- #1
Aion96
- 7
- 2
- Homework Statement
- Suppose that a man runs into a wall and dies while experiencing 50 g for 0,002 seconds. How much force was exerted on him during impact, and how big was the impulse? How fast did he run, what was the stopping distance, contact time
- Relevant Equations
- Horizontal number of g¨'s = ##\frac{a_{x}}{g}## and the vertical number of g¨'s = ##\frac{F_{N}}{Fg}=\frac{F_{N}}{mg}##
My homework consists of trying to create a simple model for a collision. But I have trouble understanding a specific part of the assignment, namely what g-force is. I'm guessing that I'm allowed to make assumptions. But without understanding adequately the definition for g-force, I don't think I'll make any headway.
I've read that g-force is an object's apparent weight. I've found some formulas online that states that the experienced g-force can be separated into the horizontal and the vertical direction. That the horizontal number of g¨'s = ##\frac{a_{x}}{g}## and the vertical number of g¨'s = ##\frac{F_{N}}{Fg}=\frac{F_{N}}{mg}##. I'm guessing ##50 g## means the experienced apparent weight in the horizontal direction. And we have to assume a reasonable mass for the person. So I'll suppose the person has a mass of ##80 kg##. In either case, I'll proceed with the following reasoning
$$50=\frac{a_{x}}{g} \Rightarrow a_{x}=50g$$
By Newtons second law $$\Sigma F_{x}=ma_{x}=50mg$$
Since we are given ##\Delta t##, by the impulse law, it follows that
$$I=\Sigma F_{x}\Delta t =m\Delta v=\Delta p=ma_{x}\Delta t=50mg\Delta t$$
I'll assume that the person after the collision has zero momentum since the final velocity is zero. This simplifies ##\Delta p## into
$$\Delta p=p_{f}-p{i}=-p_{i}=-mv_{i}=50mg\Delta t \iff v_{i}=-50g\Delta t$$
Since we've already assumed that the final velocity is zero, this implies that the final kinetic energy is also zero. By the work-energy theorem we can calculate the displacement, ##d##, the person traveled before the collision. Hence
$$\Sigma F_{x} \cdot d=\Delta W_{k}=W_{k_{f}}-W_{k_{i}}=-W_{k_{i}}=-\frac{1}{2}mv_{i}^2\iff d=\frac{-m_{v_{i}}^2}{2\Sigma F_{x}}=-25g\Delta t^2$$
And the average speed is given by
$$v_{avg}=\frac{d}{\Delta t}=\frac{-25g\Delta t^2}{\Delta t}=-25g\Delta t$$
I'm moderately sure I'm wrong, but this was my attempt anyway. Any help would be appreciated!
P.S. Apologies for my grammar.
I've read that g-force is an object's apparent weight. I've found some formulas online that states that the experienced g-force can be separated into the horizontal and the vertical direction. That the horizontal number of g¨'s = ##\frac{a_{x}}{g}## and the vertical number of g¨'s = ##\frac{F_{N}}{Fg}=\frac{F_{N}}{mg}##. I'm guessing ##50 g## means the experienced apparent weight in the horizontal direction. And we have to assume a reasonable mass for the person. So I'll suppose the person has a mass of ##80 kg##. In either case, I'll proceed with the following reasoning
$$50=\frac{a_{x}}{g} \Rightarrow a_{x}=50g$$
By Newtons second law $$\Sigma F_{x}=ma_{x}=50mg$$
Since we are given ##\Delta t##, by the impulse law, it follows that
$$I=\Sigma F_{x}\Delta t =m\Delta v=\Delta p=ma_{x}\Delta t=50mg\Delta t$$
I'll assume that the person after the collision has zero momentum since the final velocity is zero. This simplifies ##\Delta p## into
$$\Delta p=p_{f}-p{i}=-p_{i}=-mv_{i}=50mg\Delta t \iff v_{i}=-50g\Delta t$$
Since we've already assumed that the final velocity is zero, this implies that the final kinetic energy is also zero. By the work-energy theorem we can calculate the displacement, ##d##, the person traveled before the collision. Hence
$$\Sigma F_{x} \cdot d=\Delta W_{k}=W_{k_{f}}-W_{k_{i}}=-W_{k_{i}}=-\frac{1}{2}mv_{i}^2\iff d=\frac{-m_{v_{i}}^2}{2\Sigma F_{x}}=-25g\Delta t^2$$
And the average speed is given by
$$v_{avg}=\frac{d}{\Delta t}=\frac{-25g\Delta t^2}{\Delta t}=-25g\Delta t$$
I'm moderately sure I'm wrong, but this was my attempt anyway. Any help would be appreciated!
P.S. Apologies for my grammar.
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