A ball with a mass of $5 k g$ $5 kg$ is rolling at $1 \frac{m}{s}$ $1 m/s$ 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 $1 \frac{m}{s}$ $1 m/s$ 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 · 2018-06-28T09:36:03

The post collisions velocities are $= 011 m s^{- 1}$ $=011ms^-1$ and $= 1.411 m s^{- 1}$ $=1.411ms^-1$

Explanation:

Since the collision is elastic, we have conservation of momentum

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

And conservation of kinetic energy

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

The mass the first ball is $m_{1} = 5 k g$ $m_1=5kg$

The velocity of the first ball before the collision is $u_{1} = 1 m s^{- 1}$ $u_1=1ms^-1$

The mass of the second ball is $m_{2} = 4 k g$ $m_2=4kg$

The velocity of the second ball before the collision is $u_{2} = 0 m s^{- 1}$ $u_2=0ms^-1$

The velocity of the first ball after the collision is $= v_{1} m s^{- 1}$ $=v_1ms^-1$

The velocity of the second ball after the collision is $= v_{2} m s^{- 1}$ $=v_2ms^-1$

Therefore,

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

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

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

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

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

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

Substituting in equation $(2)$ $(2)$

$5 v_{1}^{2} + 4 \cdot {(\frac{5 (1 - v_{1})}{4})}^{2} = 5$ $5v_1^2+4*((5(1-v_1))/4)^2=5$

$4 v_{1}^{2} + 5 + 5 v_{1}^{2} - 10 v_{1} = 4$ $4v_1^2+5+5v_1^2-10v_1=4$

$9 v_{1}^{2} - 10 v_{1} + 1 = 0$ $9v_1^2-10v_1+1=0$

$v_{1} = \frac{10 \pm \sqrt{({(- 10)}^{2}) - 4 (9) (1)}}{18} = \frac{10 \pm 8}{18}$ $v_1=(10+-sqrt(((-10)^2)-4(9)(1)))/(18)=(10+-8)/18$

Therefore,

$v_{1} = 1 m s^{- 1}$ $v_1=1ms^-1$ or $v_{1} = \frac{1}{9} = 0.11 m s^{- 1}$ $v_1=1/9=0.11ms^-1$

$v_{2} = 0 m s^{- 1}$ $v_2=0ms^-1$ or $v_{2} = \frac{10}{9} = 1.11 m s^{- 1}$ $v_2=10/9=1.11ms^-1$

The first solutions are the initial conditions.

The post collisions velocities are $= 011 m s^{- 1}$ $=011ms^-1$ and $= 1.411 m s^{- 1}$ $=1.411ms^-1$

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A ball with a mass of $5 k g$ $5 kg$ is rolling at $1 \frac{m}{s}$ $1 m/s$ 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 $1 \frac{m}{s}$ $1 m/s$ 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?