A ball with a mass of #2kg# moving at #1 m/s# hits a still ball with a mass of #6 kg#. If the first ball stops moving, how fast is the second ball moving?

Question

A ball with a mass of #2kg# moving at #1 m/s# hits a still ball with a mass of #6 kg#.  If the first ball stops moving, how fast is the second ball moving?

Layla Ahmed · Accepted Answer

#~~0.33 \ "m/s"#

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

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

#m_1,m_2# are the masses of the two objects

#u_1,u_2# are the initial velocities of the two objects

#v_1,v_2# are the final velocities of the two objects

Since we want to know how fast is the second ball moving after the collision, we need to solve for #v_2#, and we rearrange the equation into:

#v_2=(m_1u_1+m_2u_2-m_1v_1)/m_2#

Now, we can plug in values into the equation and we get,

#v_2=(2 \ "kg"*1 \ "m/s"+6 \ "kg"*0 \ "m/s"-2 \ "kg"*0 \ "m/s")/(6 \ "kg")#

#=(2 \ "kg m/s"+0 \ "kg m/s"-0 \ "kg m/s")/(6 \ "kg")#

#=(2color(red)cancelcolor(black)"kg""m/s")/(6color(red)cancelcolor(black)"kg")#

#=1/3 \ "m/s"#

#~~0.33 \ "m/s"#

Nora Arzate · Answer

undefined Use the law of conservation of momentum:

$$ \text{Initial momentum} = \text{Final momentum} $$

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

$$ (2 \, \text{kg} \times 1 \, \text{m/s}) + (6 \, \text{kg} \times v_2) = 0 $$

$$ v_2 = -\frac{2}{6} \, \text{m/s} $$