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I still need someone smart to explain 2 me momentum and Impulse throughly .

  1. Nov 28, 2008 #1
    I still need someone smart to explain 2 me momentum and Impulse throughly.....

    1. A hunter has rifle that can fire a.05 kg bullets with a speed of 900 m/s. A 40 kg leapord springs at him at 10 mm/s. How many bullets would the hunter need to stop the tiger dead in its tracks? How do you find the answer? Can someone please explain this problem step by step to me
     
  2. jcsd
  3. Nov 28, 2008 #2
    Re: I still need someone smart to explain 2 me momentum and Impulse throughly.....

    The leopard has momentum, p=m*v, so 40*10=400 [kg*m/s] (I'm assuming the leopard is going 10 m/s, not 10 millimeter per second, which would be awfull slow).

    Everytime the hunter shoots a bullet, it's momentum is added to the leopard's. However, since p=m*v is actually a vector equation, we have to account for a minus sign since because the bullets are shot in the opposite direction in which the leopard is moving. So, effectively, each bullet takes some momentum from the leopard, subtracting it from the leopard's total momentum.

    Now, each bullet has momentum p=m*v=0.05*900=45 [kg*m/s]. How many bullets should the hunter fire to reduce the leopard's momentum to zero?
     
  4. Nov 28, 2008 #3
    Re: I still need someone smart to explain 2 me momentum and Impulse throughly.....

    yea that was a typo 10 m/s not 10 mm/s.
     
  5. Nov 28, 2008 #4
    Re: I still need someone smart to explain 2 me momentum and Impulse throughly.....

    I'd just like to add that you can think of the bullet and the leopard as an inelastic collision.

    By the way, this is kind of a gruesome physics question :biggrin:
     
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