A 218- kg projectile, fired with a speed of 104 m/s at a 52.0 degree angle break

Question

A 218- kg projectile, fired with a speed of 104 m/s at a 52.0 degree angle breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally.


a) Determine the magnitude of the velocity of the third fragment immediately after the explosion. (in m/s)

b)Determine the direction of the velocity of the third fragment immediately after the explosion. (in degrees)

c)Determine the energy released in the explosion. (in joules)

Explanation / Answer

The mass of the projecctile is M = 218 kg

The speed of the projectile is u = 104 m/s

The angle of projection is = 52

At the heighest point the mass will have only horizontal velocity given by

ucos = 104*cos52 = 64.028 m/s

The mass of the each fragment is m = 218/3 = 72.67 kg

In Horizontal direction:

The linear momentum of the projectile just before the collision is Pi = Mu = 218*64.028 = 13958.104 kgm/s

The linear momentum of the fragment moving horizontally is P1 = mucos = 72.67*64.028 = 4652.91476 kgm/s

The linear momentum of the fragment moving vertically is P2 = 0

The linear momentum of the third fragment is P3 = mvcos = 72.67 vcos

Where v is the velocity of the third particle

is the angle made by the third particle with the horizontal

According to the conservation of linear momentum in horizontal direction:

Pi = P1 + P2 + P3

13958.104 = 4652.91476 + 0 + 72.67 vcos

vcos = 128.047 m/s

In vertical direction:

The linear momentum of the projectile just before the collision is Pi = 0

The linear momentum of the fragment moving horizontally is P1 = 0

The linear momentum of the fragment moving vertically is P2 = mucos = 72.67*64.028 = 4652.91476 kgm/s

The linear momentum of the third fragment is P3 = mvsin = 72.67 vsin

According to the conservation of linear momentum in vertical direction:

Pi = P1 + P2 + P3

0 = 0 + 4652.91476 + 72.67 vsin

vsin = -64.028 m/s

1)

The magnitude of the velocity of the third fragment is

v = (vcos^2+vsin^2)

v = 143.16 m/s

2)

The direction of the velocity of the third fragment is

= tan^-1(-64.028/128.047)

= 26.57o below the horizontal

3)

The initial kinetic energy of the system is

KEi = 0.5*M*u^2

KEi = 0.5*218*64.028^2

KEi = 446854.741456 J

The final kinetic energy of the system is

KEf = 0.5*m*ucos^2 + 0.5*m*ucos^2 + 0.5*m*v^2

KEf = 0.5*72.67*64.028^2 + 0.5*72.67*64.028^2 + 0.5*72.67*143.16^2

KEf = 1042594.86102928

the energy released in the explosion = The change in kinetic energy of the system = 595740.12 J

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