ft/s (since the wagon is at rest initially)

Initial momentum = 0 (before throwing first brick)

Mass after throwing first brick = 4.043-0.311 slug

From conservation of momentum

0.311x10 + (4.043-0.311)V₁ = 0

Hence V₁ = -0.833 ft/s

=-0.91 ft/s

Second throw

Similarly let V₂  be the wagon velocity after 2nd throw.

Absloute velocity of brick = 10-0.833 = 9.17 ft/s

Initial momentum = -(4.043-.0.311)x0.833 = - 3.11 slug-ft/s

Hence

-3.11 = 9.17x0.311 +(4.043-2x0.311)xV₂

Hence V₂ = (-3.11-9.17x0.311)/(4.043-2x0.311) = -1.743 ft/s

Third Throw

Absolute velocity of brick = 10-1.743 = 8.257 ft/s

Initial momentum = -1.743x(4.043-2x0.311) = -5.962 slug-ft/s

Hence -5.962 = 8.257x0.311+ (4.043-3x0.311)V₂

V₂ = (-5.962-8.257x0.311)/(4.043-3x0.311)

= -2.743 ft/s(Ans)

b) Mass of 3 bricks = 3x0.311 = 0.933 slug

Let V_T be the wagon velocity when 3 bricks are thrown together

Proceeding as before

0 = 10x3x0.311 + (4.043-3*0.311)V_T

V_T= -10x3x0.311/(4.043-3x0.311)

= -3 ft/s (Ans)