Momentum and Breaking force of car

F=m*delta v/t. In summary, The conversation covers various problems involving momentum, force, and velocity. The initial momentum, impulse, and average force of a 100-kg quarterback being tackled are calculated, as well as the speed and direction of two entangled players. The thrust of a jet engine is determined using the equation F=ma. The recoil speed of a 2000-kg launcher after launching a 40-kg projectile at 400 m/s is also calculated. Finally, the velocity of a car-truck system after a collision is found, and the difficulty of starting from the back of a physics book is discussed.
  • #1
vitaly
48
0
I have some questions as homework, and for some, it would be nice to know whether or not my answers are wrong. I need guidance on one of the questions, and I have no idea where to start on other. Our physics teacher is making us start from the back of the book (nuclear physics) and move forwards. That means, I really have no concrete knowledge of anythign that comes before chapter 7 (momentum), and that's why I think I'm having a lot of problems with it. All help is appreciated. And I'm too slow with the latex, so I hope this work is okay..

A car of mass 1100 kg moves at 24 m/s. What braking force is needed to bring the car to a halt in 20 s?
Breaking force = mv/t = (1100 kg)(24 m/s)/20 s = 1.32x10^3 N

What average force is exerted on a 25-g egg by a bed sheet if the egg hits the sheet at 4 m/s and takes 0.2 s to stop?
p=mv, = (0.025 kg)(4 m/s), = 0.1 kg (m/s)
pfinal = 0
deltap = pfinal - pinitial = 0 - 0.1 kg (m/s)
f = -.01 kg (m/s) / .02
= -(1/2)kN

a 100-kg quarterback is traveling 5 m/s and is stopped by a tackler in 1 s. Calculate (a) the initial momentum of the quarterback, (b) the impulse imparted by the tackler, and (c) the average force exerted by the tackler.
(a) momentum = mv = (100kg)(5 m/s) = 500 kg (m/s)
(b) p = 500 kg (m/s), deltap = 0 - 500 kg (m/s) = -500 kg (m/s)
(c)average force - deltap/t = -500 kg (m/s)/1s = -500 kN

A 40-kg football player going through the air at 4 m/s tackles a 60-kg player who is heading toward her at 3 m/s, in the air. What is the speed and direction of the entangled players?
I'm not really sure how to solve this problem... I know they are vector forces, but are they drawn like this: ----> <------ or are they at a different angle? And to find the speed, you would use c^2 = a^2 + b^2, but this isn't a right triangle, or I don't think, so what do I use. And I know to find the angle you usually use inverse tangent, but, can you in this case?

A jet engine gets its thrust by taking in air, heating and compressing it, adn then ejecting it at a high speed. If a particular engine takes in 20 kg of air per second at 100 m/s, and ejects it at 500 m/s, calculate the thrust of the engine.
What is the equation I use to find the thrust of the engine?

A 40-kg projectile leaves a 2000-kg launcher with a speed of 400 m/s. What is the recoil speed of the launcher?
I'm not sure, but here is my work:
momentum = mv = (40kg)(400 m/s) = 16,000 kg (m/s)
mlauncher*vlauncher = -mprojectile*vprojectile = -(16,000 kg (m/s))
vlauncher = -16,000 kg (m/s) / 2000 kg = -8 m/s

A car of mass 1400 kg travels at 20 m/s and collides with a stationary truck of mass 2800 kg, with its parking brake off. The two vehicles interlock as a result of the collision and slide along the icy road. What is the velocity of the car-truck system?
(1400 kg)(20 m/s) = (1400 kg + 2800 kg)*v
28,000 kg (m/s) = 4200kg*v
v = 28,000 kg (m/s) / 4200 kg = 6.7 m/s
 
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  • #2
vitaly said:
Our physics teacher is making us start from the back of the book (nuclear physics) and move forwards.
This makes no sense whatsoever. How are you supposed to know what Kinetic energy, work, and momentum are if you don't even know about forces and accelerations?
(looking through your work now)

Problem 2:
[tex]F_{net}=\frac{J}{\Delta t}=-{\frac{0.025kg \cdot 4 m/s}{0.2 s}}=-{\frac{0.1 kg\frac{m}{s}}{0.2 s}}=-0.5 N[/tex]

not [tex]-0.5 kN[/tex]

Last two problems look right to me.
 
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  • #3
On the ones you solved, your method looks good, but it looks like you oopsed a few decimal places. Don't think an egg could stand force equivalent to about 50kg (100lb+). You slipped a k multiplier through somehow. Remember in the MKS system, the unit is the kg.

Problem 4:

Both people's speed is referenced to the ground. They are headed directly for each other. You sum the momentum after calculating each individually. The sum is the final momentum of the pair together.

Problem 5

I believe you use the delta v. Divide by one second is the acceleration. F=ma
 
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1. What is momentum and why is it important in the context of cars?

Momentum is a measure of an object's motion, determined by its mass and velocity. In the context of cars, it is important because it affects the car's ability to maintain its speed and overcome resistance.

2. How is momentum calculated for a moving car?

Momentum is calculated by multiplying the car's mass (in kilograms) by its velocity (in meters per second). The equation is: momentum = mass x velocity.

3. What is the relationship between momentum and breaking force in a car?

The relationship between momentum and breaking force is inverse. This means that as the momentum of a car increases, the breaking force required to stop it also increases. Conversely, a car with lower momentum will require less breaking force to come to a stop.

4. How does the weight of a car affect its momentum and breaking force?

The weight of a car has a direct effect on its momentum and breaking force. A heavier car will have more momentum and therefore require more breaking force to stop it compared to a lighter car. This is because the weight of an object is a factor in determining its momentum.

5. Can the momentum and breaking force of a car be changed?

Yes, the momentum and breaking force of a car can be changed. This can be achieved by altering the car's mass or velocity. For example, increasing the mass of a car or increasing its speed will result in a higher momentum and therefore require more breaking force to stop it.

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