Two 6.6 kg bodies, A and B, collide. The velocities before the collision are A = (32 i + 41 j) m/s and B = (31 i + 3.2 j) m/s. After the collision, velocity of A = (7.9i + 11 j) m/s. ?

What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?

1 Answer
A08
Apr 29, 2018

(a) Using Law of Conservation of momentum we have for #x#-components

#m_A*V_"Axi" + m_B*V_"Bxi" = m_A*V_"Axf" + m_B*V_"Bxf"#

Since #m_A=m_B=6.6\ kg#, above equation reduces to

#V_"Axi" + V_"Bxi" = V_"Axf" + V_"Bxf"#

Inserting given values we get

#32 + 31 = 7.9 + V_"Bxf"#
#=>V_"Bxf" =55.1\ ms^-1#

(b) Similarly, for #y#-components

#V_"Ayi" + V_"Byi" = V_"Ayf" + V_"Byf"#
#=>41 + 3.2 = 11 + V_"Byf"#
#=> V_"Byf"=33.2\ ms^-1#

(c) Change in kinetic energy

#DeltaKE="Total final KE"-"Total initial KE"#
#=>DeltaKE=1/2xx6.6((V_(Af)^2+V_(Bf)^2)-(V_(Ai)^2+V_(Bi)^2))#
#=>DeltaKE=1/2xx6.6((7.9^2+11^2)+(55.1^2+33.2^2)-((32^2 + 41^2)+(31^2 + 3.2^2))#
#=>DeltaKE=1/2xx6.6(1102.24-6895.66)#
#=>DeltaKE=-19118.3\ J#

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