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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?

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?
 
 

Grade:11

2 Answers

Arun
25750 Points
5 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
28 Points
5 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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