The answer is attached below.
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Answer:
B. 57.4 N horizontal (right)
<u>Explanation</u>:
using cos method:


<em>As its vertical, the direction of this force should be up.</em>
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)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5Bm_2v_2Sin%20%5Ctheta_2%5D%7D%7Bm_1v_1%20-%20%28m_2v_2cos%20%5Ctheta_2%29%7D)
![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)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B0.875%5Ctimes%2010.5%20%5Ctimes%20Sin%2021.5%5E0%5D%7D%7B7.25%5Ctimes%209.85%20-%20%280.875%5Ctimes%2010.5%20%5Ctimes%20cos%2021.5%5E0%29%7D)
![tan \theta_1 = \dfrac{-[3.3672]}{62.86}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B3.3672%5D%7D%7B62.86%7D)


b) magnitude of final velocity


v = 8.68 m/s