# a boy and boat were initially at rest before the boy stepped onto the dock. The mass of hte boy is 70kg and the mass of the boat is 200kg. The boy stepped forward with a speed of 3 m/s. How fast...

a boy and boat were initially at rest before the boy stepped onto the dock. The mass of hte boy is 70kg and the mass of the boat is 200kg. The boy stepped forward with a speed of 3 m/s. How fast did the boat go? If the boat came to rest in 15 seconds, how much frictional force did the water exert on the boat?

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This is a problem of linear momentum conservation. It can be written mathematically as

`m_(boy)*v_(boy)+m_(boat)*v_(boat)=0` .

Or, equivalently

`m_(boy)*v_(boy)=-m_(boat)*v_(boat)` .

Plugging into the equation the given values we have

`70kg*3 m/s=-200kg*v_(boat) rArrv_(boat)=-1.05 m/s` .

This is the initial velocity of the boat and the minus sign simply means that the direction of it is opposite to the direction of the velocity of the boy.

Now, from this velocity, we can calculate its initial linear momentum, that is

`P_(0)^(boat) = m_(boat)*v_(boat) =200kg*1.05 m/s=210 (kg*m)/s` .

When the boat stops due to frictional forces (mainly due to water, but also in a very small way due to air), its final velocity is zero, and so the final linear momentum of the boat is also zero. So for the boat until it stops, there is a net momentum change of

`Delta P_(boat)=210 (kg*m)/s=210 N*s`

In the last step, we used the unit of momentum in terms of newtons times seconds. Now from the impulse theorem we know that the Impulse is equal to the variation of linear momentum, and impulse is defined as force times time variation, i.e.

`I=FDelta t=Delta P`

The impulse due to friction forces acting on the boat is therefore

`F*15s = 210 N*s`

so that the friction force is

`F=210/15 = 14 N.`

Here we can use momentum balance. Since there is no external force momentum of boy and boat should be cancel out.

Let;

m1 = mass of boy = 70kg

m2 = mass of boat = 200kg

v1 = velocity of boy = 3m/s

v2 = velocity of boat = ?

Using momentum balance;

`m1v1 = m2v2`

`70*3 = 200*v2`

`v2 = 210/200 = 21/20 = 1.05m/s`

*So the velocity of the boat when the boy step in to the dock is 1.05m/s.*

With this initial velocity the boat starts to go along the water. But the friction of water act against this motion and ultimately the boat will stop.

Using velocity equation `v = u+at` ;

`0 = 1.05+a*15`

`a = -0.07m/(s^2)` (Negative because of deceleration)

Using `F = ma` to the opposite direction of motion;

`F = 200*(-(-0.07)) N`

`F = 14N`

**So the frictional force of water is 14N.**

The boy and his boat were initially at rest before the boy stepped onto the dock. The boy has a mass 70 kg and the mass of the boat is 200 kg.The boy steps forward with a speed of 3 m/s. To determine the speed of the boat as a result of this, use the law of conservation of momentum. In a closed system, if their are no external forces, the total momentum remains constant.

Here, the boy and the boat are taken as a closed system. When he steps onto the dock, the momentum of the boy is 70*3 = 210 kg*m/s. This should equal the momentum of the boat. If the speed of the boat is v, 200*v = 210 or v = 1.05 m/s. Note that the boat moves in the opposite direction to that in which the boy moves.

The speed of the boat is reduced to 0 in 15 seconds. The deceleration of the boat is (1.05 - 0)/15 = .07 m/s^2. The force of friction that results in the boat coming to a halt is equal to 0.07*200 = 14 N