Question

A 7.25-kg bowling ball moving at 8.65 m/s collides with a 0.75-kg bowling pin, which is scattered at an angle of θ = 85.5° from the initial direction of the bowling ball, with a speed of 14 m/s. calculate the direction, in degrees of the final velocity of the bowling ball. B) Calculate the magnitude of the final velocity in meters per second of the bowling ball.

Answer 1

mass of the bowling ball, m₁ = 7.25 kg

initial speed of the bowling ball, v₀ = 8.65 m/s

mass of the bowling ball, m₂ = 0.75 kg                                            (at angle, = 85.5⁰)

final speed of the bowling ball, v_2f = 14 m/s

The direction, in degree of the final velocity of the bowling ball which is given as :

using conservation of momentum, we have

on X-axis : m₁ v₀₁ + m₂ v₀₂ = (m₁ v_1f + m₂ v_2f) cos    { eq.1 }

where, v₀₂ = 0

inserting the values in eq.1,

(7.25 kg) (8.65 m/s) = (7.25 kg) v_1f cos + (0.75 kg) (14 m/s) cos 85.5⁰

(62.7 kg.m/s) = (7.25 kg) v_1f cos + (0.82 kg.m/s)

(61.8 kg.m/s) = (7.25 kg) v_1f cos

v_1f cos = (8.52 m/s)                                                                               { eq.2 }

on Y-axis :    m₁ v₀₁ + m₂ v₀₂ = (m₁ v_1f + m₂ v_2f) sin { eq.3 }

inserting the values in eq.3,

(7.25 kg) (8.65 m/s) = (7.25 kg) v_1f sin + (0.75 kg) (14 m/s) sin 85.5⁰

(62.7 kg.m/s) = (7.25 kg) v_1f sin + (10.4 kg.m/s)

(52.3 kg.m/s) = (7.25 kg) v_1f sin

v_1f sin = (7.21 m/s)                                                                                  { eq.4 }

dividing eq.4 by eq.2, we get

tan = (7.21 m/s) / (8.52 m/s)      

= tan^-1 (0.8462)

= 40.2 degree

And

magnitude of the final velocity in meters per second of the bowling ball which is given as :

using eq.4, v_1f sin = (7.21 m/s)                

v_1f = (7.21 m/s) / sin                                                                                    { eq.5 }

v_1f = (7.21 m/s) / sin 40.2⁰

v_1f = 11.1 m/s