A 204-kg projectile, fired with a speed of 131 m/sat a 65.0 angle, breaks into t

Question

A 204-kg projectile, fired with a speed of 131 m/sat a 65.0   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.

B. Determine the direction of the velocity of the third fragment immediately after the explosion.

C. Determine the energy released in the explosion.

Explanation / Answer

At the heighest point the mass has only horizontal velocity:

ucos = 131 ( cos65o ) = 55.36 m/s

Mass of the each fragment is

m = 204/3 = 68 kg

In Horizontal direction:

Linear momentum of the projectile just before the collision:

Pi = Mu = 218 ( 55.36 m/s ) = 12069.13 kg. m/s

Linear momentum of the fragment moving horizontally:

P1 = mucos = 68 ( 55.36 m/s ) = 3764.48 kg. m/s

Linear momentum of the fragment moving vertically: P2 = 0

Linear momentum of the third fragment:

P3 = mvcos = 68 vcos

Here 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

12069.13 kg. m/s = 3764.48 kg. m/s + 0 + 68 vcos

vcos = 122.12 m/s

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In vertical direction:

Linear momentum of the projectile just before the collision: Pi = 0

Linear momentum of the fragment moving horizontally is P1 = 0

Linear momentum of the fragment moving vertically:

P2 = mucos = 68 ( 55.36 m/s ) = 3764.48 kg. m/s

Linear momentumm of the third fragment:

P3 = mvsin = 68 vsin

According to the conservation of linear momentum in vertical direction:

Pi = P1 + P2 + P3

0 = 0 + 3764.48 kg. m/s + 68 vsin

vsin = - 55.36 m/s

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(a)

The magnitude of the velocity of the third fragment is

v = (vcos^2+vsin^2)

v = 134.08 m/s

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(b)

Direction of the velocity of the third fragment:

= tan-1(- 55.36 m/s/122.12)

= 24.38o below the horizontal

= 24.4o below the horizontal ( after round off to 3 significant digits )

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(c)

Find initial kinetic energy of the system:

KEi = 0.5 M u2

= 0.5 (218 ) ( 55.36 m/s )2

= 334055.52 J

Final kinetic energy of the system:

KEf = 0.5 m (ucos)2 + 0.5 m (ucos)2 + 0.5 ( m)v2

= 0.5 ( 68 ) ( 55.36 m/s )2 + 0.5 ( 68 ) ( 55.36 m/s )2 + 0.5 ( 68 ) (134.08 m/s )2

KEf = 819634.79

Energy released in the explosion is equal to the change in kinetic energy of the system:

KEf - KEi =  485579 J

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