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Physics Problem Please Help!!!?
1) A ball of mass 0.255kg that is moving with a speed of 7.7m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.7m/s .
A) Calculate the velocity of the target ball after the collision.
2) The jet engine of an airplane takes in 120 kg of air per second, which is burned with 4.6kg of fuel per second. The burned gases leave the plane at a speed of 600m/s (relative to the plane). If the plane is traveling 290m/s (650mi/hr ), determine the following quantities.
A) The thrust due to accelerated air passing through the engine.
1 Answer
- 8 years agoFavorite Answer
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
If u need more help email me at g_zayo@yahoo.com
Source(s): civil engineer