zarkam21:
At a race track, a car of mass 1150 kg crashes into a concrete wall at a speed of 85 m/s. a. If the car comes to a stop when it hits the wall, what is the magnitude of the impulse applied to the car? b. The car crashes into the wall, stopping in 0.1 seconds. What force is applied to the car?
zarkam21:
A. 0 B. F=m*a F=1150*9.8 F=11270
Shadow:
Why is (a), 0?
zarkam21:
because it has come to a stop
Shadow:
What is impulse?
zarkam21:
change in momentum
Shadow:
Do you think there was a change in momentum/
zarkam21:
Well the part of the question where it says, "If the car comes to a stop when it hits the wall" makes me think that there is a zero somewhere in the answer or formula
Shadow:
Yes, the final velocity is 0. But it was traveling at 85m/s. Therefore what do we have?
zarkam21:
well the change in momentum so 85-0?
Shadow:
Well, it's technically 0 - 85 \[\Delta V = V _{f} - V _{i} = 0 - 85 = -85\] We don't want to play with negatives, but this helps you realize that an impulse was delivered that countered the velocity of the car, bringing it's velocity to zero.
Shadow:
Technically we do need the negative in this case though.
zarkam21:
oh okay so if it is negative it just equals to 0?
Shadow:
What equals to zero?
zarkam21:
no I'm saying it the change is negative in every case does it just mean its zero
Shadow:
The impulse?
zarkam21:
Yes
Shadow:
Impulse is equal to the change in momentum. If an object has an impulse of 0 applied to it, and therefore 0 change in moment, it should still be moving, with some velocity.
zarkam21:
In this case the answer is -85 right?
zarkam21:
and you said it is zero
zarkam21:
so i assumed that every negative velocity just means it is 0
Shadow:
The final velocity is zero because at that point the car has been stopped. But in order for a car to be stopped, there must be a change in velocity, thus a change in momentum, which is also an impulse delivered (of some magnitude).
zarkam21:
I think i got it
zarkam21:
Sorry I just wanted the concept to be clear :/
Shadow:
It's okay. If you have more questions or areas to explore, bring them up. I enjoy the topic and it's important that you understand it. This is applied physics, but curriculum nowadays is moving towards abstract (especially if you plan on taking AP Physics).
Shadow:
Basically it's good to understand the formulas now.
Shadow:
So do you know the magnitude of the applied impulse?
zarkam21:
wait so a is 0 we concluded that
zarkam21:
we are on b now right
Shadow:
If the wall applied an impulse of zero, what would happen to the car?
zarkam21:
well it wouldn't move
Shadow:
Nothing would happen to the car. It would keep moving.
zarkam21:
so same right like everything would stay the same
Shadow:
Yes
Shadow:
But it doesn't stay the same, since the velocity goes from 85 to 0. What must that mean?
zarkam21:
well 85-0=85
Shadow:
What are you saying when you write that
zarkam21:
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow Well, it's technically 0 - 85 \[\Delta V = V _{f} - V _{i} = 0 - 85 = -85\] We don't want to play with negatives, but this helps you realize that an impulse was delivered that countered the velocity of the car, bringing it's velocity to zero. \(\color{#0cbb34}{\text{End of Quote}}\)
zarkam21:
Well it stays the same so it would be like this??
Shadow:
Do you know what Vf stands for?
zarkam21:
Final velocity
Shadow:
What do you think I am saying through that post? Or how are you interpreting it?
zarkam21:
you are subtracting the initial velocity from the final velocity Hence, why it is 0-85 not 85-0
Shadow:
And from that, how are you concluding that the impulse has a magnitude of 0?
zarkam21:
well the answer is 0 :/
zarkam21:
Ugh I'm a little confused
Shadow:
Yeah, that is why I am trying to understand your thought process, of how you get to 0
zarkam21:
Well because i know you said it would be the same as A. and a was 0 because we discussed that it is vf-vi
zarkam21:
that is how I got that b is zero
Shadow:
\[J = F \times \Delta T = m \times \Delta V\] There is a change in velocity due to the car going from 85m/s to 0m/s. The car has some mass, therefore \[\Delta V = V _{f} - V _{i} = 0 - 85 = -85\] \[J = m \times \Delta V = 1150 \times -85 = -97750\]
Shadow:
The car's momentum prior to crashing was \[P = m \times V = 1150 \times 85 = 97750\]
Shadow:
A car with no velocity has a momentum of zero \[P = m \times v = 1150 \times 0 = 0\]
Shadow:
It has a final momentum of 0, with an initial momentum of 97750, but the change in momentum is -97750, thus how it came to zero. That change in momentum, is equal to what?
zarkam21:
the change of momentum is equal to zero?
Shadow:
The final momentum is zero because it has a velocity of zero
Shadow:
But a car going at 85m/s does not reach a velocity of zero, without an impulse being applied with a magnitude higher than 0
Shadow:
Any confusion?
zarkam21:
could you jst explain the magnitude concept. like why does it need to be higher than zero to be a velocity of zero
Shadow:
Because if the impulse is 0, that means there is 0 change in momentum. If there is zero change in momentum, then the car would maintain it's initial momentum, since there would be no change. If this is no change in p, then the car would still be in motion at 85m/s
Shadow:
But in our problem, it says that it has been stopped by the wall. Stopped = 0m/s, therefore an impulse must have been applied, since there was a change in momentum (as I illustrated above).
zarkam21:
okay i think i kindda get it
zarkam21:
well that means it would be 85 then right?
zarkam21:
it cant be 0 because if it was it would mean it is still in 85 motion but it is not it is in 0 motion at the final state
Shadow:
85 is the initial velocity, not the impulse.
zarkam21:
yes im just saying that if it is ending up at 0 it wont be 85
zarkam21:
we have the info that the ending is 0
Shadow:
Yes, the final velocity is 0
Shadow:
What wont be 85?
zarkam21:
change in momentum
Shadow:
Yeah, 85 is the change in V
zarkam21:
so is that the answer for b
Shadow:
No
Shadow:
I thought we work working on a xD
Shadow:
Since it isn't 0, but you have it as 0
zarkam21:
wait i thought we already solved a that it is 85
Shadow:
Impulse equals the change in momentum, not the change in velocity
Shadow:
\[\Delta V = V _{f} - V _{i} = 0 - 85 = -83\] \[\Delta p = p _{f} - p _{i}\] \[p _{f} = m \times v = 1150 \times 0 = 0\] \[p _{i} = m \times v = 1150 \times 85 = 97750\] \[\Delta p = 0 - 97750 = -97750\]
Shadow:
You can also solve for the impulse this way \[J = F \times \Delta T = m \times \Delta V\] \[J = impulse\] \[J = m \times \Delta V\] \[J = 1150 \times -85 = -97750\]
Shadow:
If a car has a momentum of 97750, how does it stop and achieve a final momentum of 0 (due to zero velocity). Well, the change in momentum must be equal in magnitude to the initial momentum, in order to cancel it out.
zarkam21:
0
Shadow:
what is 0
zarkam21:
actually if the momentum is 97750 and we want to achieve 0 wouldnt it just be 97750
Shadow:
The impulse?
zarkam21:
Yes
Shadow:
\[\Delta p = p _{f} - p _{i}\] Do we agree that pf = 0 and pi = 97750?
zarkam21:
yes
zarkam21:
it would be -97750
Shadow:
mhm
zarkam21:
so that is it :/
Shadow:
Yep
Shadow:
Why the face? xD
zarkam21:
okay so a is 85 and b is -977509
zarkam21:
97750*
Shadow:
What is (a) asking for?
zarkam21:
because i feel as if i stressed you out
Shadow:
Impulse = change in momentum, therefore (a) is 97750 It actually isn't negative, since it says "magnitude" therefore they just want the value, they don't care about vectors.
zarkam21:
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow \[\Delta V = V _{f} - V _{i} = 0 - 85 = -83\] \[\Delta p = p _{f} - p _{i}\] \[p _{f} = m \times v = 1150 \times 0 = 0\] \[p _{i} = m \times v = 1150 \times 85 = 97750\] \[\Delta p = 0 - 97750 = -97750\] \(\color{#0cbb34}{\text{End of Quote}}\)
zarkam21:
okay and b is he same just as a negative
Shadow:
(b) is force, not impulse or change in momentum
Shadow:
"b. The car crashes into the wall, stopping in 0.1 seconds. What force is applied to the car?" It's easy just use this \[F \times \Delta t = m \times \Delta v\]
zarkam21:
-99750 * 0.1=1150*85
Shadow:
Impulse is not force
zarkam21:
we are solving for force?
Shadow:
"What force is applied to the car?"
zarkam21:
UGh ;/
Shadow:
F stands for what?
zarkam21:
force
Shadow:
\[F \times \Delta t = m \times \Delta v\]
Shadow:
Solve
zarkam21:
I dont know what the force is in the problem
Shadow:
It's the force the wall applies to the car, in order to stop it.
zarkam21:
so in other words answer a?
Shadow:
No, that's the impulse, the product of mass times the change in velocity
zarkam21:
ughhhhh
zarkam21:
Wait so the force i will find my doing mass times the change in velocity
Shadow:
No, use this: \[F \times \Delta t = m \times \Delta v\]
zarkam21:
F*0.1=1150*85
Shadow:
-85
zarkam21:
F*0.1=1150*-85
Shadow:
mhm
zarkam21:
-97750
Shadow:
That's the change in momentum, what did you get for Force?
zarkam21:
well i got f=-97750
Shadow:
Did you divide by 0.1
zarkam21:
Yup
zarkam21:
well thats what im saying i got -97750
Shadow:
You're missing a 0
zarkam21:
Ohhhhhhhh :/
Shadow:
That number is what makes care accidents devastating. It's also why cars are designed to crumple in the front, as it makes the car stop over a longer period of time as the front crumples in ward. Try making the time a larger number, such as 0.2, and you see a trend in the value of the force.
zarkam21:
Thank you so the answer is 977500
Shadow:
I would apply the negative sign, as it shows you understand Newton's Third Law. The car impacts the wall, applying a force to it. The wall then applies a force equal in magnitude, and OPPOSITE in direction to the car. That negative sign shows the opposite aspect of Newton's Third Law.
zarkam21:
okay so a is 85 b is -977500
zarkam21:
Cheeseburger i am a pain
zarkam21:
right? :/
zarkam21:
@Shadow
zarkam21:
?
Shadow:
A is the magnitude of the impulse, which is the magnitude of the change of momentum, which is 97750
Shadow:
85 is initial velocity
Shadow:
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow Impulse = change in momentum, therefore (a) is 97750 It actually isn't negative, since it says "magnitude" therefore they just want the value, they don't care about vectors. \(\color{#0cbb34}{\text{End of Quote}}\)
zarkam21:
Cool so a is 97750 b is -977500
zarkam21:
and the units wouldbe kg m/s
Shadow:
Force is in Newtons, and yeah a is in kg times m/s
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