A 1000kg plane is trying to make a forced landing on the deck of a 2000 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck, and this braking force is constant and equal to 1/4 of the plane's weight What must the minimum length of the barge be, in order that the plane can stop safely on the deck, if the plane touches down at the rear end of the deck with a velocity of 50m/s towards the front of the barge?(adsbygoogle = window.adsbygoogle || []).push({});

My work:

m1 = 1000kg

m2 = 2000kg

Ff = 1/4Fg

V1' = V2'

Fg = mg

=(1000)(9.81)

=9810N

Ff = 1/4Fg

= 1/4(9810)

=2452.5N

Fnet=Ff

P1(plane) = mv

= (1000)(50)

= 50000

P2(plane) = mv

= (1000)(0)

= 0

deltaP = (Fnet)t

50000 = 2452.5t

t = 20.387s

m1v1 + m2v2 = m1v1' + m2v2'

(1000)(50) + 2000(0) = 0 + (3000)v2'

50000 = 3000v2'

v2' = 16.667 m/s

a = V2-V1 / T

= (16.667-50) / 20.387

= -1.635m/s²

d = v1t + 1/2at²

= (50)(20.387) + 1/2(-1.635)(20.387)²

= 1019.35 - 339.777

=679.573m

Thats what i did but the answer is 340m...wondering what I did wrong...

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# Conservation of momentum

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