A ball with a mass of $3 k g$ $3 kg$ is rolling at $2 \frac{m}{s}$ $2 m/s$ 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 $3 k g$ $3 kg$ is rolling at $2 \frac{m}{s}$ $2 m/s$ 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 · 2017-06-26T10:13:42

The velocity of the first ball is $= - \frac{2}{7} m s^{- 1}$ $=-2/7ms^-1$ and the second ball is $= \frac{12}{7} m s^{- 1}$ $=12/7ms^-1$

Explanation:

Since the collision is elastic, there is conservation of linear momentum and conservation of kinetic energy.

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

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$ $1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2mv_2^2$

$m_{1} = 3 k g$ $m_1=3kg$

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

$u_{1} = 2 m s^{- 1}$ $u_1=2ms^-1$

$u_{2} = 0 m s^{- 1}$ $u_2=0 ms^-1$

So,

$3 \cdot 2 + 4 \cdot 0 = 3 \cdot v_{1} + 4 \cdot v_{2}$ $3*2+4*0=3*v_1+4*v_2$

$3 v_{1} + 4 v_{2} = 6$ $3v_1+4v_2=6$ ............................... $(1)$ $(1)$

$\frac{1}{2} \cdot 3 \cdot 2^{2} + \frac{1}{2} \cdot 4 \cdot 0 = \frac{1}{2} \cdot 3 \cdot v_{1}^{2} + \frac{1}{2} \cdot 4 \cdot v_{2}^{2}$ $1/2*3*2^2+1/2*4*0=1/2*3*v_1^2+1/2*4*v_2^2$

$3 v_{1}^{2} + 4 v_{2}^{2} = 12$ $3v_1^2+4v_2^2=12$ ........................ $(2)$ $(2)$

Solving for $v_{1}$ $v_1$ and $v_{2}$ $v_2$ in equations $(1)$ $(1)$ and $(2)$ $(2)$

$v_{2} = \frac{6 - 3 v_{1}}{4}$ $v_2=(6-3v_1)/4$

$3 v_{1}^{2} + 4 \cdot {(\frac{6 - 3 v_{1}}{4})}^{2} = 12$ $3v_1^2+4*((6-3v_1)/4)^2=12$

$12 v_{1}^{2} + 36 - 36 v_{1} + 9 v_{1}^{2} = 48$ $12v_1^2+36-36v_1+9v_1^2=48$

$21 v_{1}^{2} - 36 v_{1} - 12 = 0$ $21v_1^2-36v_1-12=0$

$7 v_{1}^{2} - 12 v_{1} - 4 = 0$ $7v_1^2-12v_1-4=0$

$v_{1} = \frac{12 \pm \sqrt{12^{2} + 4 \cdot 4 \cdot 7}}{14}$ $v_1=(12+-sqrt(12^2+4*4*7))/(14)$

$= \frac{12 \pm 16}{14}$ $=(12+-16)/14$

$v_{1} = 2 m s^{- 1}$ $v_1=2ms^-1$ or $v_{1} = - \frac{2}{7} = - 0.28 m s^{- 1}$ $v_1=-2/7=-0.28ms^-1$

We keep the second result $v_{1} = 0.28 m s^{- 1}$ $v_1=0.28ms^-1$

$v_{2} = \frac{6 + 3 \cdot 0.28}{4} = \frac{12}{7} = 1.71$ $v_2=(6+3*0.28)/4=12/7=1.71$

Verification

$4 \cdot \frac{12}{7} - 3 \cdot \frac{2}{7} = 6$ $4*12/7-3*2/7=6$

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A ball with a mass of $3 k g$ $3 kg$ is rolling at $2 \frac{m}{s}$ $2 m/s$ 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 $3 k g$ $3 kg$ is rolling at $2 \frac{m}{s}$ $2 m/s$ 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?