Impulse and momentum / work-energy

In summary, impulse and momentum are related concepts in physics that describe the effect of forces acting over time. Impulse is the product of force and the time duration over which it acts, resulting in a change in momentum, which is the product of an object's mass and its velocity. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Both concepts highlight the relationship between forces, motion, and energy transfer in physical systems.
  • #1
GoonP
2
1
Homework Statement
A car moving at 70km/h collides rams into an immobile steel wall. Its front of the is
compressed by 0.94m . What average force must a seat belt exert in order to restrain
a 75-kg passenger?
Relevant Equations
F = \frac{dp}{dt}
Hello, I am wondering why this take at a solution is wrong.

So basically, $F = \frac{dp}{dt} \Rightarrow F = m\frac{dv}{dt} \Rightarrow F_{\text{avg}}\Delta t = m \Delta v$. Quick conversions show that 70 km/hr = 19.4 m/s, and therefore $\Delta t = \frac{0.94}{19.4} = 0.04845$. Therefore, the answer I am getting is $\frac{75 \cdot 19.4}{0.04845} \approx 30 \text{kN}$ which is exactly double the correct answer.

The solution I have access too uses the fact that $F_{\text{avg}} = \frac{\Delta E}{\Delta x}$ which i can understand is a basic consequence of work energy theorem, and substituting kinetic energy for $E$ it is easy to see mathematically why it is half my answer.

Any guidance as to why my initial approach is incorrect?

Thanks!
 
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  • #2
Welcome!

Consider that while the front of the car is being compressed by 0.94 m, its velocity is simultaneously and uniformly reduced from 70 km/h to zero.
Therefore, your calculated time is too short.
 
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  • #3
Oh ok that makes sense. Thanks!
 
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  • #4
Please use two \$$ to bracket LaTeX expressions. Click on "LaTeX Guide", lower left for more useful informaton about using LaTeX on this site.
The expression $$F = \frac{dp}{dt} \rightarrow F = m\frac{dv}{dt} \rightarrow F_{\text{avg}}\Delta t = m \Delta v$$ is appropriate for finding the average force if you know ##\Delta t## which you don't. In that case you have to make the additional assumption that the force is constant. The solution you have does that and calculates the work done by that force and averages over distance.

You could have gotten the same answer if you found ##\Delta t## by dividing ##\Delta x## not by the initial velocity ##v_0## but by the average velocity ##v_{\text{avg}}=\frac{1}{2}(v_0+0).## That's where your missing factor of 2 is. The answers are the same because the average velocity is calculated under the assumption that the acceleration (and hence the force) is constant.
 
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  • #5
One can dispense with the uniform acceleration requirement if one decides to creatively interpret the "average" force as an average over displacement rather than as an average over time. The two averages can differ in general. If the deceleration is constant, they will match.

We have a force that acts on the passenger over a known displacement while doing a known amount of work.

The average (over displacement) force is defined as $$\frac {\int f(s) \cdot ds}{\int ds} = \frac{\int f(s) \cdot ds}{\Delta s}$$ The work done is given by ##\int f(s) \cdot ds##. We know initial and final kinetic energies for the passenger, so we know the work done. That gives us the numerator of the fraction. We are given the total displacement, ##\Delta s##. So we know the denominator of the fraction.

With this method, the factor of ##\frac{1}{2}## comes from the formula for kinetic energy: ##KE=\frac{1}{2}mv^2##.
 
  • #6
jbriggs444 said:
One can dispense with the uniform acceleration requirement if one decides to creatively interpret the "average" force as an average over displacement rather than as an average over time.
While it is technically true that one can choose to define an average wrt one or another variable, it is highly misleading to intend average force as being wrt displacement without saying so. The default meaning of "average force" has to be average wrt time in order to be consistent with the default meaning of average acceleration. When did you last see a question asking for average acceleration wrt distance? Educators should know better.
That said, I cannot think what the practical value could be of computing an average force. In the real world, we care about momentum, energy and forces exceeding thresholds. So maybe no great harm done.
 

FAQ: Impulse and momentum / work-energy

What is the relationship between impulse and momentum?

Impulse is defined as the change in momentum of an object when a force is applied over a period of time. Mathematically, impulse (J) is the product of the average force (F) applied and the time duration (Δt) over which it acts: J = F × Δt. This relationship shows that the impulse applied to an object is equal to the change in its momentum (Δp), where Δp = p_final - p_initial.

How is work defined in physics?

In physics, work is defined as the process of energy transfer that occurs when an object is moved over a distance by an external force. Mathematically, work (W) is calculated as the product of the force (F) applied to the object and the distance (d) over which the force is applied, in the direction of the force: W = F × d × cos(θ), where θ is the angle between the force and the direction of motion.

What is the work-energy theorem?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This can be expressed as W = ΔKE, where ΔKE is the change in kinetic energy, defined as KE_final - KE_initial. This theorem illustrates the direct relationship between the work performed on an object and its resulting motion.

How can we calculate momentum?

Momentum (p) of an object is calculated as the product of its mass (m) and its velocity (v): p = m × v. Momentum is a vector quantity, meaning it has both magnitude and direction. The total momentum of a system is conserved in the absence of external forces, making it a crucial concept in analyzing collisions and interactions between objects.

What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, where no external forces act, the total momentum of the system remains constant over time. This means that the total momentum before an event (such as a collision) is equal to the total momentum after the event. This principle is fundamental in analyzing collisions and interactions in physics.

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