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Momentum and Kinematics

  1. Nov 9, 2007 #1
    1. A 10g bullet moving at 1000m/s collides inelastically with a 30 kg wooden block at rest. The block is 20 cm thick and can slide freely on a frictionless table. After fully penetraing the block, the bullet comes to rest within the block.
    a) calculate the time required for the bullet to penetrate the box.
    b) Find the average force exerted by the bullet on the block during impact.

    2. impact time= 2(impact distance)/impact velocity, mv=mv, v=v0 + at, x=x0 +v0t + .5at^2

    3. (a) I tried imputing it into the 1st equstion. the impact distance is 20cm and the impact
    velocity is 1000m/s and i got 0.0004 secs.
    (b) favg= delta P/impact time. so i did (mvfinal - mvinitial)0.0004 secs. I got
    [0-(0.01)(10000)]/(0.0004 secs) and then i got -25,000N.

    Last edited: Nov 9, 2007
  2. jcsd
  3. Nov 9, 2007 #2
    anyone have any ideas.... i think its not really a momentum problem but mroe of a kinematics and sum of forces
  4. Nov 10, 2007 #3


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    When the bullet hits the wooden block, it transfers its KE to the wooden block.And they move togather. Equating their KEs you can find the velocity V of the wooden block. Now the bullet is inside the moving wooden block and it is retarding due to the resistance of the wood.Now the relative velocity of bullet with respect the block is (1000 - V)m/s. This is the initial velocity u of the bullet. Final velocity = 0. From these values find retardation of the bullet. Now find force f and time t
  5. Nov 10, 2007 #4
    thanks, when i did that I got t=4x10^-4 secs but by TA said something about it being 4x10^-5 secs for time. 0.3333 is the velocity of the bullet and block combo after collision

    What I did was v=vo + at. 0=999.666 +at. solved for a and got a= -999.666/t.

    I then did x=x0 + voxt +.5at^2 and plugged in values.... 0.2=999.666t + .5at^2.
    I pluged in the value for a (a= -999.666/t) and got t=4x10^-4 secs.

    Can anyone quickly calculate the answer/check my work to see if what i did is right?

  6. Nov 10, 2007 #5


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    You cannot apply the conservation of linear momentum for inelastic collision. You have to use only conservation of energy. Accordingly 1/2*10*10^-3*1000^2 = 1/2*30.01*v^2. with this the combined velocity of bullet and block = 18.25 m/s. The initial velocity of the bullet with respect to the block = 1000 - 18.25 =981.75 m/s. Any way you get t = 4.1*10^-4.
  7. Nov 11, 2007 #6
    thanks for the reply but im pretty sure thats not correct from what i learned in class. Conservation of momentum is always conserved but kinetic energy isnt conserved in inelastic collisions. Also, this is the chapter before conservation of energy in my book anyway so we shouldnt be able to use those equations.

    At least it doesnt really change the answer
  8. Nov 11, 2007 #7


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    Yes. You are right.
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