Projectile Motion and Momentum

  • Thread starter blayman5
  • Start date
  • #1
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Homework Statement


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.






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?
 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,826
6
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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