A ball with a mass of $3 k g$ $3 kg$ is rolling at $4 m s^{- 1}$ $4 ms^-1$ 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 $4 m s^{- 1}$ $4 ms^-1$ 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 · 2016-01-15T05:24:28

In an elastic collision, both momentum and kinetic energy are conserved. The velocity of the $3 k g$ $3kg$ ball is $0.9 or 2.7 m s^{- 1}$ $0.9 or 2.7 ms^-1$ and the velocity of the $4 k g$ $4kg$ ball is $2.7 or 1.1 m s^{- 1}$ $2.7 or 1.1 ms^-1$ .

Explanation:

Momentum is conserved in all collisions. Kinetic energy is conserved in elastic collisions but not in inelastic or partially elastic collisions.

The initial momentum of the whole system is $p = m v$ $p = mv$ for the $3 k g$ $3kg$ ball: the $4 k g$ $4kg$ ball has zero momentum because its has zero velocity.

$p = m v = 3 \cdot 4 = 12 k g m s^{- 1}$ $p=mv=3*4=12 kgms^-1$

The total momentum after the collision will be the same after the collision. If we call the $3 k g$ $3kg$ ball 1 and the $4 k g$ $4kg$ ball 2, the final momentum will be given by:

$p = m_{1} v_{1} + m_{2} v_{2} = 3 v_{1} + 4 v_{2} = 12$ $p = m_1v_1 +m_2v_2 = 3v_1 +4v_2 = 12$ (call this Equation 1)

The total kinetic energy before the collision will be $E_{k} = \frac{1}{2} m v^{2}$ $E_k=1/2mv^2$ for the $3 k g$ $3 kg$ ball only - the $4 k g$ $4 kg$ ball has zero kinetic energy because it has zero velocity.

$E_{k} = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 3 \cdot 4^{2} = 24 J$ $E_k = 1/2mv^2 = 1/2*3*4^2 = 24 J$

Since kinetic energy is conserved, the final kinetic energy will be the same, and will be given by:

$E_{k} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{3}{2} v_{1}^{2} + \frac{4}{2} v_{2}^{2} = 24$ $E_k = 1/2m_1v_1^2 + 1/2m_2v_2^2 = 3/2v_1^2+4/2v_2^2 = 24$

Multiply both sides by 2 to make it tidier:

$3 v_{1}^{2} + 4 v_{2}^{2} = 48$ $3v_1^2 + 4v_2^2 = 48$ (call this Equation 2)

We now have two equations and two unknowns, so we can solve them as simultaneous equations. Rearranging Equation 1 to express $v_{2}$ $v_2$ in terms of $v_{1}$ $v_1$ :

$v_{2} = \frac{12 - 3 v_{1}}{4}$ $v_2 = (12-3v_1)/4$

Substituting this into Equation 2:

$3 v_{1}^{2} + 4 {(\frac{12 - 3 v_{1}}{4})}^{2} = 48$ $3v_1^2+4((12-3v_1)/4)^2 = 48$

I'll leave the algebra as an exercise for the reader, but this solves to give $v_{1} = 0.9 m s^{- 1} or 2.5 m s^{- 1}$ $v_1=0.9ms^-1 or 2.5 ms^-1$ . Substituting this back into Equation 1 gives $v_{2} = 2.7 m s^{- 1} or 1.1 m s^{- 1}$ $v_2=2.7 ms^-1 or 1.1 ms^-1$ , respectively.

Substituting these values should confirm that both momentum and kinetic energy were conserved.

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A ball with a mass of $3 k g$ $3 kg$ is rolling at $4 m s^{- 1}$ $4 ms^-1$ 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 $4 m s^{- 1}$ $4 ms^-1$ 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?