A ball with a mass of $2 k g$ $2 kg$ is rolling at $4 \frac{m}{s}$ $4 m/s$ 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 $2 k g$ $2 kg$ is rolling at $4 \frac{m}{s}$ $4 m/s$ 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 · 2017-05-09T08:25:43

The velocity of the first ball is $= - 1.33 m s^{- 1}$ $=-1.33ms^-1$
The velocity of the second ball is $= 2.67 m s^{- 1}$ $=2.67ms^-1$

Explanation:

In an elastic collision, we have conservation of momentum and conservation of kinetic energy.

The velocities before the collision are $u_{1}$ $u_1$ and $u_{2}$ $u_2$ .

The velocities after the collision are $v_{1}$ $v_1$ and $v_{2}$ $v_2$ .

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

and

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

Solving the above 2 equations for $v_{1}$ $v_1$ and $v_{2}$ $v_2$ , we get

$v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \cdot u_{1} + \frac{2 m_{2}}{m_{1} + m_{2}} \cdot u_{2}$ $v_1=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2$

and

$v_{2} = \frac{2 m_{1}}{m_{1} + m_{2}} \cdot u_{1} + \frac{m_{2} - m_{1}}{m_{1} + m_{2}} \cdot u_{2}$ $v_2=(2m_1)/(m_1+m_2)*u_1+(m_2-m_1)/(m_1+m_2)*u_2$

Taking the direction as positive $\to^{+}$ $rarr^+$

$m_{1} = 2 k g$ $m_1=2kg$

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

$u_{1} = 4 m s^{- 1}$ $u_1=4ms^-1$

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

Therefore,

$v_{1} = - \frac{2}{6} \cdot 4 + \frac{8}{6} \cdot (0) = - \frac{4}{3} = - 1.33 m s^{- 1}$ $v_1=-2/6*4+8/6*(0)=-4/3=-1.33ms^-1$

$v_{2} = \frac{4}{6} \cdot 4 - \frac{2}{6} \cdot (0) = \frac{8}{3} = 2.67 m s^{- 1}$ $v_2=4/6*4-2/6*(0)=8/3=2.67ms^-1$

Verificaition

$m_{1} u_{1} + m_{2} u_{2} = 2 \cdot 4 + 4 \cdot 0 = 8$ $m_1u_1+m_2u_2=2*4+4*0=8$

$m_{1} v_{1} + m_{2} v_{2} = - 2 \cdot \frac{4}{3} + 4 \cdot \frac{8}{3} = 8$ $m_1v_1+m_2v_2=-2*4/3+4*8/3=8$

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A ball with a mass of $2 k g$ $2 kg$ is rolling at $4 \frac{m}{s}$ $4 m/s$ 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?

1 Answer

Explanation:

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A ball with a mass of 2 kg2kg2 kg is rolling at 4 m/s4ms4 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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A ball with a mass of $2 k g$ $2 kg$ is rolling at $4 \frac{m}{s}$ $4 m/s$ 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?