A 45.0-kg girl is standing on a 161-kg plank. The plank, originally at rest, is

Question

A 45.0-kg girl is standing on a 161-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.54 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.) (a) What is her velocity relative to the surface of ice? Correct: Your answer is correct. . m/s (b) What is the velocity of the plank relative to the surface of ice?

Explanation / Answer

Momentum is conserved.

Initial momentum = 0 (because she's at rest).
Final momentum = 0 (because it has to equal initial momentum)

Since the final momentum (while she's walking) is zero, this means the girl's momentum in one direction must exactly cancel out the plank's momentum in the opposite direction.

Let v_g = girl's velocity relative to ice.
Let v_p = plank's velocity relative to ice.

total momentum = 0
(45kg)(v_g) + (161kg)(v_p) = 0

We also know that the girl's velocity relative to the PLANK is 1.54 m/s. You get relative velocity by subtracting one velocity from another. So:

Girl's velocity relative to plank = 1.54 m/s
v_g v_p = 1.54 m/s

To recap, you have these two equations:

(45kg)(v_g) + (161kg)(v_p) = 0 ……….equation 1
v_g v_p = 1.54 m/s …….equation 2

That's 2 equations in 2 unknowns; so you can use algebra to solve for v_g and v_p. (v_g is the answer to Part 1; v_p is the answer to Part 2.)

Multiply equation 2 with 45 and substract from equation 1

(45kg)(v_g) + (161kg)(v_p) = 0

-(45kg)(v_g) - (45kg)(v_p) = - 69.3

We get,

(116kg)(v_p) = 69.3

So, (v_p) = 0.5974 m/s

Put this value in equation 2

v_g v_p = 1.54 m/s

v_g – 0.5974 = 1.54 m/s

So,

v_g = 1.54 + 0.5974 m/s

v_g = 2.1374 m/s

Therefore answers are,

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