A ball with a mass of #8 kg# moving at #14 m/s# hits a still ball with a mass of #4 kg#. If the first ball stops moving, how fast is the second ball moving?

Question

A ball with a mass of #8 kg# moving at #14 m/s# hits a still ball with a mass of #4 kg#.  If the first ball stops moving, how fast is the second ball moving?

Emily Abbate · Accepted Answer

See explanation.

Two balls should have the same momentum both before and after they collide. Prior to the collision, we have: #v_1=14# and #v_2=0#. Given that the masses are #m_1=8# and #m_2=4# we can write that: #m_1v_1+m_2v_2=m_1v_1'+m_2v_2'# The first ball stops, so the velocity #v_1'=0#. If we replace the information, we obtain: #8*14+4*0=8*0+4*v_2'# #8*14=4*v_2'# #v_2'=(8*14)/4=2*14=28# Answer: The velocity of the second ball is #v_2'=28 m/s#

Julia Aguilar · Answer

The velocity of the second ball after the collision is #=28ms^-1#

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=8kg#

The velocity of the first ball before the collision is #u_1=14ms^-1#

The mass of the second ball is #m_2=4kg#

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,

#8*14+4*0=8*0+4*v_2#

#4v_2=112#

#v_2=112/4=28ms^-1#

The velocity of the second ball after the collision is #v_2=28ms^-1#

Benjamin Asiedu · Answer

undefined To find the velocity of the second ball, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Initial momentum = final momentum

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

$8 \times 14 + 4 \times 0 = (8 + 4) \times v_f$

$112 = 12 \times v_f$

$v_f = \frac{112}{12} = 9.33 \, \text{m/s}$

So, the second ball is moving at approximately 9.33 m/s after the collision.