Question

Two ice skaters, Daniel (mass = 65.0 kg) and Rebecca (mass = 45.0 kg) are practicing on the ice. Daniel stops to tie his lace, and while at rest he is struck by Rebecca, who is moving 12.0 m/s before she collides with him. After the collision, Rebecca is moving forward at 8.50 m/s at a 53 degree angle with respect to her initial direction. What is the velocity vector (magnitude and direction, or x- and y-components) of Daniel after the collision? Assume frictionless ice. Hint: Momentum is conserved separately in both the x- and y-directions.

Answer #1

Let, M_{r}= Rebecca's mass = 45 kg

M_{d} = Danial's mass = 65 kg

V_{ri} = Rebecca's initial velocity =12 m/s

V_{rf} = Rebecca's final velocity = 8.50 m/s

_{r}
= 53°

Initial momentum situation is:

1) L_{xi} = M_{r}×V_{ri}.

2) Ly_{i} =0

Final momentum equation is:

3) L_{xf} = M_{r}×V_{rf}Cos_{r}
+M_{d}×V_{dxf}

4) L_{yf} = M_{r}V_{rf}Sin_{r}
+ M_{d}V_{dyf}

Conservation of momentum in each of the two dimensions implies following:

L_{xi} =L_{xf}

i.e 5)M_{r}V_{ri} =
M_{r}V_{rf}Cos_{r}
+M_{dxf}

L_{yi} = L_{yf}

i.e 6)0= M_{r}V_{rf}Sin_{r}
+ M_{d}V_{dyf}

solving (5) for V_{dxf}:

V_{dxf} = (M_{r}V_{ri}-
M_{r}V_{rf}Cos_{r})÷
M_{d}

= (45×12- 45×8.50×cos53)÷65

= 4.77 m/s

Solving (6) for V_{dyf} :

V_{dyf} = - (45×8.50×Sin53)÷65

= -4.69 m/s

Danial's final angle of motion :

_{d}
= tan^{-1}(-4.699/4.77) = 44.57°

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at rest, is struck by Rebecca, who is moving at 12.0 m/s before she
collides with him. After the collision, Rebecca has a velocity of
magnitude 8.00 m/s at an angle of 54.1 ∘ from her initial
direction. Both skaters move on the frictionless, horizontal
surface of the rink.
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direction. Both skaters move on the frictionless, horizontal
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