How Much Force Does a Man Feel in a Train-Truck Collision?

In summary: A collision that lasts 1/1000th of a second would still feel very significant.In summary, the collision between the train and truck would cause the truck to move forward with a velocity of 22 m/s and the man to move forward with a velocity of 22 m/s. The force felt by the man is about 2640 N.
  • #1
centrix
2
0

Homework Statement


A 15000 kg train traveling at 30m/s collides with a 2500 truck with a 90 kg man going at 22m/s in the opposite direction. The time of collision is 1.5 seconds. After collision, both objects are moving in the same direction with a velocity of 22 m/s. What is the size of the force felt by the man

Homework Equations


ft=mΔv
f=ma

The Attempt at a Solution


I tried to use the impulse equation, but can't seem to get a reasonable answer from it. I tried to find the acceleration then the next force, but still couldn't find a reasonable answer for the force.

I first tried to find acceleration using Vf=Vi+at.
22=-22+1.5a
a=29.2 m/s/s

f=90*29.2
f=2640 N

that force seems to small for a collision like this. would it be reasonable to assume the force felt by the car is about equal to what the person will feel?
 
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  • #2
centrix said:

Homework Statement


A 15000 kg train traveling at 30m/s collides with a 2500 truck with a 90 kg man going at 22m/s in the opposite direction. The time of collision is 1.5 seconds. After collision, both objects are moving in the same direction with a velocity of 22 m/s. What is the size of the force felt by the man


Homework Equations


ft=mΔv
f=ma

The Attempt at a Solution


I tried to use the impulse equation, but can't seem to get a reasonable answer from it. I tried to find the acceleration then the next force, but still couldn't find a reasonable answer for the force.

I first tried to find acceleration using Vf=Vi+at.
22=-22+1.5a
a=29.2 m/s/s

f=90*29.2
f=2640 N

that force seems to small for a collision like this. would it be reasonable to assume the force felt by the car is about equal to what the person will feel?

Hi centrix, welcome to Physics Forums.

You should obtain the same result from either the impulse or acceleration methods. Be sure to keep track of the velocity and momentum directions.

The man was lucky that the collision occurred over a relatively long timespan (1.5 seconds is practically an eternity in a collision. His truck must have excellent crumple zones!). What force would he experience if the time were only 1/10 second?
 
  • #3
The force felt would definitely increase because the acceleration would increase while mass stays constant.

How long does a collision normally last?
 
  • #4
centrix said:
The force felt would definitely increase because the acceleration would increase while mass stays constant.

How long does a collision normally last?

It depends upon the nature of the bodies involved. The more rigid the bodies, the smaller the deformation and the more brief the interaction. Vehicles are often designed to lengthen the collision time by the use of crumple zones which absorb energy as they 'fail', lengthening the time. Think in terms of fractions of a second.
 
  • #5


I would approach this problem by first considering the conservation of momentum principle. This states that the total momentum before a collision is equal to the total momentum after the collision. In this case, the total momentum before the collision is (15000 kg * 30 m/s) + (2500 kg * 22 m/s) = 465000 kg*m/s. After the collision, the total momentum is (15000 kg + 2500 kg) * 22 m/s = 396000 kg*m/s. Since the total momentum before and after the collision must be equal, we can set these two values equal to each other and solve for the final velocity of the combined objects.

465000 kg*m/s = (15000 kg + 2500 kg) * vf
vf = 22.5 m/s

We can then use this final velocity to calculate the change in momentum of the man, which is equal to the impulse experienced by the man.

Δp = mvf - mvi
Δp = 90 kg * (22.5 m/s - 22 m/s)
Δp = 45 kg*m/s

Finally, we can use the impulse equation to calculate the force experienced by the man.

F = Δp/t
F = 45 kg*m/s / 1.5 s
F = 30 N

This result does seem small, but it is important to remember that the man's mass is much smaller than the train and truck, so he will experience a much smaller force. Additionally, the force experienced by the man may also be distributed over a larger area (such as the seat or airbag), which would further decrease the perceived impact force. It is also possible that the man's body may have some resistance or ability to absorb the force, which would also reduce the perceived impact force. Overall, while the calculated force may seem small, it is still a reasonable result given the variables and assumptions of the problem.
 

FAQ: How Much Force Does a Man Feel in a Train-Truck Collision?

What factors affect the amount of force felt in an impact?

There are several factors that can affect the amount of force felt in an impact, including the mass and velocity of the object, the surface area of the impact, and the materials involved.

How is the amount of force felt measured?

The amount of force felt in an impact is typically measured in units of Newtons (N) using tools such as force sensors or strain gauges. These tools can provide a quantitative measurement of the force experienced during an impact.

What is the relationship between force and impact?

Force and impact are closely related, as the force experienced during an impact is directly related to the mass and velocity of the objects involved. The greater the mass and velocity, the greater the force felt during impact.

How can the amount of force felt in an impact be reduced?

There are several ways to reduce the amount of force felt in an impact, including using protective equipment, increasing the surface area of impact, and decreasing the velocity of the objects involved.

Can the amount of force felt in an impact be predicted?

While the exact amount of force felt in an impact can be difficult to predict, there are mathematical models that can estimate the force based on factors such as mass, velocity, and surface area of impact.

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