Question

A ball with a mass of `12 kg` moving at `3 m/s` hits a still ball with a mass of `21 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

`vec v_1^'=-9/11 " "m/s`
`v_2^'=24/11 " "m/s`

Explanation:

`"Momentums before collision :"`

`vec P_1=m_1*vec v_1" (for the first object)"`
`vec P_1=12*3=" "36 kg.m/s`

`vec P_2=m_2*vec v_2" (for the second object)"`
`vec P_2=21*0=0`

`Sigma vec P_b=vec P_1+ vec P_2" (total momentum before collision)"`
`Sigma vec P_b=36+0=36" " kg*m/s`

`"Momentums after collision :"`

`vec P_1^'=m_1*v_1^'" (for the first object)"`
`vec P_1^'=12*v_1^'`

`vec P_2^'=m_2*v_2^'" (for the second object)"`
`vec P_2^'=21*v_2^'`

`Sigma vec P_a=vec P_1^'+vec P_2^'" (total momentum after collision)"`
`Sigma vec P_a=12*v_1^'+21*v_2^'`

`Sigma vec P_b=Sigma vec P_a" (conservation of momentum")`
`36=12*vec v_1^'+21*vec v_2^'" (1)"`

`"We can fallow two ways to solve problem:"`

`"1 )............................................................"`

`v_1+v_1^'=v_2+v_2^' " (using the conservation of kinetic energy)"`
`3+v_1^'=0+v_2^'`
`v_2^'=3+v_1^' " (2)"`

`"let' use (1)"`
`36=12*vec v_1^'+21*(3+vec v_1^')`
`36=12*vec v_1^'+63+21*vec v_1^'`
`36-63=33*vec v_1^'`
`-27=33*vec vec v_1^'`

`vec v_1^'=-27/33" "vec v_1^'=-9/11 " "m/s`

`"now ,let's use (2)"`

`vec v_2^'=3+vec v_1^'`
`vec v_2^'=3-9/11" "vec v_2^'=(33-9)/11" "v_2^'=24/11 " "m/s`

`"2)............................................................."`

`v_1^'=(2*Sigma vec P_b)/(m_1+m_2)-v_1`

`v_1^'=(2*36)/(12+21)-3`

`v_1^'=72/33-3" "v_1^'=(72-99)/33 " "v_1^'=-27/33=-9/11 " m/s"`

`v_2^'=(2*Sigma vec P_b)/(m_1+m_2)-v_2`

`v_2^'=(2*36)/(12+21)-0`

`v_2^'=72/33=" "v_2^'=24/11 " "m/s`