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.

Answer 1

At the heighest point the mass has only horizontal velocity:

ucosθ = 131 ( cos65^o ) = 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 = mucosθ = 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 = mvcosφ = 68 vcosφ

Here v is the velocity of the third particle

φ is the angle made by the third particle with the horizontal

According to the conservation of linear momentum in horizontal direction:

P_i = P₁ + P₂ + P₃

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

vcosφ = 122.12 m/s

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In vertical direction:

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

Linear momentum of the fragment moving horizontally is P₁ = 0

Linear momentum of the fragment moving vertically:

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

Linear momentumm of the third fragment:

P3 = mvsinφ = 68 vsinφ

According to the conservation of linear momentum in vertical direction:

P_i = P₁ + P₂ + P₃

0 = 0 + 3764.48 kg. m/s + 68 vsinφ

vsinφ = - 55.36 m/s

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(a)

The magnitude of the velocity of the third fragment is

v = √(vcosφ^2+vsinφ^2)

v = 134.08 m/s

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(b)

Direction of the velocity of the third fragment:

φ = tan^-1(- 55.36 m/s/122.12)

φ = 24.38^o below the horizontal

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

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(c)

Find initial kinetic energy of the system:

KE_i = 0.5 M u²

= 0.5 (218 ) ( 55.36 m/s )²

= 334055.52 J

Final kinetic energy of the system:

KE_f = 0.5 m (ucosθ)² + 0.5 m (ucosθ)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