Question

A ball with a mass of `5 kg` moving at `2 m/s` hits a still ball with a mass of `3 kg`. If the first ball stops moving, how fast is the second ball moving?

Answer 1

`3.333\ \text{m/sec}`

Explanation:

One ball of mass `m_1=5`kg moving at `u_1=2\ m/s` hits another ball of mass `m_2=3` kg at rest `u_2=0`. After collision, the first ball stops i.e. `v_1=0` & second ball moves with a velocity of `v_2` in the initial direction of first ball then

By the conservation of momentum in moving direction of first ball

`\text{momentum before collision}=\text{momentum after collision}`

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`5(2)+3(0)=5(0)+3v_2`

`3v_2=10`

`v_2=10/3`

`=3.333\ \text{m/sec}`

Answer 2

`~~3.33` meters per second

Explanation:

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

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

where:

• `m_1,m_2` are the masses of the two balls

• `u_1,u_2` are the initial velocities of the two balls

• `v_1,v_2` are the final velocities of the two balls

So, we get:

`5 \ "kg"*2 \ "m/s"+3 \ "kg"*0 \ "m/s"=5 \ "kg"*0 \ "m/s"+3 \ "kg"*v_2`

`v_2=(10color(red)cancelcolor(black)"kg""m/s")/(3color(red)cancelcolor(black)"kg")`

`~~3.33 \ "m/s"`