Two ice skaters, Daniel (mass 69.1 kg) and Rebecca (mass 50.7 kg) are practicing

Question

Two ice skaters, Daniel (mass 69.1 kg) and Rebecca (mass 50.7 kg) are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 12.9m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 6.9 m/s at an angle of 50.7° from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

(a) What are the magnitude and direction of Daniel's velocity after the collision?
________________ m/s
_______________° (clockwise from Rebecca's original direction of motion)

(b) What is the change in total kinetic energy of the two skaters as a result of the collision?
_______________ J

Explanation / Answer

Let initially rebecca is moving in +x direction.

before collsion,

m1 = 50.7 kg(Rebecca), u1 = 12.9 m/s

m2 = 69.1 kg(Daniel), u2 = 0

after the collsion,

v1 = 6.9 m/s

v2 = ?

let theta is the angle made by Daniel with +x axis

a) Apply conservation of moemntum in x-direction

m1*u1x + m2*u2x = m1*v1x + m2*v2x

50.7*12.9 + 0 = 50.7*6.9*cos(50.7) + 69.1*v2x

v2x = (50.7*12.9 - 50.7*6.9*cos(50.7) )/69.1

= 6.26 m/s

Apply conservation of moemntum in y-direction

m1*u1y + m2*u2y = m1*v1y + m2*v2y

0 + 0 = 50.7*6.9*sin(50.7) + 69.1*v2x

v2y = - 50.7*6.9*sin(50.7)/69.1

= -3.92 m/s

V2 = sqrt(v2x^2 + v2y^2)

= sqrt(6.26^2 + 3.92^2)

= 7.38 m/s <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<------------Answer

Direction : theta = tan^-1(v2y/v2x)

= tan^-1(3.92/6.26)

= 32 degrees <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<------------Answer

b) delta_KE = 0.5*m1*u1^2 - (0.5*m1*v1^2 + 0.5*m2*v2^2)

= 0.5*50.7*12.9^2 - (0.5*50.7*6.9^2 + 0.5*69.1*7.38^2)

= 1130 J <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<------------Answer

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