Question

A 45.0 kg girl is standing on a 159 kg plank. The plank, originally at rest,...

A 45.0 kg girl is standing on a 159 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 1.38 m/s relative to the plank.
(a) What is her velocity relative to the surface of ice?
1 m/s
(b) What is the velocity of the plank relative to the surface of ice?
2 m/s
0 0
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Answer #1
Concepts and reason

The problem deals with the concept of the momentum conservation which is a direct consequence of the newton’s third law and the concept of the relative velocity as the relative velocity of the girl is to be determined with respect to the ice and the velocity of the plank relative to the surface of ice.

Fundamentals

According to the law of conservation of momentum the initial and final momentum of an object both in magnitude and direction remains same until or unless the object is acted upon by some external force.

According to the concept of relative velocity if two bodied are moving in opposite direction then the relative velocity will be the addition of their individual velocity and if two bodied are moving in same direction then the relative velocity will be the subtraction of their individual velocity.

The momentum is defined as the quantity of motion of a moving body, measured as a product of its mass and velocity i.e.

P=mxy

Here is the momentum, is the mass of the object and is the velocity.

(a)

Mass of the plank is M = 159kg

Mass of the girl is m=45kg

Velocity of the girl relative to the plank is v = 1.38 m/sec

The velocity of the plank plus girl is .

From the conservation of momentum the momentum of the plank plus girl in addition to the momentum of girl will be zero as:

(Momentum of plank + girl) +(Momentum of girl)= 0)
(M + m)v, +(m)v = 0
(M +m)v2 =-(m)
v2 = (M +m)

Therefore substitute the values mass of plankM = 159kg
, mass of the girlm=45kg
, velocity of girl relative to plank v = 1.38 m/sec
in the equation as:

-(45kg)1.38m/sec
v2 =
(159+45)
-(45kg) 1.38 m/sec
(204)
=-0.304 m/sec

Since the ice surface is frozen lake and girl is moving on it so the relative velocity will get added. Therefore the velocity of the girl relative to the ice surface is as:

vi +v= 1.38+(-0.304)
= 1.076m/sec

(b)

The velocity of the plank plus girl isv2 =-0.304m/sec
.

Since the ice surface is frozen lake and plank is moving with girl on it so the relative velocity will get added. Therefore the velocity of the plank relative to the ice surface is as:

v2 =-0.304m/sec

Ans: Part a

The velocity of the girl relative to the ice surface is1.076 m/sec
.

Part b

The velocity of the plank relative to the ice surface is-0.304 m/sec
.

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