Question

Two ice skaters, Daniel (mass 60.0kg ) and Rebecca (mass 45.0kg ), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0m/sbefore she collides with him. After the collision, Rebecca has a velocity of magnitude 6.00m/s at an angle of 52.1? from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

What is the magnitude of Daniel's velocity after the collision?

What is the direction of Daniel's velocity after the collision?

What is the change in total kinetic energy of the two skaters as a result of the collision?

Answer #1

rabecca initial veocity=vi=13 m/s

initial KE=(45*13*13)/2

initial momentum=45*13

Daniel initial veocity=vi=0

initial KE=0

initial momentum=0

total initial KE and momentum is due to only rebacaa

let initial KE and momentum is along x axis

after collision

for rebacca

velocity of rebacca along x axis=6sin52.1

velocity of rebacca along y axis=6cos52.1

KE along x=(45*6sin52.1*6sin52.1)/2 1)

KE along y=(45*6cos52.1*6cos52.1)/2 2)

total energy=1)+2)

momentum along x=45*6sin52.1

momentum along y=45*6cos52.1

for daniel

let daniel velocity vmakes angle a with x axis in clockwise direction

such that momentum and energy remain conserved before and after collison

KE along x=60*vsina*vsina/2

KE along y=60*vcosa*vcosa/2

momentum along x axis=60*vsina

momentum along y axis=60*vcosa

equating

60*vcosa+45*6cos52.1=0

vcosa=-(45*6cos52.1)/60 3)

equating

45*6sin52.1+60*vsina=45*13

vsina=45*(13-6sin52.1)/60 4)

dividing 4)/3)

tana=(13-6sin52.1)/(6cos52.1)

a =tan inv (13-6sin52.1)/(6cos52.1)

putting it in 3)

we will get

v=-(45*6cos52.1)/(60*cosa)

change in KE of rebecca=(KEf-KEi)of rebecca

change in KE of rebecca=(KEf-KEi)of daneil

KEf=KE along x axis+KE along y axis

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