In: Physics

# ((8.27)Q:7) Two ice skaters, Daniel (mass 70.0 kg ) and Rebecca (mass 45.0 kg ), are...

((8.27)Q:7)

Two ice skaters, Daniel (mass 70.0 kg ) and Rebecca (mass 45.0 kg ), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 14.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 55.1 ∘ from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

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

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

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

## Solutions

##### Expert Solution

Let initially rebecca is moving in +x direction.

before collsion,

m1 = 45 kg(Rebecca), u1 = 14 m/s

m2 = 70 kg(Daniel), u2 = 0

after the collsion,

v1 = 8 m/s at 55.1 degrees

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

45*14 + 0 = 45*8*cos(55.1) + 70*v2x

v2x = (45*14 - 45*8*cos(55.1) )/70

= 6.06 m/s

Apply conservation of moemntum in y-direction

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

0 + 0 = 45*8*sin(55.1) + 70*v2x

v2y = - 45*8*sin(55.1)/70

= - 4.22 m/s

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

= sqrt(6.06^2 + 4.22^2)

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

= tan^-1(4.22/6.06)

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

= 0.5*55*14^2 - (0.5*45*8^2 + 0.5*70*7.38^2)

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