Question

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?

Answer 1

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

Explanation:

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=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

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

Explanation:

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"`