How can you find the rate of change of momentum?

[Synxervious]

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?

 

[collinsmark]

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?

 

[PhysicMeister]

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)?

 

[cupid.callin]

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

momentum is not zero

 

[asteriatic]

I would first start by clarifying the question and making sure I understand what is being asked. In this case, we are trying to find the rate of change of momentum of the bus, which is a measure of how much the momentum of the bus changes over time.

To find this, we can use the formula for momentum, p=mv, where p is momentum, m is mass, and v is velocity. In this case, the bus has a constant velocity of 6.2 m/s, so we can use this value for v. The mass of the bus is given as 8.5 x 10^3 kg, so we can plug this into the equation and solve for p.

p = (8.5 x 10^3 kg)(6.2 m/s) = 52,700 kg*m/s

This is the momentum of the bus at a given instant. To find the rate of change of momentum, we need to divide this value by the time interval over which the momentum changes. However, in this problem, we are not given a specific time interval. Therefore, we can assume that the rate of change of momentum is constant, and we can use the average velocity over the entire time period instead of a specific time interval.

So, the rate of change of momentum (dp/dt) is equal to the change in momentum (p) divided by the change in time (t).

dp/dt = (p2 - p1)/(t2 - t1)

In this problem, the bus is traveling at a constant speed, so the change in velocity is zero. Therefore, the rate of change of momentum is also zero.

For part d, we can use the formula for work, W=Fd, where W is work, F is force, and d is distance. In this case, we are trying to find the magnitude of the resistive forces acting on the bus. The work done by these forces would be equal to the change in kinetic energy of the bus, which can be calculated using the formula KE=1/2mv^2.

So, we can set up an equation:

Work done by resistive forces = Change in KE

W = 1/2mv^2

To solve for the resistive force, we need to know the distance traveled by the bus. This information is not given in the problem, so we cannot calculate the magnitude of the resistive force without