# How Do You Calculate Force and Acceleration in a Crash Test Scenario?

• JB83
In summary, the dummy moves a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. The average force which acts on the dummy during that time is 5.0 N.
JB83
I can not work out how to do the first part of this problem. I know that I am supposed to use impulse to find the average force, but F*dt= J. and J= mvf-mvo. I don't know how to work this problem out without using time. The distance part of the equation is also throwing me off. I don't know where to start. Any help would be greatly appreciated.

A set of crash tests consists of running a test car moving at a speed of 10.0 m/s (22.0 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 65.0 kg (143.0 lbs) dummy is found to move a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).

In a different car, the distance the dummy moves while being stopped is reduced from 0.630 m to 0.210 m calculate the average force on the dummy as that car stops.

JB83 said:
I can not work out how to do the first part of this problem. I know that I am supposed to use impulse to find the average force, but F*dt= J. and J= mvf-mvo. I don't know how to work this problem out without using time. The distance part of the equation is also throwing me off. I don't know where to start. Any help would be greatly appreciated.

A set of crash tests consists of running a test car moving at a speed of 10.0 m/s (22.0 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 65.0 kg (143.0 lbs) dummy is found to move a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).

In a different car, the distance the dummy moves while being stopped is reduced from 0.630 m to 0.210 m calculate the average force on the dummy as that car stops.
You don't have the time, but you have the distance and the average speed during the collision. delta t = distance/avg speed. Solve for the average collision force. Then its Newton 2 to solve for a.

so if delta t = distance/ave speed... delta t = 0.630/10. Using that in the J= ave. Force * delta t would be ave Force = J/delta t. If J= mvf-mvo. Since there is no final velocity J = -mvo. When I did these equations, the answer that I got was incorrect.

JB83 said:
so if delta t = distance/ave speed... delta t = 0.630/10. Using that in the J= ave. Force * delta t would be ave Force = J/delta t. If J= mvf-mvo. Since there is no final velocity J = -mvo. When I did these equations, the answer that I got was incorrect.
That's on account of you did not compute the average speed correctly. Starts at 10. Ends up at 0. Average speeed is ?

Ok I feel somewhat silly, and I know you are trying your best to explain it to me, but I have no idea what you are trying to lead me to do. Average speed is distance/change in time... I don't know what to do. Thank you for your help though.

JB83 said:
Ok I feel somewhat silly, and I know you are trying your best to explain it to me, but I have no idea what you are trying to lead me to do. Average speed is distance/change in time... I don't know what to do. Thank you for your help though.
Yes, that is correct, you are trying to find the time, given the distance and the average speed.
$$v_{average} = (v_i + v_f)/2 = 10/2 = 5.$$ Now if that equation disturbs you, why not use $$v_0^2 = 2a\Delta x$$, solve for a and use F=ma to solve for the avg force. More than one way to skin a cat, you know.

Thank you so so much. I got it, and I still feel kinda dumb.. but you were right I need to not limit myself to the equations that are in the chapter that the homework is from. Thanks again

## 1. What is the difference between impulse and momentum?

Impulse and momentum are both concepts related to the motion of an object, but they are not the same thing. Momentum is a property of an object that is defined as its mass multiplied by its velocity. On the other hand, impulse is a measure of the change in an object's momentum over time. In other words, momentum is a measure of an object's motion, while impulse is a measure of the force that causes a change in motion.

## 2. How are impulse and momentum related?

Impulse and momentum are closely related because impulse causes a change in momentum. The more impulse an object experiences, the greater the change in its momentum will be. This relationship is described by the equation FΔt = mΔv, where F is the force applied, Δt is the time period over which the force is applied, m is the mass of the object, and Δv is the change in velocity. This equation shows that a larger force or a longer duration of force will result in a greater change in momentum.

## 3. What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. In other words, the total amount of momentum in the system remains constant, regardless of any internal changes that may occur. This principle is based on the law of inertia, which states that an object will remain in its state of motion unless acted upon by an external force. The conservation of momentum is a fundamental concept in physics and is used to explain various phenomena, such as collisions and explosions.

## 4. How do you solve impulse/momentum problems?

To solve impulse/momentum problems, you can use the equation FΔt = mΔv. First, determine the initial and final momentum of the objects involved. Then, calculate the change in momentum (Δp) by subtracting the final momentum from the initial momentum. Next, determine the time period over which the force acted (Δt). Finally, plug in the values into the equation FΔt = mΔv to find the force (F) applied during the event.

## 5. Can impulse and momentum be negative?

Yes, both impulse and momentum can be negative. Negative impulse means that the force applied caused a decrease in momentum, while negative momentum means that the object is moving in the opposite direction of the defined positive direction. In physics, positive and negative signs are simply used to indicate direction, so a negative impulse or momentum does not necessarily mean a "bad" or "wrong" value. It is important to pay attention to the signs and directions when using equations involving impulse and momentum to accurately interpret the results.

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