A 2.97-kg projectile is fired with an initial speed of 133 m/s at an angle of 28
ID: 2007503 • Letter: A
Question
A 2.97-kg projectile is fired with an initial speed of 133 m/s at an angle of 28° with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.97 kg and 2 kg. At 3.87 s after the explosion, the 2-kg fragment lands on the ground directly below the point of explosion.(a) Determine the velocity of the 0.97-kg fragment immediately after the explosion.
V1 = m/s i + 66.73 m/s j
(b) Find the distance between the point of firing and the point at which the 0.97-kg fragment strikes the ground.
? km
(c) Determine the energy released in the explosion.
? kJ
Explanation / Answer
Mass of the projectile M = 2.97 Kg Initial speed of the projectile U = 133 m/s Angle of projection = 280 Speed of the projectile at the top of the trajectory just before explosion = U cos28 i = 117.43 i m/s Mass of the first fragment m1 = 0.97 Kg velocity of m1 immediately after the explosion v1 = x i + 66.73 j m/s Here v1x = x and v1y = 66.73 m/s Mass of the second fragment m2 = 2 Kg Distance covered by m2 before landing S = (U sin28)^2/(2g) = 198.9 m velocity of m2 immediately after the explosion v2 = [(S/t) - (gt/2)] j = [51.39 - 18.963] j = 32.42 m/s j Here v2x = 0 and v2y = 34.42 m/s (a) According to law of conservation of momentum Along the X axis MU = m1v1x + m2v2x (2.97)(117.43) = (0.97)(x) + (2)(0) 348.7671 = 0.97x x = 359.55 m/s (b) Range R = [Ucos28*Usin28/g] + [x/g] = 748.2 + 179.775 = 927.975 m (c) Energy released E = [(1/2)m1v1^2 + (1/2)m2v2^2] - (1/2)MU^2 = [64858.611 + 1051.0564] -26268.165 = 39641.50 J = 39641.50 JRelated Questions
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