A ball with a mass of #18 kg# moving at #2 m/s# hits a still ball with a mass of #25 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?

1 Answer
Jan 13, 2016

The second ball is moving at a velocity of #1.44 m/s#.
The kinetic energy lost as heat is #10.08" Joules"#.

Explanation:

  • If the momentum of an object of mass #m# moving at a velocity #v# is #p# , then
    #p=m*v#

  • All still objects have no velocity and thus have a momentum of zero.

  • The law of conservation of momentum states that In an isolated or closed system, the total momentum of the objects after a certain change equals the total momentum of the objects before that change, so
    #sump_"initial"=sump_"final"# (GCSEscience, 2015)

  • Since, in the given case, the two balls are hitting each other and their velocities are moving in opposing directions, the total momentum will be calculated by subtracting the two momentums from each other.

  • The given information is given in the following table:

Objects------------- #p_"initial"# in #kg*m/s#--------------- #p_"final"# in #kg*m/s#

#Ball_A#---------------- #18*2=36# ------------------- #0# (since ball is still)

#Ball_B#---------------- #0# (since ball is still)----------- #25*v_B#

#sump#----------------- #36# -----------------------------------#25v_B#

  • According to the law of conservation of momentum:
    #36=25v_B#

#v_B=36/25#
#v_B=1.44 m/s#

  • For the second half of the question we need to calculate the total kinetic energy of the system before and after the collision, then find the difference between the two.

  • Kinetic Energy (#KE#) = #1/2mv^2#
    where #m#and #v# are as defined earlier.
    #KE_"initial"=(1/2*18*2^2)+(0)# = #36 " Joules"#
    #KE_"final"=(0)+(1/2*25*1.44^2)# = #25.92" Joules"#

  • The difference between the two kinetic energies that turned into heat #=36-25.92#
    #=10.08 " Joules"#

  • References:
    GCSEscience, 2015. What is the Law of Conservation of Momentum? . [Online]. Available from:http://www.gcsescience.com/pfm39.htm. [Accessed: 13th Jan 2016].