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Perfect Inelastic collision with string tension

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 0.0132 kg is fired at the block with a horizontal velocity v-_i. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.92 N . What is the v_i of the bullet?

    2. Relevant equations
    Tension=mg+ma
    a=mv^2/r
    (m_1)(v_1)=(m_1+m_2)v_2

    3. The attempt at a solution
    First I input the centripetal acceleration equation into the tension equation so T=mg+m(((m_1+m_2)(v_2)^2)/r) then plugged in the numbers given in the equation so the formula began to look like 4.92N=(0.8132)(-9.8)+0.8132((0.8132(v_2)^2)/1.6) and when I solved for v_2 I got approximately 5.584. Then I used the 5.584 in the conservation of motion equation so v_1=(0.8132*5.584)/0.0132. That gives me 344m/s for velocity of the bullet but mastering physics says that wrong.
     
  2. jcsd
  3. Mar 12, 2017 #2

    BvU

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    Hello erm, :welcome:

    Good start ! Good use of the template. Clear post. Pleasure to assist.

    You want to take into account that the tension in the cord has a different direction than the ##mg##: it's all vectors.

    Furthermore, your ##v_2## is at the bottom; if the block swings up, the speed will decrease. Exercise doesn't give that it hangs still after rising 0.8 m -- that would be too easy I suppose.

    This is a nice, but not so easy exercise. You'll have to work it out in a few steps. Momentum balance for the initial inelastic collision is a good start. What do you want to get out of it (in the form of an expression with symbols) ?
     
  4. Mar 12, 2017 #3
    So for the tension of the cord part would I have to include the cos(theta) into the work so T=mg(cos(theta)+ma? I don't understand what you mean by what I want to get out of the initial inelastic collision? If I understand you correctly I want to get the velocity of the bullet from the equality so v_1=((m_1+m_2)v_2)/m_1
     
  5. Mar 12, 2017 #4

    BvU

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    OK, so you have a relationship between ##v_2## and the speed of the bullet. That's part of what I meant. For the next step you have to relate this ##v_2## to something else.

    For the other part, I think you'll make things much clearer for yourself with a free-body diagram of the block at the point when it's risen 0.8 m.
     
  6. Mar 12, 2017 #5
    Ok I figured it out. I had to find out the velocity at the height of 0.8m then use the KE+PE=Total energy equation so that 1/2 mv^2+mgh=1/2mv^2+mgh and one side is the velocity at 0.8 and the other (the velocity im looking for) is calculated at h=0. so 1/2(0.8132)(v^2)=1/2(0.8132)(v at 0.8)+0.8132(9.8)(0.8)
     
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