A ball with a mass of #8 kg# moving at #7 m/s# hits a still ball with a mass of #16 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
Oct 5, 2016

#"velocity of the second object after collision : "v_2^'=3.5" "m/s#

#"the lost kinetic energy is 100 Joules"#

Explanation:

#"Momentum and Kinetic energy before collision :"#
#".........................................................................."#

#vec P_1:"momentum of the first object before collision"#

#m_1:"mass of the first object"#

#v_1:"velocity of the first object before collision"#

#vec P_1=m_1*v_1#

#vec P_1=8*7=56 " "kg*m/s#

#vec P_2:"momentum of the second object before collision"#

#m_2:"mass of the second object"#

#v_2:"velocity of the second object before collision"#

#vec P_2=m_2*v_2#

#vec P_2=16*0=0#

#Sigma vec P_b:"the vectorial sum of the momentums before collision"#

#Sigma vec P_b=vec P_1+vec P_2#

#Sigma vec P_b=56+0=56 " "kg*m/s#

#Sigma E_k:"Total kinetic energy"#

#Sigma E_k=1/2*m_1*v_1^2+0" (the second object hasn't kinetic energy)"#

#Sigma E_k=1/2*8*7^2=4*49=196" Joules"#

#"Momentum and Kinetic energy after collision :"#
#".........................................................................."#

#P_1^'=m_1*v_1^'#

#"So "v_1^'=0" ; "P_1^'=8*0=0#

#P_2^'=m_2*v_2^'#

# Sigma vec P_a=P_1^'+P_2^'" total momentum after..."#

#Sigma vec P_a=0+16*v_2^'#

#Sigma vec P_b=Sigma vec P_a#

#56=16*v_2^'#

#v_2^'=56/16=3.5 " "m/s#

#E_k=1/2*m_2*v_2^('2)+0=1/2*16*(3.5)^2=8*12.25#

#E_k=98" Joules"#

#Delta E_k=196-96#

#Delta E_k=100" Joules ( lost kinetic energy)" #