A ball with a mass of $5 k g$ $5 kg$ is rolling at $12 \frac{m}{s}$ $12 m/s$ and elastically collides with a resting ball with a mass of $4 k g$ $4 kg$ . What are the post-collision velocities of the balls?

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A ball with a mass of $5 k g$ $5 kg$ is rolling at $12 \frac{m}{s}$ $12 m/s$ and elastically collides with a resting ball with a mass of $4 k g$ $4 kg$ . What are the post-collision velocities of the balls?

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Answer 1 · 2017-11-23T13:18:47

The velocities are $\frac{4}{3} m s^{- 1}$ $4/3ms^-1$ and $\frac{40}{3} m s^{- 1}$ $40/3ms^-1$

Explanation:

Here it's an elastic collision with no loss of kinetic energy.

We have the conservation of momentum.

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ $m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_{1} = 5 k g$ $m_1=5 kg$

$u_{1} = 12 m s^{- 1}$ $u_1=12 ms^(-1)$

$m_{2} = 4 k g$ $m_2=4 kg$

$u_{2} = 0$ $u_2=0$

$v_{1} = ?$ $v_1=?$

$v_{2} = ?$ $v_2=?$

$5 \cdot 12 + 4 \cdot 0 = 5 \cdot v_{1} + 4 \cdot v_{2}$ $5*12+4*0=5*v_1+4*v_2$

$5 v_{1} + 4 v_{2} = 60$ $5v_1+4v_2=60$ ............ $(1)$ $(1)$

There is also conservation of kinetic energies

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ $1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2$

$5 \cdot 12^{2} + 4 \cdot 0 = 5 \cdot v_{1}^{2} + 4 \cdot v_{2}^{2}$ $5*12^2+4*0=5*v_1^2+4*v_2^2$

$5 v_{1}^{2} + 4 v_{2}^{2} = 720$ $5v_1^2+4v_2^2=720$ ......... $(2)$ $(2)$

From $(1)$ $(1)$ , we get

$v_{1} = \frac{60 - 4 v_{2}}{5}$ $v_1=(60-4v_2)/5$

Plugging this value in $(2)$ $(2)$

$5 \cdot {(\frac{60 - 4 v_{2}}{5})}^{2} + 4 v_{2}^{2} = 720$ $5*((60-4v_2)/5)^2+4v_2^2=720$

$\frac{{((60 - 4 v_{2}))}^{2}}{5} + 4 v_{2}^{2} = 720$ $((60-4v_2))^2/5+4v_2^2=720$

$16 {(15 - v_{2})}_{2}^{2} + 20 v_{2}^{2} = 720 \cdot 5 = 3600$ $16(15-v_2)^2+20v_2^2=720*5=3600$

$4 {(15 - v_{2})}_{2}^{2} + 5 v_{2}^{2} = 900$ $4(15-v_2)^2+5v_2^2=900$

$4 (225 - 30 v_{2} + v^{2}) + 5 v_{2}^{2} = 900$ $4(225-30v_2+v^2)+5v_2^2=900$

$900 - 120 v_{2} + 4 v_{2}^{2} + 5 v_{2}^{2} = 900$ $900-120v_2+4v_2^2+5v_2^2=900$

$9 v_{2}^{2} - 120 v_{2} = 0$ $9v_2^2-120v_2=0$

$v_{2} (9 v_{2} - 120) = 0$ $v_2(9v_2-120)=0$

$v_{2} = 0$ $v_2=0$ or $v_{2} = \frac{120}{9} = \frac{40}{3} = 13.3 m s^{- 1}$ $v_2=120/9=40/3=13.3ms^-1$

$v_{1} = \frac{60 - 4 \cdot \frac{40}{3}}{5} = \frac{4}{3} m s^{- 1}$ $v_1=(60-4*40/3)/5=4/3ms^-1$

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A ball with a mass of $5 k g$ $5 kg$ is rolling at $12 \frac{m}{s}$ $12 m/s$ and elastically collides with a resting ball with a mass of $4 k g$ $4 kg$ . What are the post-collision velocities of the balls?

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Explanation:

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A ball with a mass of 5 kg5kg 5 kg is rolling at 12 m/s12ms12 m/s and elastically collides with a resting ball with a mass of 4 kg4kg4 kg. What are the post-collision velocities of the balls?

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Explanation:

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A ball with a mass of $5 k g$ $5 kg$ is rolling at $12 \frac{m}{s}$ $12 m/s$ and elastically collides with a resting ball with a mass of $4 k g$ $4 kg$ . What are the post-collision velocities of the balls?