Math / Algebra

Question4. A car moving at $10 \mathrm{~m} / \mathrm{s}$ crashes into a tree and stops in 0.26 s . Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg . Would the answer to this question be different if the car with the $70-\mathrm{kg}$ passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed?

Studdy Solution

STEP 1

What is this asking? How much force does a seatbelt exert on a 70 kg person to stop them in 0.26 seconds from an initial speed of 10 m/s?
Also, would the force be different if the car hit another car head-on, both going the same speed? Watch out! Don't forget to consider Newton's Laws and how impulse relates to momentum!

STEP 2

1. Calculate the change in momentum.
2. Determine the impulse.
3. Calculate the force.
4. Analyze the car-on-car collision scenario.

STEP 3

Alright, let's start with the momentum!
Remember, momentum is how much "oomph" an object has when it's moving.
It's calculated as mass times velocity: $p = mv$ .

STEP 4

The initial momentum of the passenger is $70 \text{ kg} \cdot $10 \text{ m/s}$ = $700 \text{ kg} \cdot \text{m/s}$ $.
Since the passenger comes to a complete stop, their final momentum is zero!

STEP 5

So, the change in momentum, often written as $\Delta p$ , is $0 \text{ kg} \cdot \text{m/s} - $700 \text{ kg} \cdot \text{m/s}$ = $-700 \text{ kg} \cdot \text{m/s}$ $.
The negative sign just means the momentum decreased.

STEP 6

Now, let's talk about impulse!
Impulse is the change in momentum, and it's equal to the force applied multiplied by the time it's applied for: $J = F \cdot t = \Delta p$ .

STEP 7

We already know the change in momentum is $-700 \text{ kg} \cdot \text{m/s}$ , and we know the time the force is applied is 0.26 seconds.

STEP 8

Now we can solve for the force: $F = \frac{\Delta p}{t}$ .

STEP 9

Plugging in our values, we get $F = \frac{-700 \text{ kg} \cdot \text{m/s}}{0.26 \text{ s}} \approx -2692.3 \text{ N}$ .
The negative sign indicates the force is in the opposite direction of the initial motion, which makes sense since the seatbelt is stopping the passenger.

STEP 10

Now, what if the car hit another car head-on, both going the same speed?
The change in momentum for the passenger would still be the same, assuming they still come to a stop in the same amount of time.

STEP 11

Since the change in momentum and the time are the same, the force exerted by the seatbelt would also be the same!
It doesn't matter what the car crashes into, only how quickly the passenger's momentum changes.

STEP 12

The force the seatbelt exerts on the passenger is approximately $\textbf{2692.3 N}$ .
The force would be the same if the car collided head-on with another car of equal mass and speed.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Calculate the change in momentum.
2. Determine the impulse.
3. Calculate the force.
4. Analyze the car-on-car collision scenario.

STEP 3

Alright, let's start with the momentum!
Remember, momentum is how much "oomph" an object has when it's moving.
It's calculated as mass times velocity: $p = mv$ .

STEP 4

The initial momentum of the passenger is \(70 \text{ kg} \cdot $10 \text{ m/s}$ = $700 \text{ kg} \cdot \text{m/s}$ \).
Since the passenger comes to a complete stop, their final momentum is zero!

STEP 5

So, the change in momentum, often written as $\Delta p$ , is \(0 \text{ kg} \cdot \text{m/s} - $700 \text{ kg} \cdot \text{m/s}$ = $-700 \text{ kg} \cdot \text{m/s}$ \).
The negative sign just means the momentum decreased.

STEP 6

Now, let's talk about impulse!
Impulse is the change in momentum, and it's equal to the force applied multiplied by the time it's applied for: $J = F \cdot t = \Delta p$ .

STEP 7

We already know the change in momentum is $-700 \text{ kg} \cdot \text{m/s}$ , and we know the time the force is applied is 0.26 seconds.

STEP 8

Now we can solve for the force: $F = \frac{\Delta p}{t}$ .

STEP 9

Plugging in our values, we get $F = \frac{-700 \text{ kg} \cdot \text{m/s}}{0.26 \text{ s}} \approx -2692.3 \text{ N}$ .
The negative sign indicates the force is in the opposite direction of the initial motion, which makes sense since the seatbelt is stopping the passenger.

STEP 10

Now, what if the car hit another car head-on, both going the same speed?
The change in momentum for the passenger would still be the same, assuming they still come to a stop in the same amount of time.

STEP 11

Since the change in momentum and the time are the same, the force exerted by the seatbelt would also be the same!
It doesn't matter what the car crashes into, only how quickly the passenger's momentum changes.

STEP 12

The force the seatbelt exerts on the passenger is approximately $\textbf{2692.3 N}$ .
The force would be the same if the car collided head-on with another car of equal mass and speed.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Calculate the change in momentum.2. Determine the impulse.3. Calculate the force.4. Analyze the car-on-car collision scenario.

STEP 3

Alright, let's **start** with the *momentum*!Remember, momentum is how much "oomph" an object has when it's moving.It's calculated as mass times velocity: p=mvp = mvp=mv.

STEP 4

The **initial momentum** of the passenger is \(70 \text{ kg} \cdot 10 m/s10 \text{ m/s}10 m/s = 700 kg⋅m/s700 \text{ kg} \cdot \text{m/s}700 kg⋅m/s\).Since the passenger comes to a complete stop, their **final momentum** is *zero*!

STEP 5

So, the **change in momentum**, often written as Δp\Delta pΔp, is \(0 \text{ kg} \cdot \text{m/s} - 700 kg⋅m/s700 \text{ kg} \cdot \text{m/s}700 kg⋅m/s = −700 kg⋅m/s-700 \text{ kg} \cdot \text{m/s}−700 kg⋅m/s\).The negative sign just means the momentum *decreased*.

STEP 6

Now, let's talk about *impulse*!Impulse is the change in momentum, and it's equal to the force applied multiplied by the time it's applied for: J=F⋅t=ΔpJ = F \cdot t = \Delta pJ=F⋅t=Δp.

STEP 7

We already know the **change in momentum** is −700 kg⋅m/s-700 \text{ kg} \cdot \text{m/s}−700 kg⋅m/s, and we know the time the force is applied is **0.26 seconds**.

STEP 8

Now we can **solve for the force**: F=ΔptF = \frac{\Delta p}{t}F=tΔp​.

STEP 9

STEP 10

Now, what if the car hit another car head-on, both going the same speed?The change in momentum for the passenger would still be the same, assuming they still come to a stop in the same amount of time.

STEP 11

Since the change in momentum and the time are the same, the force exerted by the seatbelt would *also* be the same!It doesn't matter *what* the car crashes into, only how quickly the passenger's momentum changes.

STEP 12

The force the seatbelt exerts on the passenger is approximately 2692.3 N\textbf{2692.3 N}2692.3 N.The force would be the *same* if the car collided head-on with another car of equal mass and speed.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Calculate the change in momentum.
2. Determine the impulse.
3. Calculate the force.
4. Analyze the car-on-car collision scenario.

Alright, let's start with the momentum!
Remember, momentum is how much "oomph" an object has when it's moving.
It's calculated as mass times velocity: $p = mv$ .

The initial momentum of the passenger is \(70 \text{ kg} \cdot $10 \text{ m/s}$ = $700 \text{ kg} \cdot \text{m/s}$ \).
Since the passenger comes to a complete stop, their final momentum is zero!

So, the change in momentum, often written as $\Delta p$ , is \(0 \text{ kg} \cdot \text{m/s} - $700 \text{ kg} \cdot \text{m/s}$ = $-700 \text{ kg} \cdot \text{m/s}$ \).
The negative sign just means the momentum decreased.

Now, let's talk about impulse!
Impulse is the change in momentum, and it's equal to the force applied multiplied by the time it's applied for: $J = F \cdot t = \Delta p$ .

We already know the change in momentum is $-700 \text{ kg} \cdot \text{m/s}$ , and we know the time the force is applied is 0.26 seconds.

Now we can solve for the force: $F = \frac{\Delta p}{t}$ .

Now, what if the car hit another car head-on, both going the same speed?
The change in momentum for the passenger would still be the same, assuming they still come to a stop in the same amount of time.

Since the change in momentum and the time are the same, the force exerted by the seatbelt would also be the same!
It doesn't matter what the car crashes into, only how quickly the passenger's momentum changes.

The force the seatbelt exerts on the passenger is approximately $\textbf{2692.3 N}$ .
The force would be the same if the car collided head-on with another car of equal mass and speed.