A ball with a mass of #12 kg# moving at #1 m/s# hits a still ball with a mass of #8 kg#. If the first ball stops moving, how fast is the second ball moving?

Answer 1

#v_2'=1,5 m/s#

#P_1:" momentum of object with a mass of 12 kg before event"# #P_1^':" momentum of object with a mass of 12 kg after event"# #P_2:" momentum of object with a mass of 8 kg before event"# #P_2^':" momentum of object with a mass of 8 kg after event"# #P_1+P_2=P_1'+P_2'# #"the momentum of an object equal zero if it has a velocity of 0"# #m_1*v_1+0=0+m_2*v_2'# #12*1=8*v_2'# #v_2'=12/8# #v_2'=1,5 m/s#
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Answer 2

Using the principle of conservation of momentum:

[ m_1 \cdot v_1 + m_2 \cdot v_2 = 0 ]

Given ( m_1 = 12 , \text{kg} ), ( v_1 = 1 , \text{m/s} ), ( m_2 = 8 , \text{kg} ), and ( v_2 = ? ):

[ 12 \cdot 1 + 8 \cdot v_2 = 0 ]

[ 12 + 8 \cdot v_2 = 0 ]

[ 8 \cdot v_2 = -12 ]

[ v_2 = -\frac{12}{8} ]

[ v_2 = -1.5 , \text{m/s} ]

The second ball is moving at 1.5 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.

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.

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.

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.

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