Question

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?

Answer 1

`"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)"`