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?

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?

Lily Adams · Accepted Answer

#3.333\ ext{m/sec}# 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 using momentum conservation while the first ball is moving # ext{momentum before collision}= ext{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\ ext{m/sec}#

Benjamin Asiedu · Answer

#~~3.33# meters per second

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"#

Parker Abbott · Answer

undefined To find the final velocity of the second ball after the collision, you can use the principle of conservation of momentum:

$$ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} $$

where $ m_1 $ and $ m_2 $ are the masses of the first and second balls, $ v_{1i} $ and $ v_{2i} $ are the initial velocities of the first and second balls, and $ v_{1f} $ and $ v_{2f} $ are the final velocities of the first and second balls.

Given that the first ball stops moving ($ v_{1f} = 0 $), and the initial velocities are $ v_{1i} = 2 \, \text{m/s} $ and $ v_{2i} = 0 $, and the masses are $ m_1 = 5 \, \text{kg} $ and $ m_2 = 3 \, \text{kg} $, we can solve for $ v_{2f} $:

$$ 5 \cdot 2 + 3 \cdot 0 = 5 \cdot 0 + 3 \cdot v_{2f} $$

$$ 10 = 3 \cdot v_{2f} $$

$$ v_{2f} = \frac{10}{3} $$

Therefore, the final velocity of the second ball after the collision is $ \frac{10}{3} \, \text{m/s} $.