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

Linear momentum and collisions

  1. Oct 21, 2003 #1
    I'm trying to do these problems by only reading about them in the book and, having not discussed them in class at all, I'm a little weary about some questions. So bear with me if some of the questions seem a little bit stupid .. I'm just starting out with this stuff. There are more questions than I usually post, but I think you'll find that almost all of them have answers that would take 2 minutes to explain to me because I haven't quite put all the pieces of the chapter together yet.

    1) A 3.04 kg particle has a velocity of (2.90 i - 4.04 j) m/s. Find the magnitude and direction of its momentum (counterclockwise from the positive x axis).

    I found the magnitude, but for some reason I'm having problems with the direction. I found the angle to be 54.3 degrees by using cosine and then added 180 degrees to it because it is in the third quadrant.

    2) Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.6). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. Find the original elastic energy in the spring if M = 0.300 kg.

    This one I am officially confused on. Don't you need to use (1/2)kx^2 to get the energy of a spring? What does mass have to do with this? An earlier part to this problem asked for the speed of the block of mass M, which I found to be -6 m/s. I don't know if that has anything to do with this.

    3) A garden hose full of motionless water is held in the manner shown in Figure P9.13. What additional force is necessary to hold the nozzle stationary if the discharge rate is 0.592 kg/s with a speed of 23.4 m/s?

    I multiplied the change in momentum by the velocity and got the correct answer (13.9 N), but I'm not quite sure why it worked. Isn't linear momentum mass times velocity and impulse force times the change in time? I didn't have a change in time and the two formulas didn't seem to go together with the information provided.

    4) A 7.00 kg bowling ball collides head-on with a 2.00 kg bowling pin. The pin flies forward with a speed of 3.01 m/s. If the ball continues forward with a speed of 1.77 m/s, what was the initial speed of the ball? Ignore rotation of the ball.

    Here I thought I could use the formula for an elastic collision. I got
    (1/2)(7)v^2 + (1/2)(2)(0) = (1/2)(7)(1.77^2) + (1/2)(2)(3.01^2)
    and got 2.39 m/s for the velocity, but I guess that isn't right.

    5) Consider a frictionless track ABC with a height of 5.00 m (A) going down to a height of 0 m (B). A block of mass m1 = 4.98 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 = 9.10 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

    I got the velocity to be 33.3 m/s, but I don't know if that helps at all. I wasn't quite sure how to attack this problem because I didn't understand why m1 would reach a different height at all .. would it be zero??

    6) A 89.5 kg fullback running east with a speed of 5.06 m/s is tackled by a 95.7 kg opponent running north with a speed of 3.08 m/s. If the collision is perfectly inelastic, calculate the speed and direction of the players just after the tackle.

    I got the speed to be 2.92 m/s. I'm just having problems with the angle again. I got, using sine, the angle to be 33.0 degrees, so I subtracted that from 90 and then added 90 to it considering the direction the magnitude was facing and got 147 degrees.

    7) During the battle of Gettysburg, the gunfire was so intense that several bullets collided in mid-air and fused together. Assume a 4.90 g Union musket ball was moving to the right at a speed of 244 m/s, 19.6° above the horizontal, and that a 3.07 g Confederate ball was moving to the left at a speed of 279 m/s, 14.1° above the horizontal. Immediately after they fuse together, what is their velocity?

    I thought I could use vf = (m1v1i + m2v2i) / m1+m2, and got vf to be 246 m/s by incorporating the cosine of the angles into their corresponding velocities. The problem is, I need a velocity on the x axis and on the y axis and I was confused as to how to split them up.

    8) Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.20 m/s. After the collision, the orange disk moves along a direction that makes an angle of 39.0° with its initial direction of motion and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision). Determine the final speed of each disk.

    This one plain confuses the hell out of me.
  2. jcsd
  3. Oct 21, 2003 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You sure it's the third quadrant?

    We know other ways of dealing with energy other than computing them directly... try attacking the problem using the principle of conservation of energy.

    0.592 kg/s is not a change in momentum; it's how fast mass is being discharged from the hose.

    You are correct that you're not given a change in time... but isn't 0.592 kg/s telling you the change in mass divided by the change in time?

    Try conservation of momentum.

    Some of the kinetic energy in block A is transferred to block B... A doesn't have enough energy to return to its starting position.

    The resultant vector is pointing northeast, isn't it? What's wrong with 33.0 degrees for the answer?

    Just grab the x and y components of the two velocity vectors and apply conservation of momentum to the x components and to the y components independantly.

    Apply some geometry; draw the picture and figure out the angles at which all the vectors are pointing. (You have some freedom to choose some of the parameters yourself) Then break it up into x and y worlds like problem 7.
  4. Oct 23, 2003 #3
    Unfortunately I'm still lost on some of these. I tried some different methods and got a few of them, but I can't get all of them.

    1) 54.3 degrees doesn't work in ANY Of the quadrants.

    2) The conservation of energy? Alright, so
    E = KE(I)+PE(I)=KE(F)+PE(F).
    Which box do you use for the values? And I'm assuming you use (1/2)kx^2 for the potential evergy of the spring .. without a k value or x value? If you add the potential of the box in, there is still no x value.

    3) ok.

    4) ok.

    5) alright, I understand that, but now what? try to find the initial/final velocity of m1? Use the conservation of energy theorem to somehow get these velocities and then get the potential energy (final) thus leading to the height?

    6) ok.

    7) alright, I understand that, so where did I go wrong?
    m1v1+m1v1 = m2v2 + m2v2
    (.0049)(244)(cos19.6) + (.00307)(279)(cos14.1) = x values?

    8) I didn't want to tackle this until I understood 7.

    Oh, and here's one I forgot.

    A 1260 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 10000 kg truck moving in the same direction at 20.0 m/s (Fig. P9.21). The velocity of the car right after the collision is 18.0 m/s to the east. How much mechanical energy is lost in the collision?

    I know this should be easy, but I just can't get the right answer of 8.68 kj. I tried the conservation of momentum (m1v1+m1v1=m2v2+m2v2) and kept getting 180 j.
  5. Oct 23, 2003 #4


    User Avatar
    Science Advisor

    The momentum is, of course, mass times velocity so the momentum vector is (3.04*2.90i- 3.04*4.04j)= 8.816i- 12.2816j.

    The magnitude is √(8.8162+ 12.28162)
    = √(228.559)= 15.18 kg m/s. Is that what you got?

    If you have set up your coordinate system correctly and drawn the vector is should be obvious that the vector is NOT in the third quadrant. In the third quadrant both the i component and j components are negative.

    You also say that "I found the angle to be 54.3 degrees by using cosine". Since the momentum vector, together with its i and j components separately forms a right triangle with "near side" (i component) 8.816 and hypotenuse of length 15.18, you should have
    cos(θ)= 8.816/15.18= 0.5807 or θ= 54.3 degrees- okay, that's what you have. If you know what the "quadrants" are, it should be easy to determine the true angle (hint: it's much better, IMO, to use tangent- tan(θ)= -12.2816/8.816= -1.3931 so θ= ?

    Well, first of all, they are "blocks", not "boxes". And the answer of course, is that you use both of them. Initially the two blocks were sitting still right next to each other. The only energy that stored in the spring. At the end, the 3M= 0.9kg block is moving "to the right with speed 2.0m/s" so it has momentum 0.9*2= 1.8 kg m/s. Momentum is conserved in this experiment (there is no "outside" force) so for the other block, 0.3*v= -1.8: v= -0.6 m/s (negative so it is moving to the left). Now that you know its speed you can find the kinetic energy of both blocks and thus the total energy. That total energy had to come from the spring's elastic energy.

    Initially, the block with mass 4.98 kg has potential energy (relative to point B) of mgh= (4.98)(9.8)(5)= 244 Joules. When it goes down to point B, its potential energy is 0 so it must have kinetic energy (1/2)mv2= (1/2)(4.98)v2= 244 so
    v2= 98 and v= √(98)= 9.9 m/s.
    It makes an elastic collision with a 9.10 kg block. Since the collision is "elastic" both momementum and energy are conserved.
    The momentum of the first mass before the collision was (4.98)(9.9)= 49 kg m/s. After the collision we must have
    4.98v1+ 9.10v2= 49 where v1 is the speed, after the collision, of the 4.98 kg mass and v2 is the speed, after the collision, of the 9.10 kg mass. Conservation of energy give (1/2)(4.98)v12+ (1/2)(9.10)v22= 244 Joules (still). Using those two equations, you can solve for both v1 and v2. After you know v1 you can find the (new) kinetic energy of the 4.98 mass and you know that must be the potential energy at the highest point (where it speed is 0) so you can find "h".

    Almost. Since the two bullets are moving toward each other one of the x-velocities is negative. You are welcome to decide if toward the Union or Confederate lines is positive.
    Then do the same with the y coordinates: use sine instead of cosine and note that both y components are positive.

    For 8, carefully set up a coordinate system, draw a picture and look at x and y coordinates of momentum.
  6. Oct 24, 2003 #5
    Great, I have everything ironed out except for 5 and 8. For number 5, using the systems of equations with those two equations, I got v2 to equal -.0586 and thus v1 to be 9.95. Using the formula for KE, I got KE to be 246. Converting that to PE, my height was 5.05, which is impossible!! What did I do wrong?

    And 8. I figure with inelastic glancing collisions, each shuffleboard will get have of the original velocity of the first shuffleboard, or 2.6 for each. The angle of the one glancing off the other would be 51 degrees. So to find the velocity of each shuffleboard, do I multiply 2.6 times the cosine of 51 degrees or what?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook