A ball with a mass of $3 k g$ $3 kg$ is rolling at $8 \frac{m}{s}$ $8 m/s$ 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 $8 \frac{m}{s}$ $8 m/s$ 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 · 2018-01-05T09:37:18

The post collision velocities are $= - 1.14 m s^{- 1}$ $=-1.14ms^-1$ and $= 6.86 m s^{- 1}$ $=6.86ms^-1$

Explanation:

In an elastic collision, there is conservation of momentum and conservation of kinetic energies.

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

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

Here,

The masses are

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

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

The initial velocities are

$u_{1} = 8 m s^{- 1}$ $u_1=8ms^-1$

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

$3 \cdot 8 + 4 \cdot 0 = 3 \cdot v_{1} + 4 \cdot v_{2}$ $3*8+4*0=3*v_1+4*v_2$ , $\Leftrightarrow$ $<=>$ , $3 v_{1} + 4 v_{2} = 24$ $3v_1+4v_2=24$ ........ $(1)$ $(1)$

$\frac{1}{2} \cdot 3 \cdot 8^{2} + \frac{1}{2} \cdot 4 \cdot 0^{2} = \frac{1}{2} \cdot 3 \cdot v_{1}^{2} + \frac{1}{2} \cdot 4 \cdot v_{2}^{2}$ $1/2*3*8^2+1/2*4*0^2=1/2*3*v_1^2+1/2*4*v_2^2$

$3 v_{1}^{2} + 4 v_{2}^{2} = 192$ $3v_1^2+4v_2^2=192$ ............................. $(2)$ $(2)$

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

From $(1)$ $(1)$ , $v_{2} = \frac{24 - 3 v_{1}}{4}$ $v_2=(24-3v_1)/4$

Substituting is $(2)$ $(2)$

$3 v_{1}^{2} + 4 \cdot {(\frac{24 - 3 v_{1}}{4})}^{2} = 192$ $3v_1^2+4*((24-3v_1)/4)^2=192$

$12 v_{1}^{2} + {(24 - 3 v_{1})}_{1}^{2} = 768$ $12v_1^2+(24-3v_1)^2=768$

$12 v_{1}^{2} + 576 - 144 v_{1} + 9 v_{1}^{2} = 768$ $12v_1^2+576-144v_1+9v_1^2=768$

$21 v_{1}^{2} - 144 v - 192 = 0$ $21v_1^2-144v-192=0$

$7 v_{1}^{2} - 48 v - 64 = 0$ $7v_1^2-48v-64=0$

$v_{1} = \frac{48 \pm \sqrt{{(- 48)}^{2} - 4 \cdot 7 \cdot (- 64)}}{2 \cdot 7}$ $v_1=(48+-sqrt((-48)^2-4*7*(-64)))/(2*7)$

$= \frac{48 \pm \sqrt{4096}}{14}$ $=(48+-sqrt(4096))/(14)$

$= \frac{48 \pm 64}{14}$ $=(48+-64)/(14)$

So,

$v_{1} = (\frac{48 + 64}{14}) = 8 m s^{- 1}$ $v_1=((48+64)/14)=8ms^-1$ , $\Rightarrow$ $=>$ , $v_{2} = 0 m s^{- 1}$ $v_2=0ms^-1$

This is the initial conditions.

or

$v_{2} = (\frac{48 - 64}{14}) = - 1.14 m s^{- 1}$ $v_2=((48-64)/14)=-1.14ms^-1$ , $\Rightarrow$ $=>$ , $v_{2} = 6.86 m s^{- 1}$ $v_2=6.86ms^-1$

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A ball with a mass of $3 k g$ $3 kg$ is rolling at $8 \frac{m}{s}$ $8 m/s$ 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 3kg 3 kg is rolling at 8ms8 m/s and elastically collides with a resting ball with a mass of 4kg4 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 $8 \frac{m}{s}$ $8 m/s$ 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?