A 26 kg body is moving through space in the positive direction of an x axis with
ID: 2010231 • Letter: A
Question
A 26 kg body is moving through space in the positive direction of an x axis with a speed of 190 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 14 kg, moves away from the point of explosion with a speed of 160 m/s in the positive y direction. A second part, with a mass of 5.6 kg, moves in the negative x direction with a speed of 300 m/s. What are the (a)x-component and (b)y-component of the velocity of the third part? (c) How much energy is released in the explosion? Ignore effects due to the gravitational force.Explanation / Answer
Total mass of the body M = 26 KgIt initial speed U = 190i m/s After explosion : Mass of the first piece m1 = 14 Kg Its speed v1 = 160 j m/s Mass of the second piece m2 = 5.6 Kg Its speed v2 = - 300 i m/s Mass of the third piece m3 = M - (m1+m2) = 26 - (14 + 5.6) = 6.4 Kg Let v3 be the speed of the third part. From the conservation of momentum MU = m1 v1 + m2 v2 + m3 v3 26 Kg*190i m/s = (14 Kg*160 j m/s) + (5.6 Kg*- 300 i m/s) + (6.4 Kg * v3) 4940 i = 2240 j - 1680 i + 6.4 v3 v3 = [6620 i - 2240 j]/6.4 = 1034.37 i - 350 j m/s = 1091.98 m/s (a) X component of velocity of third peice = 1034.37 m/s (b) Y component of velocity of third peice = -350 m/s (c) Energy before explosion U1 = (1/2)MU^2 = 0.5 * 26 Kg * (190i m/s)^2 = 469300 J Energy after the collision U2 = (1/2)m1 v1^2 + (1/2)m2 v2^2 v+ (1/2) m3 v3^2 = 0.5*14*(160)^2 + 0.5*5.6*(300)^2 + 0.5* 6.4*(1091.98)^2 = 179200 + 252000 + 3815745.02528 = 4246945.02 J Energy released in the explosion U = U1 - U2 = 4246945.02 J - 469300 J = 3777645.02 J Energy released in the explosion U = U1 - U2 = 4246945.02 J - 469300 J = 3777645.02 J
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