Two ice skaters, Daniel (mass 70.0 kg ) and Rebecca (mass 45.0 kg ), are practic

Question

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.

Part A

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

Part B

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

Part C

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

Explanation / Answer

m1 = 70 kg , m2 =45 kg , u1 =0 , u2 = 14 m/s

v2x = 8cos(55.1) =4.58 m/s

v2y = 8 sin(55.1) = 6.56 m/s

From conservaion of momentum along x direction

m1u1x+m2u2x =m1v1x+m2v2x

0 + (45*14) = (70*v1x) +(45*4.58) ... (1)

From conservation of momentum along y direction

m1u1y +m2u2y = m1v1y+m2v2y

0 = (70*v1y) +(45*6.56) ..(2)

By solving (1) and (2) we get

v1x =6.056 m/s

v1y = -4.217 m/s

(a) v1 = ((6.056)^2+(-4.217)^2)^0.5

v1 = 7.38 m/s

(b) direction tan(theta) = 4.217/6.056

theta = 34.85 degrees below the initial direction

(or) theta = 360 -34.85 = 325.15 degrees with initial direction

(c) Initial kinetic energy K1 =(1/2)m1u1^2 +(1/2)m2u2^2

K1 = 0 +(0.5*14*14*45) =4410 J

K2 = (1/2)m1v1^2 +(1/2)m2v2^2

K2 = (0.5*70*7.38*7.38) +(0.5*45*8*8)

K2 =3346.25 J

Change in kinetic energy

K1 -K2 = 4410 -3346.25

=1063.75 J

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