How can you find the rate of change of momentum?

AI Thread Summary
The discussion focuses on calculating the rate of change of momentum for a bus traveling at a constant speed of 6.2 m/s on a 6-degree incline, concluding that the rate of change of momentum is zero due to constant speed. The input power to the bus's engine is determined to be 200,000 W, given its output power of 70 kW and 35% efficiency. The rate of increase of gravitational potential energy (GPE) is calculated as 54,000 J, although it is noted that this should be expressed as a rate (e.g., J/s). The conversation also explores estimating the magnitude of resistive forces acting on the bus, emphasizing the relationship between power, force, and velocity. The discussion highlights the importance of correctly interpreting units and understanding energy transfer in the context of the bus's motion.
Synxervious
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Hi guys. Need a little help with an IB paper(Just two questions!) and would really appreciate your help.

Homework Statement



A bus is traveling at a constant speed of 6.2m/s along a section of road that is inclined at an angle of 6.0 degrees to the horizontal.

State the value of the rate of change of momentum of the bus.

b) The total output power of the engine of the bus is 70kW and efficiency of the engine is 35%. Calculate the input power to the engine.

c) The mass of the bus is 8.5 x 10^3 kg. Determine the rate of increase of gravitational potential energy of the bus.

d) Using the answer to c and data in b, estimate the magnitude of resistive forces acting on the bus.

Homework Equations



p=mv.

a= g sin theta.

F= ma.

GPE = mgh.

SUVAT equations.

The Attempt at a Solution



a) As it is constant speed, zero net force. Zero net force, zero momentum. Answer is zero.

b) Output power is 70000/x * 100% = 35%. Solving for x yields x=200000W.

c) Velocity is 6.2 m/s at a 6 degree incline. 6.2 sin 6 gives me the vertical component of velocity, 0.648 m/s. Rate of increase of GPE is mgh/t, so 8.5*10^3*9.8*0.648 = 54000 J.

(Is this right?)

d) No idea. @_@ Can someone lead me in the right direction?
 
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Synxervious said:
Hi guys. Need a little help with an IB paper(Just two questions!) and would really appreciate your help.

Homework Statement



A bus is traveling at a constant speed of 6.2m/s along a section of road that is inclined at an angle of 6.0 degrees to the horizontal.

State the value of the rate of change of momentum of the bus.

b) The total output power of the engine of the bus is 70kW and efficiency of the engine is 35%. Calculate the input power to the engine.

c) The mass of the bus is 8.5 x 10^3 kg. Determine the rate of increase of gravitational potential energy of the bus.

d) Using the answer to c and data in b, estimate the magnitude of resistive forces acting on the bus.

Homework Equations



p=mv.

a= g sin theta.

F= ma.

GPE = mgh.

SUVAT equations.

The Attempt at a Solution



a) As it is constant speed, zero net force. Zero net force, zero momentum. Answer is zero.

b) Output power is 70000/x * 100% = 35%. Solving for x yields x=200000W.

c) Velocity is 6.2 m/s at a 6 degree incline. 6.2 sin 6 gives me the vertical component of velocity, 0.648 m/s. Rate of increase of GPE is mgh/t, so 8.5*10^3*9.8*0.648 = 54000 J.

(Is this right?)
'Looks okay to me, with one exception. Your answer for c) isn't going to be Joules. A Joule is a unit of energy. You're looking for a unit of rate of change of energy.
d) No idea. @_@ Can someone lead me in the right direction?
From part b), you know that the bus's engine outputs a certain amount of power. From part c) (after you correct the units) you know that part of that power is going into getting the bus up the hill, thus increasing its gravitational energy as time goes on. So where does the rest of the power go? The bus's momentum isn't changing, so it's not going into that.

Once you figure out the above hint, here's another hint.

W = \vec F \cdot \vec d

Where W is work (a measure of energy), \vec F is force, and \vec d is displacement.

Putting things in terms of "deltas" (changes):

\Delta W = \vec F \cdot \Delta \vec d

Divide both sides by change in time, Δt.

Look at the left hand side of the equation. What's change in energy per unit time?

Look at the right hand side of the equation. What's change in displacement per unit time?
 
Last edited:
Yes, so P = Fv, and that power is wasted as resistive forces. However, I am just confused about one thing. When solving for F do we substitute the vertical velocity (0.648) into the equation or the velocity along the plane as given (6.2)?
 
Synxervious said:
a) As it is constant speed, zero net force. Zero net force, zero momentum change. Answer is zero.

momentum is not zero
 
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