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April 19, 2014

April 19, 2014

Posted by **sjeba** on Friday, August 30, 2013 at 5:54am.

- science -
**Devron**, Friday, August 30, 2013 at 6:12amThis is a momentum problem and momentum is conserved:

M1V+M2V=M3V

Let

M1=75 kg

M2=25 kg

M3=300 kg

V1=3 m/s

V2= 3 m/s

V3=?

(3m/s)*(75 kg +25 kg)/300 kg=V3

v=1 m/s

Second problem:

Answer Should be the same; momentum is conserved.

But I am not 100% sure.

- science -
**Graham**, Friday, August 30, 2013 at 7:06amSecond Problem:

Momentum is still conserved on each leap (of course) but note that when the boy jumps off at 3m/s _relative_ to the boat, the boat is no longer at rest.

M1 = 75 kg mass of man

M2 = 25 kg mass of boy

M3 = 300 kg mass of boat

V1 = 3 m/s launch velocity (relative to boat)

V2 = ? velocity of boat (and boy) after man leaps.

V3 = ? velocity of boat after boy leaps.

First jump:

0 = M1 V1 + (M2+M3) V2

or

V2 = -M1 V1 / (M2+M3)

Second jump:

(M2+M3) V2 = M2 (V2+V1) + M3 V3

or

V3 = V2 - M2 V1 / M3

Substituting from the first jump:

V3 = -M1 V1/(M2+M3) - M2 V1/M3

Thus using the values:

V3 = -(75)(3)/(75+300) - (25)(3)/(300)

Finally:

V3 = -0.85 m/s

The boat does not move back as fast in this case, because some of the impulse from the man's leap has been imparted to the boy.

- science -
**Devron**, Friday, August 30, 2013 at 5:27pmThis is a momentum problem and momentum is conserved:

M1V+M2V=M3V

Let

M1=75 kg

M2=25 kg

M3=300 kg

V1=3 m/s

V2= 3 m/s

V3=?

(3m/s)*(75 kg +25 kg)/300 kg=V3

v=1 m/s

Second problem:

Answer Should be the same; momentum is conserved.

But I am not 100% sure.

- science -
**Devron**, Friday, August 30, 2013 at 10:37pmI do not know how my answer earlier was reposted, but I think I should have went through the math. There is agreement that the velocity will be 1m/s for the first question. However, there is disagreement concerning the velocity for the second question.

Let

M1=300 kg

V1i=0

V1f=?

M2=25kg

V2i=0

V2f=?

M3=75kg

V3i=0

V3f=-3m/s

***I don't know what direction that each is traveling, but I will let the jumpers' velocity be negative when they jump, since that will imply in the opposite direction.

Momentum Initial=Momentum Final

M1V1i + M2V2i+M3V3i=M1V1f + M2V2f+M3V3f

300kg(0) + 25kg(0) + 75kg(0)=300kg(V1f) + 25kg(V2f) + 75kg(-3m/s)

0=300kg*(V1f) + 25 kg*(V2f) + (75kg)*(-3m/s)

V1f=V2f, so both equal V.

0=300kg*(V) + 25 kg*(V)+(75kg)*(-3m/s)

0=-225J + V(300kg+ 25 kg)

0=-225J + V(325kg)

225J/(325kg)=V

0.692m/s=V

Now, when the boy jumps off, the boat and the boy have a initial velocity of 0.692m/s, but the final velocity of the boy will be -3m/s in the opposite direction.

So let

M1=300 kg

V1i=0.692 m/s

V1f=?

M2=25kg

V2i=0.692 m/s

V2f=3 m/s

M1V1i+M2V2i=M1V1f+M2V2f

V1f=V2f, so both equal V, which is equal to 0.692 m/s

M1V+M2V=M1V1f+M2V2f

Solve for V2f

(300kg*0.692m/s)+ (25kg*0.692m/s)=(300kg)V1f + (25kg*-3m/s)

0.692m/s*(300kg+25kg)=(300kg)V1f + (25kg*-3m/s)

225J=300kg*V1f + (-75J)

225J+75J=300kg*V1f

300J=300kg*V1f

300J/300kg=V1f

1 m/s=V1f

It doesn't make a difference if they jump at the same time or right after each other. The only way that the velocity of the boat will change is if there is a long protracted wait after the initial jump.

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