A ball with a mass of #4 kg# moving at #3 m/s# hits a still ball with a mass of #26 kg#. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

2 Answers

Answer:

#6/13=0.461\ \text{m/s}# & loss of K.E. #=15.231\ J#

Explanation:

Ball of mass #m_1=4kg # moving with #u_1=3\ m/s# strikes another ball of mass #m_2=26\ kg# at rest #u_2=0#. After collision, first ball stops #v_1=0# & second ball moves with the velocity #v_2# then by law of conservation of momentum we have

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

#4(3)+26(0)=4(0)+26v_2#

#v_2=12/26#

#=6/13\ \text{m/s}#

Loss of kinetic energy

#=\text{initial K.E.}-\text{final K.E.}#

#=1/2m_1u_1^2-1/2m_2v_2^2#

#=1/2(4)(3)^2-1/2(26)(6/13)^2#

#=15.231\ J#

Jul 28, 2018

Answer:

The speed is #=0.46ms^-1# and the loss in kinetic energy is #=15.23J#

Explanation:

There is conservation of momentum

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The mass #m_1=4kg#

The velocity #u_1=3ms^-1#

The mass #m_2=26kg#

The velocity #u_2=0ms^-1#

The velocity #v_1=0ms^-1#

Therefore,

#4*3+26*0=4*0+26*v_2#

#v_2=12/26=0.46ms^-1#

The loss in kinetic energy is

#DeltaKE=KE_1-KE_2#

#=1/2m_1u_1^2-1/2m_2v_2^2#

#=1/2*4*3^2-1/2*26*0.46^2#

#=15.23J#