Question

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 1

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

$v = \sqrt{32.44^2 + (-1.8)^2}$

v = 32.49 m/s

$tan \theta = \dfrac{-1.8}{32.44}$

$\theta = tan^{-1}(\dfrac{-1.8}{32.44})$

θ = 3.176° (downward with horizontal)