Grade 11Mechanics

A 50kg man is standing at one end of a 25m long boat.he starts running towards at the other end.On reaching the other end his velocity is 2m/s. If the mass of the boat is 200 kg , final velocity of the boat is?

Vishwareddy

7 Years agoGrade 11

2 Answers

Arun

7 Years ago

Momentum Before = Momentum After
0 = (50 Kg) • (2 m/sec) + (50 + 200 Kg) • (Vв)
Vв = - 2/5 m/sec

... where the boat travels in the opposite direction of the runner.

SIDDHI AMRALE

7 Years ago

Here the external unbalanced force is 0, so the law of conservation of momentum can be applied.

Mass of man (Mm)=50

Mass of boat(Mb)=200

Velocity of man(Vm)=2

Velocity of boat(Vb)=?

External force=0

Hence, initial momentum is equal to final momentum

Pi=Pf

boat and man initially at rest

Pi=0

Pf= Pm + Pb

Here, final momentum will be 0 as initial momentum is 0

0= MmVm +( Mb+Mm)Vb

As mass of boat will also contain mass of man as man is standing on the boat

0=50×2+(200+50)Vb

-250 Vb = 100

Vb= -2/5

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A 50kg man is standing at one end of a 25m long boat.he starts running towards at the other end.On reaching the other end his velocity is 2m/s. If the mass of the boat is 200 kg , final velocity of the boat is?

Vishwareddy

2 Answers

Arun

SIDDHI AMRALE

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