Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Momentum and Breaking force of car

  1. Mar 2, 2005 #1
    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
  2. jcsd
  3. Mar 2, 2005 #2
    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.
    Last edited: Mar 2, 2005
  4. Mar 2, 2005 #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
    Last edited: Mar 2, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook