A ball with a mass of $4 k g$ $4 kg$ and velocity of $3 \frac{m}{s}$ $3 m/s$ collides with a second ball with a mass of $2 k g$ $2 kg$ and velocity of $- 1 \frac{m}{s}$ $- 1 m/s$ . If $75 %$ $75%$ of the kinetic energy is lost, what are the final velocities of the balls?

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A ball with a mass of $4 k g$ $4 kg$ and velocity of $3 \frac{m}{s}$ $3 m/s$ collides with a second ball with a mass of $2 k g$ $2 kg$ and velocity of $- 1 \frac{m}{s}$ $- 1 m/s$ . If $75 %$ $75%$ of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-26T15:05:44

The final velocities are $= 0.85 m s^{- 1}$ $=0.85ms^-1$ and $= 3.3 m s^{- 1}$ $=3.3ms^-1$

Explanation:

We have conservation of momentum

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

The kinetic energy is

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

Therefore,

$4 \times 3 + 2 \times (- 1) = 4 v_{1} + 2 v_{2}$ $4xx3+2xx(-1)=4v_1+2v_2$

$4 v_{1} + 2 v_{2} = 10$ $4v_1+2v_2=10$

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

and

$0.25 (\frac{1}{2} \times 4 \times 3^{2} + \frac{1}{2} \times 2 \times {(- 1)}^{2}) = \frac{1}{2} \times 4 \times v_{1}^{2} + \frac{1}{2} \times 2 \times v_{2}^{2}$ $0.25(1/2xx4xx3^2+1/2xx2xx(-1)^2)=1/2xx4xxv_1^2+1/2xx2xxv_2^2$

$4 v_{1}^{2} + 2 v_{2}^{2} = 9.5$ $4v_1^2+2v_2^2=9.5$

$2 v_{1}^{2} + v_{2}^{2} = 4.75$ $2v_1^2+v_2^2=4.75$ ................... $(2)$ $(2)$

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

$2 v_{1}^{2} + {((5 - 2 v_{1}))}^{2} = 4.75$ $2v_1^2+((5-2v_1))^2=4.75$

$2 v_{1}^{2} + 4 v_{1}^{2} - 20 v_{1} + 25 - 4.75 = 0$ $2v_1^2+4v_1^2-20v_1+25-4.75=0$

$6 v_{1}^{2} - 20 v_{1} - 20.25 = 0$ $6v_1^2-20v_1-20.25=0$

Solving this quadratic equation in $v_{1}$ $v_1$

$v_{1} = \frac{20 \pm \sqrt{20^{2} - 4 \times 6 \times - 20.25}}{12}$ $v_1=(20+-sqrt(20^2-4xx6xx-20.25))/(12)$

$v_{1} = \frac{40 \pm \sqrt{886}}{12}$ $v_1=(40+-sqrt(886))/(12)$

$v_{1} = \frac{40 \pm 29.8}{12}$ $v_1=(40+-29.8)/(12)$

$v_{1} = 5.81 m s^{- 1}$ $v_1=5.81ms^-1$ or $v_{1} = 0.85 m s^{- 1}$ $v_1=0.85ms^-1$

$v_{2} = - 6.62 m s^{- 1}$ $v_2=-6.62ms^-1$ or $v_{2} = 3.3 m s^{- 1}$ $v_2=3.3ms^-1$

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A ball with a mass of $4 k g$ $4 kg$ and velocity of $3 \frac{m}{s}$ $3 m/s$ collides with a second ball with a mass of $2 k g$ $2 kg$ and velocity of $- 1 \frac{m}{s}$ $- 1 m/s$ . If $75 %$ $75%$ of the kinetic energy is lost, what are the final velocities of the balls?

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Explanation:

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A ball with a mass of 4kg4 kg and velocity of 3ms3 m/s collides with a second ball with a mass of 2kg2 kg and velocity of −1ms- 1 m/s. If 75%75% of the kinetic energy is lost, what are the final velocities of the balls?

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A ball with a mass of $4 k g$ $4 kg$ and velocity of $3 \frac{m}{s}$ $3 m/s$ collides with a second ball with a mass of $2 k g$ $2 kg$ and velocity of $- 1 \frac{m}{s}$ $- 1 m/s$ . If $75 %$ $75%$ of the kinetic energy is lost, what are the final velocities of the balls?