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Momentum related > heat

  1. Jun 12, 2005 #1

    I have a bit of a dillema and a few problems regarding a simple excersize.

    The situation is as folows. We have a 6,00 g bullet that has a K.E. of 3.12 J. It strikes a 122 g cart that is on railway tracks. The bullet hits the cart parallel to the tracks. Now we need to find out 1) Speed of bullet before impact 2) Speed of cart after impact 3) With what impulse did the bullet act on the cart 4) How much K.E of the bullet has converted in the internal energy of the bullet 5) For how much deg. K would the cart and bullet be heated if they are from lead with Specific heat = 130 J /KG K and the heat would not be transimitted to the environment 7) For how much would the cart and bullet be heated if the bullet would hit the cart perpendicular to the railway track.

    I have difficulty with 6 and 7 and also a bit with 3.

    1) Plain enough. K.E = 1/2 m * v^2 with some basic algebra we find that v = 32,2 m/s
    2) Momentum before = Momentum after
    m(bullet before)*v(bullet before) + 0(cart has no momentum before) = m(bullet)*v + m(cart)*v
    6*32,2 = 6v + 122v with some algebra we find that v = 1.5 m/s[/quote]
    Yes, looks good. Since you used grams through out, you don't need to convert to kg.
    No. Both cart and bullet are moving at 1.5 m/s. The mass is 6+ 122= 128 g= 0.128 kg. The total kinetic energy after the bullet hits the cart is 1/2*(0.128 kg)(1.5 m/s)2= 0.62 J. That means that 3.12-0.62= 3.5 J was converted to heat.
    The correct E is 3.5 J.
    Last edited by a moderator: Jun 12, 2005
  2. jcsd
  3. Jun 12, 2005 #2

    Andrew Mason

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    Everything you have done looks fine except the answer for 5 is 2.5 J not 3.5.

    Just one observation: this is one slow bullet. I doubt that a lead bullet travelling at 30 m/sec is going to penetrate a lead block.

  4. Jun 12, 2005 #3
    Thanks for the reply.

    Just one more question: Are you sure that 1/2*(0.128 kg)(1.5 m/s)^2= 0.62 J. I put it in my calc and I got 0.144 J.

    Why do we say that the bullet is moving with 1.5 m/s. Isn't the bullet still. It is like a passanger in a bus(cart). Soo no matter what angle of impact, all K.E. of bullet would go in I.E. and bullet would have 0 K.E. from the view point of the cart. The cart would have ofcorse then have 122g*(1,5m/s )^2 of K.E. and the difference between the K.E. of bullet before impact and K.E. of the cart after impact would be the I.E. Soo we have.
    K.E.(bullet, before) = K.E.(cart,after) + I.E.(bullet,after), ah...that would not be logical. Since the momentum...I c.
    K.E.(bullet, before) = K.E.(cart,after) + K.E.(bullet, after) + I.E.(bullet, after)
    we have
    3.12 J = 1/2((6 + 122)/1000 kg) * (1.5 m/s)^2 + I.E. we get the solution I.E. = 2.98 J.

    Hypothetical situations always make me puzzled.
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