2) The Calmarians\'first invasion failed because they still could not get their

Question

2) The Calmarians'first invasion failed because they still could not get their soldiers across the river 2) a fast enough. They desigu a new wagon holding two passengers which, with passengers included, has twice the mass of the single-passenger wagon. They position a two-passenger wagon exactly half way up each ramp (10.0m from the base measured horizontally), and then launch a single passenger wagon, as before. The spring is also redesigned so that it has the same equilibrium length and compressed length, but a new spring constant. a) The first single-passenger wagon collides elastically with the second two-passenger wagon, and both wagous eventually make it safely across the river. What must the minimun spring constant now be in order for both wagons to make it safely across the river? (Assume that any other collision that occurs is also elastic.) b) The invasion still failed because the Calmarians found it difficult to reliably make springs with the required spring coustaut. So they tried one nore invasiou with a new spring of the same dimensions, but yet a different spring constant. However, this time, they also designed a latching mechanism so that when the two wagous collide, the collision is perfectly inelastic. Do they succed in this invasion? Why or why not? Does your answer make sense? Again, assume that there is no friction anywhere, and also remember to consider the fact that the first wagon is not initially at the base of the ramp due to the finite length of the compressed spring.) 1.00 m 30.0 m Figure 2 - An additional wagon holding two passengers, with twice the mass, is positioned half way up the ramp. The first single-passenger wagon is launched as before, and collides with the second wagon. Both wagons make it safely across the river .Es-1.brvrnlungin ob spengamelees Coltase Solve ali parts! deuble wegon 230 K Page 2 of 6

Explanation / Answer

2. given mass of single passenger wagon = m

mass of two passenger wagon = 2m

a. let speed of the double passenger wagon crossing the river be vi

then

range = 40 m

theta = 15 deg

then

40 = v^2sin(2*theta)/g

v = 28.014 m/s

hence speed of single passenger wagon before collision = u

after collision = v

speed of double passenger wagon after collision = v' > 28.014 m/s

hence

from conservation of momentum

2mv' + mv = m*u

also

0.5*k*x^2 = 0.5mu^2

u = sqroot(k/m)x ( x is compressed length)

also

u/(v' - v) = 1

hence

by knowing x from previous problem one can find out u in terms of k

and then we can find out k

b. for the second invasion, perfectly inelastic collsision

hence

v = 28.014 m/s

hence

from conservation of meomtnum

(m + 2m)v = m*u

u = 3v = 84.04 m/s

hence

0.5*m*u^2 = 0.5*kx^2

7063.2 = kx^2

so by knowing x, we can find k and then verify the claim

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