# A billiard ball of mass m = 0.15 kg strikes the cushion of a billiard table at \theta1 = 64 ^o...

## Question:

A billiard ball of mass m = 0.15 kg strikes the cushion of a billiard table at {eq}\theta 1 = 64 ^o {/eq} and a speed v1 = 24 m/s. It bounces off at an angle of {eq}\theta 2 = 61^o {/eq} and a velocity of v2 = 11 m/s. What is the magnitude of its change in momentum (in kg m/s)?

## Momentum of a particle

A particle in motion is said to have momentum and energy due to its motion called Kinetic energy. When moving with a velocity {eq}\vec{v} {/eq}, particle of mass {eq}m {/eq} has momentum given by:

{eq}\vec{P} \ = \ m \times \vec{v} {/eq}

## Answer and Explanation:

**Given**:

- Mass of billiard ball: {eq}m \ = \ 0.15 \ Kg {/eq}

- Speed of billiard ball before collision: {eq}v_1 \ = \ 24 \ ms^{-1} {/eq}

- Direction of motion, before collision: {eq}\theta_1 \ = \ 64^\circ {/eq}

- Speed of billiard ball after rebounding: {eq}v_2 \ = \ 11 \ ms^{-1} {/eq}

- Direction of motion after rebounding: {eq}\theta_2 \ = \ 61^\circ {/eq}

Initial momentum of the ball is:

{eq}\vec{P_1} \ = \ m \times \vec{v_1} \ = \ 0.15 \times (24 \times \cos 64^\circ \ \hat{i} \ + \ 24 \times \sin 64^\circ \ \hat{j}) \ = \ 1.578 \ \hat{i} \ + \ 3.236 \ \hat{j} \ Kg ms^{-1} {/eq}

After rebounding, the ball will now be traveling along negative x-direction. Therefore,

{eq}\vec{P_2} \ = \ m \times \vec{v_2} \ = \ 0.15 \times 11 \times (-\cos 61^\circ \ \hat{i} \ + \ \sin 61^\circ \ \hat{j}) \ = \ - 0.8 \ \hat{i} \ + \ 1.443 \ \hat{j} \ Kg ms^{-1} {/eq}

Change in momentum:

{eq}\vec{\delta P} \ = \ \vec{P_2} \ - \ \vec{P_1} \ = \ (-0.8 \ - \ 1.578 \) \ \hat{i} \ + \ (1.443 \ - \ 3.236) \ \hat{j} \ = \ -2.378 \ \hat{i} \ - \ 1.793 \ \hat{j} \ Kg ms^{-1} {/eq}

Magnitude of change in momentum:

{eq}|\vec{\delta P}| \ = \ \sqrt{2.378^2 \ + \ 1.793^2} \ = \ 2.978 \ Kg ms^{-1} {/eq}

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from AP Physics 1: Exam Prep

Chapter 9 / Lesson 1