Physics Problem Please Help!!!?

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1)This is a linear momentum problem where linear momentum is conserved.
Let P1 and P3 be the linear momenta of the 1st ball before and after collision respectively and,
Let P2 and P4 be the linear momenta of the 1st ball before and after collision respectively

=>P1+P2=P3+P4
Where P=MV (vector relationship)

=>M1V1+M2V2=M1V3+M2V4

M1=0.255kg
M2=??
V1=7.7m/s
V2=0
V3=-3.7m/s
v4=??

=>M1V1=M1V3+M2V4
=>M2V4=M1(V1-V3)
=>M2=M1(V1-V3)/V4..................1

Kinetic energy is conserved:
1/2M1V1^2+1/2M2V2^2=1/2M1V3^2+1/2M2V4^2
=M1V1^2=M1V3^2+M2V4^2......................2

replace 1 in 2
=>M1V1^2=M1V3^2+M1(V1-V3)V4
=>V4=(V1^2-V3^2)/(V1-V3)
=>V4=(7.7^2-3.7^2)/(7.7+3.7)
=>V4=4m/s

=>M2=M1(V1-V3)/V4
=>M2=0.255(7.7+3.7)/4
=>M2=0.727kg

2)This is also a linear momentum problem where linear momentum is conserved
The momentum to be considered in this case is the momentum of the air leaving the plane which is directed backwards. This momentum of the air would cause a thrust of equal amount yet in the opposite direction. This is called conservation of linear momentum. It is the same concept as when the gun repels backward while shooting.

Pplane=Pair
=>Mplane*Vplane=Mair*Vair

Pair=Mair*Vair
Mair leaving the plane is a mixture of air and gas of fuel burned.
and since mass is conserved => amount of fuel burned is equal to amount of fuel gas craeted

=>Mair=120+4.6=124.6kg
Vair=(600+290)m/s since relative to the plane
Vair=790m/s


F=Mplane*Vplane/t=Mair*Vair/t
t= 1s
=>F=124.6*790
=98434N

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