Physics Momentum Questions - Very Confused Watch

Alexandramartis
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Good afternoon, I have an array of questions concerning momentum which I ma finding rather confusing.

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?
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.

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?

I understand there is rather a lot here but to be perfectly candid momentum is not my forte, I am rather unaccustomed to the topic and have only recently been introduced to it in any detail.
I comprehend that Force = change in momentum / time taken

Therefore, change in momentum = Force * time taken.

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

I think that this answer is defiantly wrong but honestly I am trying to fathom an answer and want to learn how to solve these types of questions.

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
Accleration = 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.

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

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.

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.

I really appreciate any responses and am tremendously grateful for further explanation of the topic of momentum and answering this style of question.
Thank you ☺️
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Eimmanuel
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(Original post by Alexandramartis)
......
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
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”
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Eimmanuel
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(Original post by Alexandramartis)
......
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
....

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

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

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

Not sure why are you writing truck in your attempt.
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Eimmanuel
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(Original post by Alexandramartis)

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.
You should note that

 \dfrac{\text{change in velocity}}{\text{distance travelled}} \ne \text{acceleration}

so the following is not correct:
Acceleration = -25/50 = -0.5ms^-2

You can use the conservation of energy:

Change in KE = Work done in stopping the truck

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.
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Eimmanuel
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(Original post by Alexandramartis)
.....
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.
.....
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.
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Eimmanuel
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(Original post by Alexandramartis)
....
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.


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

For (b) is just using the formula that you stated:
change in momentum = Force * time taken
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Alexandramartis
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(Original post by Eimmanuel)
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”
Hello Eimmanuel, thank you very much for your reply and welcome. Sorry in future I will break down my queries as suggested.

Question 1: I think that you are correct about the typo, I checked the original question and that is how it is written but I think it is meant to be 4ms^-1.
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Alexandramartis
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(Original post by Eimmanuel)
For this question, it would be useful that you draw free-body to show all the forces and then write Newton’s 2nd law to solve the problem.

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

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

Not sure why are you writing truck in your attempt.
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?
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Alexandramartis
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(Original post by Eimmanuel)
You should note that

 \dfrac{\text{change in velocity}}{\text{distance travelled}} \ne \text{acceleration}

so the following is not correct:
Acceleration = -25/50 = -0.5ms^-2

You can use the conservation of energy:

Change in KE = Work done in stopping the truck

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.
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=?
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Alexandramartis
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(Original post by Eimmanuel)
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.
Question 4: Would the correct pairing instead be the gravitational force of the girl acting upwards on the Earth?
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Alexandramartis
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(Original post by Eimmanuel)
For (a),
Change in momentum = final momentum – initial momentum
which is (in this case) equal to
mass × change in velocity

For (b) is just using the formula that you stated:
change in momentum = Force * time taken
Question 6: Initial momentum = mass * velocity
Initial momentum = 100 * 10
Initial momentum = 1000 kg ms^-1

Final momentum = 100 * 6
Final momentum = 600 kg ms^-1
Change in momentum = final momentum – initial momentum
Change in momentum = 600 - 1000
Change in momentum = -400 kgms^1

Final momentum = mass * change in velocity
Final momentum = 100 * (v-u)
Final momentum = 100 *(6-10)
Final momentum = -400 kg ms^-1

b) Change in momentum = force * time taken
Force = Change in momentum / time taken
Force = -400 / 0.04
Force = -10000 N

Would this be right? Also, could you please take a look at my answer for question 5?
Thank you so much for all of your help I really appreciate it.
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RogerOxon
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(Original post by Alexandramartis)
Therefore, change in momentum = Force * time taken.
This is called impulse.

(Original post by Alexandramartis)
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?
This is about impulse (force * time) which equals the change in momentum. As F=ma, (constant) Ft=change in mv.

(Original post by Alexandramartis)
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 ?
This is a common acceleration problem. Calculate the acceleration from the net force and total mass, then consider the forces on just the car, calcuating the tow-rope tension required to give the deduced acceleration.

(Original post by Alexandramartis)
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?
SUVAT to find the decelleration, then F=ma.

(Original post by Alexandramartis)
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?
(Original post by Alexandramartis)
Question 4:The Newton's third law pair would be the normal reaction from the ground acting upwards on the girl.
No. Hint: At the centre of mass of the Earth.

(Original post by Alexandramartis)
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.
(Original post by Alexandramartis)
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.
Yes. Hint: Think about the direction of the forces (tension and gravity) on the hanging lights.

(Original post by Alexandramartis)
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?
(Original post by Alexandramartis)
Question 6: a. change in momentum = Force * time taken
Hint: Momentum is a vector. Be careful with your convention for positive and watch your minus signs. The last part is about impulse.
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(Original post by Alexandramartis)
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?
(Original post by Alexandramartis)
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=?
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)
Last edited by RogerOxon; 3 months ago
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RogerOxon
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(Original post by Alexandramartis)
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 ?
(Original post by Alexandramartis)
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?
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
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Alexandramartis
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(Original post by RogerOxon)
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.

Question 1:
Impulse = force * time taken
You state that Ft=change in mv
Do you mean that F*t = mass * velocity
Or that force * time = change in momentum (impulse)

In the first instance if ft=mv then
300 * 10 ≠ 50*4
or that change in momentum = 300 * 10 = 3000 kgms^-1
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RogerOxon
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(Original post by Alexandramartis)
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.
Fire away.

(Original post by Alexandramartis)
Do you mean that F*t = mass * velocity
Or that force * time = change in momentum (impulse)
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.
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Alexandramartis
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(Original post by RogerOxon)
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.
Thank you very much for your clarification and for your welcoming and hospitable help! 😁
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Alexandramartis
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(Original post by RogerOxon)
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
Question 2:
Hello Roger, I am still a bit confused. You state that: net force = m*a
And that to find the acceleration I assume you rearrange this to acceleration = net force/ total mass
Is this net force on the van alone? which would be 3200-1000-T = 1000 * acceleration (as you have shown)
2200 - T = 1000 * a

And then do you do the same, finding the net force and acceleration but for the car in isolation:
3200-1500-T=800*a
1700-T=800*a

Sorry I am rather perplexed 🤔
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Alexandramartis
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(Original post by RogerOxon)
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)
Dear Roger,

Question 3
Thank you for your explanation of how to find the force using suvat. I have written out your calculations as above just so I can understand how you arrived at your conclusion and see that this was a simpler approach (and preferable).
Would my method using KE=work done be correct in that I arrived at an answer of 62,500 N to bring the truck to rest (which is not a negative value as in the suvat approach since the truck is being brought to rest?)
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(Original post by RogerOxon)
No. Hint: At the centre of mass of the Earth.
Dear Roger, below are my thought on question 4.

Question 4: Are you referring to a force at the centre of the Earth? Since I thought that gravity at the centre of the Earth was zero (since there is equal force in all directions).

I thought that the Newton's third law pair to weight would be the normal force (since this would act upwards in the opposite direction but be of equal magnitude to her weight?) I actually was just revising and found some multiple choice questions online which is where I stumbled across this problem. The options were:
1. The normal reaction from the ground acting upwards on the girl.
2. The gravitational force of the girl acting upwards on the Earth.
3. The normal contact force of the girl on the ground.
4. The frictional force of the girl on the ground.


I reasoned that the answer from the options given was 1 since it had to be a force acting in the opposite direction (upwards). Moreover it could not be 4 because the girl was not said to be moving. The more I think about it the more confused I have become, I apologise for my miscomprehension.
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