A6.25-kg bowling ball moving at 9.4 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle of \theta =83.5 degrees from the initial direction of the bowling ball, with a speed of 17.5 m/s.(part a) calculate the direction, in degrees, of the final velocity of the bowling ball. this angle should be measured in the same way that \theta is.(part b) calculate the magnitude of the final velocity, in meters per second, of the bowling ball.

A)

B) v = 9.47 m/s

Explanation:

Given:

mass of the bowling ball, m₁ = 6.25 kg

Initial velocity of the ball, v₁ = 9.4 m/s

mass of bowling pin, m₂ = 0.875 kg

velocity of the pin, v₂ = 17.5 m/s

Angle of scatter, Θ = 83.5°

a)The direction of the final velocity of the ball (∅)

on substituting the values in the above equation, we get

or

or

or

B) From the concept of conservation of momentum

we have

on substituting the values

we get

hence, the magnitude of the final velocity

v = 9.47 m/s

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explanation:

i'm only doing this the points

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