A ball with a mass of #12kg# moving at #6 m/s# hits a still ball with a mass of #7 kg#. If the first ball stops moving, how fast is the second ball moving?

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Answer 1 · 2018-03-09T13:49:09

The velocity of the second ball after the collision is #=10.3ms^-1#

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The mass the first ball is #m_1=12kg#

The velocity of the first ball before the collision is #u_1=6ms^-1#

The mass of the second ball is #m_2=7kg#

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,

#12*6+7*0=12*0+7*v_2#

#7v_2=72#

#v_2=72/7=10.3ms^-1#

The velocity of the second ball after the collision is #v_2=10.3ms^-1#

Answer 2 · 2018-03-09T13:55:36

I get approximately #10.3 \ "m/s"#.

We use the law of conservation of momentum, which states that

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

We have:

#m_1=12 \ "kg", u_1= 6 \ "m/s", m_2=7 \ "kg",u_2=0 \ "m/s",v_1=0 \ "m/s"#

Plugging in the values into the equation, we get:

#12 \ "kg"*6 \ "m/s"+7 \ "kg"*0 \ "m/s"=12 \ "kg"*0 \ "m/s"*7 \ "kg"*v_2#

#=>7 \ "kg"*v_2=72 \ "kg m/s"#

#v_2=(72color(red)cancelcolor(black)"kg""m/s")/(7color(red)cancelcolor(black)"kg")~~10.3 \ "m/s"#