Question

(10%) Problem 10: A 7.25-kg bowling ball moving at 9.85 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle of θ = 21.5° from the initial direction of the bowling ball, with a speed of 10.5 m/s.

Answer 1

Answer

given,

mass of bowling ball = 7.25 Kg

moving speed of the bowling ball = 9.85 m/s

mass of bowling in = 0.875 Kg

scattered at an angle = θ = 21.5°

speed after the collision = 10.5 m/s

angle of the bowling ball

$tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}$

$tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}$

$tan \theta_1 = \dfrac{-[3.3672]}{62.86}$

$tan \theta_1 = 0.0536$

$\theta_1 =-3.066^0$

b) magnitude of final velocity

$v = \dfrac{-m_2v_2sin\theta_2}{m_1 sin\theta_1}$

$v = \dfrac{-0.875 \times 10.5 sin21.5^0}{7.25 sin(-3.066^0)}$

v = 8.68 m/s