Question

A 124 kg shell fired from cannon explodes and breaks into two equal parts at some height. after explosion, one part moves with speed of 112 m/s at an angle of 30 degrees above the horizontal, and the other part moves with a speed of 80 m/s at an angle of 45 degrees below the horizontal. What was the b=velocity of the shell before explosion?

Answer 1

consider the the shell moving with speed u which makes an angle theta with horizontal before explosion

then using law of conservation of momentum along horizontal

m*u*cos(theta) = (m1*u1*cos(30)) + (m2*u2*cos(45)

m1 = m2 = 124/2 = 62 kg

124*u*cos(theta) = (62*112*cos(30)) + (62*80*cos(45))

u*cos(theta) = 76.77 m/s...............(1)

Along vertical

m*u*sin(theta) = (m1*u1*sin(30)) + (m2*u2*sin(45))

124*u*sin(theta) = (62*112*sin(30)) +(62*80*sin(45))

u*sin(theta) = 56.28 m/s................(2)

(2)/(1) = tan(theta) = 56.28/76.77 = 0.733

theta = atan(0.733) = 36.3 degrees

u = 56.28/sin(36.3) = 95.2 m/s