Question

# 1. Two students are kayaking on the Saint John River. Initially, they are floating directly beside...

1. Two students are kayaking on the Saint John River. Initially, they are floating directly beside each other chatting and moving with the river current at 1.50 m/s downstream. Student A pushes away from Student B and sees Student B floating away from them at 1.00 m/s in the upstream direction. The combined inertia of Student A and their kayak is 100 kg and the combined inertia of Student B and their kayak is 120 kg. Assume that there is no friction between the kayaks and the water.

a. Relative to the river flow, determine the velocities of the two students once they start moving away from each other (after the push). (Define your system to justify any conservation relations you might use, provide appropriate diagrams to describe the interaction and explain your solution approach.)

b. What are the velocities of the two students once they are moving away from each other as seen from the perspective of someone on the shore of the river?

c. What source energy does Student A expend in pushing the two kayaks apart?

2. Two children are playing with a ball. One child is on a ladder and the second child is directly below at the bottom of the ladder. The child at the bottom of the ladder throws a 500 gram ball up to the second child. The initial speed of the ball when it leaves the lower child’s hand is 5.00 m/s and the second child catches it 2.00 m above the point from which it is thrown. Produce energy bar diagrams showing the energy transformations and associated energy amounts which take place.

i. Before the ball is thrown.

ii. Just before the ball is caught.

iii. After the ball is caught. Use a closed system for this analysis and identify the components in this system.

1. given

two students are kayaking

river current speed downstream, u = 1.5 m/s

A pushes B, finds B is moving away, upstream relative to it at speed d' = 1 m/s

let upstream be positive

hence

let speed of B be v upstreams, and of A be w downstream

hence

d' = w + v = 1

also,

Ma = 100 kg

Mb = 120 kg

hence

from conservaiton of moemnntum

Mbw = Ma*v

W = 1 - V

hence

12(1 - v) = 10*v

v = 0.5454545 m/s

w = 0.45454545 m/s

a. hence

vb = 0.5252 m/s upstreams

wa = 0.4545 m/s downstreams

b. with respect to ground

Vb = vb - 1.5 = 0.9545 m/s downstreams

Wa = wa + 1.5 = 1.9545 m/s downstreams

c. the student uses energy sotred in his body to push the boats apart

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