A ball of mass 3kg is thrown at an angle of 30° with the horizontal & a speed of 10m/s. At the highest point the ball breaks into two parts, having mass ratio 2:1, due to internal forces. If the heavier part falls at a distance of 10m from the start, find the x coordinate of the second part.(a) 5.98m(b) 6.34m(c) 8.66m(d) 7.99mThe question was asked in an interview for internship.The doubt is from System of Particles in chapter System of Particles and Rotational Motion of Physics – Class 11

A ball of mass 3kg is thrown at an angle of 30° with the horizontal &

1.	A ball of mass 3kg is thrown at an angle of 30° with the horizontal & a speed of 10m/s. At the highest point the ball breaks into two parts, having mass ratio 2:1, due to internal forces. If the heavier part falls at a distance of 10m from the start, find the x coordinate of the second part.(a) 5.98m(b) 6.34m(c) 8.66m(d) 7.99mThe question was asked in an interview for internship.The doubt is from System of Particles in chapter System of Particles and Rotational Motion of Physics – Class 11
Answer» The correct ANSWER is (a) 5.98m Explanation: There is no NET force on the particle in the x-direction, so the x-coordinate of centre of mass will be equal to the range of projectile motion.The parts will have masses 2kg and 1KG.The time of motion ‘t’ = 2uy/g = 1s.The range = uxt = 5√31 = 8.66m.The centre of mass will land at 8.66m.So, (m1x1 + m2x2)/(m1 + m2) = 8.66. m1 = 2kg, m2 = 1kg, x1 = 10m, we have to calculate x2. ∴ m2x2 = (8.663) – 2*10 = 25.98 – 20 = 5.98. ∴ x2= 5.98/1 = 5.98m.

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