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Projectile Motion and Momentum

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A projectile of mass 5 kg is fired with an
    initial speed of 80 m/s at an angle of 75◦ with
    the horizontal. At the top of its trajectory,
    the projectile explodes into two fragments of
    masses 3 kg and 2 kg . The 2 kg fragment
    lands on the ground directly below the point
    of explosion 2.5 s after the explosion.
    The acceleration of gravity is 9.81 m/s2 .
    Find the magnitude of the velocity of the
    3 kg fragment immediatedly after the explo-
    sion. Answer in units of m/s.


    Find the distance between the point of firing
    and the point at which the 3 kg fragment
    strikes the ground. Answer in units of km.


    How much energy was released in the explo-
    sion? Answer in units of kJ.






    3. The attempt at a solution
    ~pi = ~pf
    m1 ~v1 = m2 ~v2 + m3 ~v3
    m1 v1^{ = m2 vx2^i + m2 vy2 ^j ¡ m3 vy3 ^j

    Vx=m1/m2 (VcosO)
    Y=(vsin0)^2/(2*9.8)
    Vy3=(y-(1/2)(g)(t^2))/2.5

    Vy2=m3/m2(Vy3)

    sqrt(Vx2^2+Vy2^2)

    magnitude V2x=80.7314m/s
    Where did i go wrong?
     
  2. jcsd
  3. Nov 20, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well you know the momentum before must equal the momentum afterwards. Since the projectile is at the top of its trajectory when it explodes it has no vertical component of momentum at the particular time. What you need to do is find the velocity of the 2Kg fragment which will help you find the components of the other fragment. Remember the vector sum before must equal the vector sum afterwards.
     
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