Question

One billiard ball is shot east at 2.0 m/s. a second, identical billiard ball is shot west at 0.8 m/s. the balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.55 m/s. what are the speed and direction of the first ball after the collision?

Answer 1

speed = 1.96 m/s

direction = 52.25° south of east

Step-by-step explanation:

given that:

the velocity of the first ball
`v_1 = 2.0 \ m/s`

the velocity of the second ball
`v_2 = \ 0.8 \ m/s`

the final velocity of the second ball
`v_2f = 1.55 \ m/s`

The mass of the identical balls
`m_1 = m_2 = m`

However from law of conservation of momentum
`m_1v_1 + m_2v_2 = m_1v_1f + m_2v_2f`

along X- axis

`m_1vx_1f = m_1v_1 + m_2v_2`

`vx_1f =v_1 + v_2`

= ( 2.0 - 0.8 ) m/s

= 1.2 m/s

along Y-axis

`0 = m_1vy_1f + m_2v_2f\\m_1vy_1f = -m_2v_2f\\then vyf_1 = - v_2f \\= -1.55 \ m/s`

then the final speed of the first ball

`v_1f = \sqrt{(1.2)^2 + (1.55)^2`

`v_1f = 1.96 \ m/s`

and direction

`\theta = tan^(-1)( (-1.55)/(1.2) )`

`\theta = 52.25^0` south of east.