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Homework Help: Need someone to look at my work!

  1. Nov 6, 2004 #1
    Hey to all those that enter this thread!


    I have 3 questions, all of which I have done the work for, but don't know if it's correct, take a look and give some feedback. Thanks.

    1)A mine car, whosse mass is 440kg, rolls at a speed of 0.50m/s on a horizontal track. A 150kg chunk of coal has a speed of 0.80 m/s when it leaves the chute at an angle of 25 degrees with respect to the horizontal. Determine the velocity of the car/coal system after the coal has come to rest in the car.

    Ok, so

    mV (coal)+ mV (cart) = [m(coal)+m(cart)]Vtotal
    150kg (Vcos 25.0) +440kg (0.50m/s)= (150kg+440kg) x Vtotal

    Vtotal = (328.75kgm/s)/590 kg
    Vtotal= 0.557m/s.
    Well it sort of makes sense as the speed would slow down if the extra weight is added, but I left the Y component out of this...

    2) Two identical balls are travelling towrd each other with velocities of -4.0 and +7.0 m/s and they experience an elastic head on collision. Obtain the velocities of each ball after the collision.

    ok seems easy enough.

    mV(ball 1) + mV (ball 2) = mV (ball1 after)+ mV (ball 2 after)

    punch in all the numbers I get: V(ball1after)= +3.0m/s-V(ball2 after)

    2 variables I don't know, try to get a sys of equation. Ke initial = Ke final because it's elastic.

    1/2mv^2(ball1initial)+1/2mv^2(ball 2 intial) = 1/2mv^2+1/2mv^2

    ok all the 1/2's and the respective masses cancel? right...

    so I'm left with after I punch in the numbers: (7.0m/s)^2 +(-4.0m/s)^2 =
    V^2(ball 1 final)+ V^2 (ball2 final)

    OR 65m/s = V^2(ball 1 final)+V^2 (ball 2 final)

    So I get two equations: 3.0m/s-V(ball2 final)= V(ball 1 final) <==I square everything so I can substitute in 65m/s = V^2(ball 1 final)+V^2 (ball 2 final)
    After squaring and substituting I get a quadratic equation of this form:

    2V^2 (ball 2 final) - 2V (b2 final) -56 m/s

    I then use the x= [-b +/- (4ac)^1/2]/2a to get my V (ball2 final).

    MY QUESTION is since we square root the 4ac, my c value is negative, and my "a" is positive, so a square root of a negative value?? does that work?

    But anyway I get 6.79 m/s for ball 2 final and after sub back into previous equations I get V ball 1 final to be -3.791 m/s. The directions make sense since it'll rebound, but the magnitude...any ideas?

    Thanks all.
  2. jcsd
  3. Nov 6, 2004 #2
    whoops, only 2 questions, not three, for now anyway.
  4. Nov 6, 2004 #3
    no one? :frown:
  5. Nov 6, 2004 #4
    Number 1 looks good to me.

    For number 2, your understanding of the concepts seems fairly accurate. However, you algebra is abit off. Your two equations: 3 = v1 + v2, 65 = v1^2 + v2^2 are correct. Now the problems begin when you start solving these. I'm not exactly sure what you were doing when you were squaring. What you could do instead is isolate a variable in one equation and substitute it into the other one. In this case:
    v2 = 3 - v1
    Then you can plug this into your second equation and get:
    65 = v1^2 + (3-v1)^2

    Then expand all of that out, collect like terms and you have an equation that needs to be solved.

    Now the next problem comes in your formula. You're quadratic formula isn't correct. And just as a side note, no it is not ok to have a square root of a negative number. This makes the number a complex number.
  6. Nov 6, 2004 #5
    Hmm, I thought something was wrong...But yea thanks for pointing that out. I'll fine tune it. Appreciate the help! :cool:
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