This 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.
A man mass 75 kg and a boy mass 25 kg dives off the end of a boat of mass 300 kg so that their relative horizontal velocity with respect to the boat is 3 m/s. if initially the boat is at rest find its final velocity if (1) the two dive off simultaneously (2) the man dives first followed by the boy
3 answers
Second 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.
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.
I 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.
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.