Respuesta :
Answer:
Momentum of block B after collision =[tex]-50\ kg\ ms^{-1} [/tex]
Explanation:
Given
Before collision:
Momentum of block A = [tex]p_{A1}[/tex]= [tex]-100\ kg\ ms^{-1}[/tex]
Momentum of block B = [tex]p_{B1}[/tex]= [tex]-150\ kg\ ms^{-1}[/tex]
After collision:
Momentum of block A = [tex]p_{A2}[/tex]= [tex]-200\ kg\ ms^{-1}[/tex]
Applying law of conservation of momentum to find momentum of block B after collision [tex]p_{B2}[/tex].
[tex]p_{A1}+p_{B1}=p_{A2}+p_{B2}[/tex]
Plugging in the given values and simplifying.
[tex]-100-150=-200+p_{B2}[/tex]
[tex]-250=-200+p_{B2}[/tex]
Adding 200 to both sides.
[tex]200-250=-200+p_{B2}+200[/tex]
[tex]-50=p_{B2}[/tex]
∴ [tex]p_{B2}=-50\ kg\ ms^{-1} [/tex]
Momentum of block B after collision =[tex]-50\ kg\ ms^{-1} [/tex]
