Physics Momentum Questions - Very Confused

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1. When a force of 300 N acts for 10 s on a body of mass 50 kg moving initially at 4 ms-2. What is the body’s change of momentum?

For question 1: change in momentum= 300 * 10 = 3000 kg ms^-1

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2. A van of mass 1000 kg is towing a car of mass 800 kg. When the van has a horizontal thrust of 3200 N the total frictional force on the van is 1000 N and the total frictional force on the car is 1500 N.
Find the tension in the tow rope ?

For question 2: I am ashamedly confused. I know that Tension = mass * acceleration.
To find the acceleration of the truck:
F = m * a therefore a =F/m
a = 2200/1000
a=2.2 ms^-2
Tension of truck = 1000 * 2.2 = 2200 N
To find the acceleration of the car:
a = 1700/800=2.125
tension of the car= 800 * 2.125= 1700 N
Overall tension on the rope = 2200 - 1700 N = 500N
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3. Calculate the force required to bring a truck of mass 10 tonnes, initially moving at 25 ms-1, to rest over a distance of 50 m?



Question 3: Using Newton's second law, F=m*a

a = F/m

10 tonnes = 10000kg

Acceleration is the change in velocity divided by the change in time (dv/dt) The change in velocity is negative because we go from 25 to 0 = -25 ms^-1

Acceleration = change in velocity / time

The change in time is positive so the acceleration has a negative value:

Acceleration = -25/50 = -0.5ms^-2

Force = m * a

Force = 10000 * -0.5 = -5000N

The force found above is the force that the truck is exerting so one is required to calculate the force which is equal in magnitude but opposite in direction.

Therefore the force needed to stop the mass of 10000 kg in 50 seconds is 5,000 N.

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4. If girl is standing on the ground, her weight is the force acting downwards, therefore, find the Newton’s Third Law pair to this force?

Question 4:The Newton's third law pair would be the normal reaction from the ground acting upwards on the girl.
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6. a. A ball hits the ground moving at 10 ms-1 and then bounces back, moving upwards at 6 ms-1. Noting that the ball has a mass of 100 g, what is its change of momentum?
b. If the ball remained in contact with the ground for 0.04 s, what was the average force of the ball on the ground?

Question 6: a. change in momentum = Force * time taken
However, I do not think the above equation is necessarily required at this stage, since the change in velocity is given (as in an acceleration calculation) I am very stuck on part a and b of this question, I think I keep overthinking the problem and confusing myself.

Hi Alexandramartis,

Welcome to TSR.

Thanks for your questions. It would be appreciated that you can break down the number of questions and post them separately in future. It can be a daunting task for some helpers to go through all the questions.
For easy view for such a long list, it may be better that you post as

Question 1
Attempt

Question 2
Attempt
…..

It is very difficult to scroll up and down to view the question and your attempts.

For Question 1, I agree with your answer. I am not sure is there any typo in the question “ initially at 4 ms-2”

For this question, it would be useful that you draw free-body to show all the forces and then write Newton’s 2^nd law to solve the problem.

Combined the van and car as a single object and then write down the Newton’s 2^nd law
Forward thrust - frictional force on the van - frictional force on the car = (mass of van + mass of car) × acceleration

Then write Newton’s 2^nd law for the car or van to find the tension.

Not sure why are you writing truck in your attempt.

You should note that



so the following is not correct:


You can use the conservation of energy:



OR

A longer method is to find the acceleration of the truck over 50 m using suvat and then find the time taken for such acceleration and initial velocity using suvat again.
Finally, use F = ma.

I think you have been misled by the question. You need to note Newton’s Third Law pair of forces have the following characters.
-same nature of forces
-acting on different body
-equal in magnitude
-opposite in direction

The normal reaction from the ground is acting on the girl and her weight is also acting on her and the nature of the forces is different.

For (a),
Change in momentum = final momentum – initial momentum
which is (in this case) equal to


For (b) is just using the formula that you stated:

Therefore, change in momentum = Force * time taken.

1. When a force of 300 N acts for 10 s on a body of mass 50 kg moving initially at 4 ms-2. What is the body’s change of momentum?

2. A van of mass 1000 kg is towing a car of mass 800 kg. When the van has a horizontal thrust of 3200 N the total frictional force on the van is 1000 N and the total frictional force on the car is 1500 N. Find the tension in the tow rope ?

3. Calculate the force required to bring a truck of mass 10 tonnes, initially moving at 25 ms-1, to rest over a distance of 50 m?

4. If girl is standing on the ground, her weight is the force acting downwards, therefore, find the Newton’s Third Law pair to this force?

Question 4:The Newton's third law pair would be the normal reaction from the ground acting upwards on the girl.

5. If a man was below deck on a boat, in a cabin with no windows but lights hanging from the ceiling and marbles on the flat floor, by witnessing the lamps hanging straight down and the marbles beginning to move about, what can you tell?
i. You can tell whether you are moving or at rest.
ii. You can tell whether you accelerating or not.
iii. You can tell whether you are changing speed, but cannot tell whether you are changing direction.
iv. You can tell whether you are changing direction, but cannot tell whether you are changing speed.

Question 5: I think that iv "You can tell whether you are changing direction, but cannot tell whether you are changing speed may be the answer. However, I am not certain and actually believe that a sufficient solution. would justify their response with evidence using Newton's laws, but I cannot situate a firm answer from which to construct any such argument.

6. a. A ball hits the ground moving at 10 ms-1 and then bounces back, moving upwards at 6 ms-1. Noting that the ball has a mass of 100 g, what is its change of momentum?
b. If the ball remained in contact with the ground for 0.04 s, what was the average force of the ball on the ground?

Question 6: a. change in momentum = Force * time taken

3. Calculate the force required to bring a truck of mass 10 tonnes, initially moving at 25 ms-1, to rest over a distance of 50 m?

Question 3: Sorry no, acceleration = change in velocity/change in time

Kinetic energy= 1/2 *mass* velocity^2
Kinetic energy = 1/2 *10000*25^2
Kinetic energy =3,125,000 J

Kinetic energy =1/2*10000*0^2
Kinetic energy= 0J

Change in KE = 3,125,000 J - 0J = 3,125,000J

Work done = force * distance moved
3,125,000 J = Force * 50 m
Force = Work done/ distance moved
Force = 3,125,000 / 50
Force = 62,500 N

Would this be correct?

Using SUVAT I am not sure how to accomplish this?
S=50m
U=25
V=0
A=?
T=?

2. A van of mass 1000 kg is towing a car of mass 800 kg. When the van has a horizontal thrust of 3200 N the total frictional force on the van is 1000 N and the total frictional force on the car is 1500 N.
Find the tension in the tow rope ?

Question 2 : I have draw and attached a free body diagram, though I do not know whether this is correct.

I calculated Weight by W=m*g so W = 1800 * 9.81= 17658 N
Inputing the values into Newton's 2nd Law:
Force = mass * acceleration
Force = 3200 N - (1000N + 1500N)
Force = 3200 N -2500N = 700N
700 N = (1000kg + 800kg) * a
700 N = 1800 * a
a = F/m
a=700/ 1800
a = 0.39 ms^-2 (2.d.p)
F = 1800 * 0.39 ~ 700N
So is this the tension of the rope?

This is called impulse.

Dear Roger, thank you very much for your replies and your explination/hints. I will address each question, though admittedly I am still confused about some questions.

Do you mean that F*t = mass * velocity
Or that force * time = change in momentum (impulse)

Fire away.


The latter - the change in mv, not its absolute value.

Impulse is the change in momentum. The starting point is irrelevant. They have deliberately given you extraneous information (the mass and speed), which you need to ignore.

No. The car (horizontally) has a frictional force (1500N) and the tow-rope tension. It would be going backwards with that tension

You get a from the net force and total mass, but then need to consider the car in isolation. The net force on is its ma.

You should check your answer (should be in the range of 1800 - 1900N) by considering the forces on the van: F=ma, so 3200 - 1000 - T = 1000a

Kinetic energy and work done is a great approach - I didn't propose it because it's often seen as a more advanced concept than SUVAT. Yes, it's right.

Using SUVAT, you have s=50m, u=25m/s, v=0, so use $v^2=u^2+2as$ to get a. That gives:

$a=\frac{v^2-u^2}{2s}=-6.25 ms^{-2}$

Then F=ma:

$F=ma=-62,500 N$ (i.e. against the direction of travel)

No. Hint: At the centre of mass of the Earth.