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A ball with a mass of #12kg# moving at #4 m/s# hits a still ball with a mass of #2 kg#. If the first ball stops moving, how fast is the second ball moving?

Answer 1

#v_2=24ms^-1#

This is a simple case of conservation of linear momentum. This means that #m_1v_1=m_2v_2#. Reason why is because they said that the first ball stops, so the velocity of the first ball after collision is #0# and that reduces our equation so.

We have, #m_1=12kg#, #m_2=2kg#, #v_1=4ms^-1#, so we have to find #v_2#.

So, #12*cancel{4}^2=cancel{2}^1*v_2#

From here on, I'm sure it becomes easy.

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Answer 2

Adrian Ayala

Using the principle of conservation of momentum, we can find the final velocity of the second ball.

Initial momentum = Final momentum

( m_1 \times v_1 + m_2 \times v_2 = 0 )

Where ( m_1 = 12 ) kg, ( v_1 = 4 ) m/s, ( m_2 = 2 ) kg (mass of the still ball), and ( v_2 ) is the final velocity of the second ball.

( 12 \times 4 + 2 \times v_2 = 0 )

( 48 + 2v_2 = 0 )

( 2v_2 = -48 )

( v_2 = \frac{-48}{2} )

( v_2 = -24 ) m/s

So, the second ball is moving at 24 m/s in the opposite direction.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.