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A ball with a mass of #13 kg# moving at #7 m/s# hits a still ball with a mass of #15 kg#. If the first ball stops moving, how fast is the second ball moving?

Answer 1

Lily Adams

The velocity of the second ball is #=6.07ms^-1#

We have conservation of momentum

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The mass the first ball is #m_1=13kg#

The velocity of the first ball before the collision is #u_1=7ms^-1#

The mass of the second ball is #m_2=15kg#

The velocity of the second ball before the collision is #u_2=0ms^-1#

The velocity of the first ball after the collision is #v_1=0ms^-1#

Therefore,

#13*7+15*0=13*0+15*v_2#

#15v_2=91#

#v_2=91/15=6.07ms^-1#

The velocity of the second ball after the collision is #v_2=6.07ms^-1#

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Answer 2

William Audy

To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

$m_1v_1 + m_2v_2 = m_1v_{1f} + m_2v_{2f}$

Given:
$m_1 = 13 \, \text{kg}$ ,
$v_1 = 7 \, \text{m/s}$ ,
$m_2 = 15 \, \text{kg}$ ,
$v_{1f} = 0 \, \text{m/s}$ (as the first ball stops)

Substituting the values:

$13 \times 7 + 15 \times 0 = 13 \times 0 + 15 \times v_{2f}$

$91 = 15 \times v_{2f}$

$v_{2f} = \frac{91}{15} \, \text{m/s}$

$v_{2f} ≈ 6.07 \, \text{m/s}$

Therefore, the second ball is moving at approximately $6.07 \, \text{m/s}$ after the collision.

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Answer from HIX Tutor

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