Several large firecrackers are inserted into the holes of a bowling ball, and the 6.3 kg ball is then launched into the air with a speed of 10.4 m/s at an angle of 43° from the horizontal. the firecrackers explode at the peak of the trajectory, breaking the ball into three pieces. a 1.8 kg piece travels straight back horizontally with a speed of 2.2 m/s. a 1.6 kg piece travels straight up with a speed of 1.8 m/s.

Answer

given,

mass of the ball = 6.3 kg

speed of the ball = 10.4 m/s

angle made with horizontal = 43°

m_a = 1.8 kg               v_a = 2.2 m/s

m_b = 1.6 kg               v_b = 1.8 m/s

mass of third particle = 6.3 - 1.8 - 1.6

                                   = 2.9 kg

u cos θ = 10.4 x cos 43° = 7.61 m/s

by using conservation momentum along x-axis

6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ

V₃ₓ = 32.44 m/s (toward right)

by using conservation momentum along y-axis

0 = 0 + 1.6 x 1.8 + 1.6 × V₃y

V₃y = -1.8 m/s (indicate downward)

velocity of the third particle

v = 32.49 m/s

θ = 3.176° (downward with horizontal)

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach