What Is the Velocity of the 3 kg Fragment After Explosion?

[blayman5]

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?

 

[Kurdt]

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.

 

[thomate1]

I would like to provide a response to the above content by saying that the solution provided seems to be on the right track but there are a few errors in the calculations. Firstly, the initial velocity of the 3 kg fragment should be the same as the initial velocity of the entire projectile, which is 80 m/s at an angle of 75 degrees with the horizontal. Secondly, the time taken for the 3 kg fragment to reach the ground should be calculated from the point of explosion, not from the top of the trajectory. These corrections would result in a slightly different magnitude of the velocity of the 3 kg fragment, which can be calculated using the conservation of momentum equation as m1v1 = m2v2 + m3v3.

Furthermore, to find the distance between the point of firing and the point at which the 3 kg fragment strikes the ground, we can use the equation for horizontal displacement, x = v0xt, where v0x is the initial horizontal velocity and t is the time taken for the 3 kg fragment to reach the ground. This distance can then be converted to kilometers.

As for the energy released in the explosion, we can use the equation for kinetic energy, KE = 1/2mv^2, where m is the mass of the 2 kg fragment and v is the velocity of the 2 kg fragment just before it hits the ground. This energy can then be converted to kilojoules.

In conclusion, while the attempt at a solution shows an understanding of the concepts of projectile motion and momentum, there are a few errors in the calculations that need to be corrected in order to get the correct answers.