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 design 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 cach 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 b. 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 wagons eventually make it safely across the river. What must the minimum 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 constant. So they tried one more invasion with a new spring of the same dimensions, but yet a different spring constant. However, this time, they also designed a latching mechanis so that when the two wagons collide, the collision is perfectly inelastic. Do they succeed 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.) 1.00m 15.0 30.0 m 20.0 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 waon. Both wagons make it safely across the river

Explanation / Answer

2. mass of two [assenger wagon = 2m]

mass of single passenger wagon = m

a, let initial compressiotn of the spring be xo, spring constant be k

then initial speed of single passenger wagon just before collision = u1

initial speed of this wagon after collision = v1

initial speed of the other wagon after collision = v2

then from the given data

0.5mu1^2 = 0.5kxo^2 - mg(10*tan(15) - sin(15))

from conservation of momentum

mu1 = 2mv2 + mv1

also

u1 = v2 - v1

hene we can find v1 and v2 in terms of k

now, speeds of wagons at the top points

single passenger wagon = v1'

double passenger wagon = v2'

0.5mv1'^2 = 0.5mv1^2 - mg(10*tan(15))

0.5(2m)v2'^2 = 0.5(2m*v2^2) - 2mg(10*tan(15))

also, for safely crossing the river

for speed v of the cart

vcos(15)*t = 30 m

-20tan(15) = vsin(15)*t - 0.5gt^2

-20*tan(15) = vsin(15)*30/vcos(15) - 0.5g*900/v^2cos^2(15)

for v1

-5.35898 = 8.038475 - 4731.4468/v^2

v1' = 18.792 m/s

v1 = 20.1427 m/s

v2- v1 = 2v2 + v1

v2 = -2v1

v2 = -40.2855 m/s

b. for inelastic collsision

let initial compressiotn of the spring be xo, spring constant be k

then initial speed of single passenger wagon just before collision = u1

initial speed of this wagon after collision = v1

initial speed of the other wagon after collision = v2

then from the given data

0.5mu1^2 = 0.5kxo^2 - mg(10*tan(15) - sin(15))

from conservation of momentum

mu1 = 3mv'

now, speeds of wagons at the top points

0.5(3m)v^2 = 0.5(3m)v'^2 - 3mg(10tan(15))

and for safe crossing

v = 18.792 m/s

v' = 20.14 m/s

u1 = 60.426 m/s

by knowing xo we can find k and comment on success of the raid

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