Question

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s. What is the momentum of block B afterward? PLEASE HELP

Answer 1

Momentum of block B after collision =
`-50\ kg\ ms^(-1)`

Step-by-step explanation:

Given

Before collision:

Momentum of block A =
`p_(A1)`=
`-100\ kg\ ms^(-1)`

Momentum of block B =
`p_(B1)`=
`-150\ kg\ ms^(-1)`

After collision:

Momentum of block A =
`p_(A2)`=
`-200\ kg\ ms^(-1)`

Applying law of conservation of momentum to find momentum of block B after collision
`p_(B2)`.

`p_(A1)+p_(B1)=p_(A2)+p_(B2)`

Plugging in the given values and simplifying.

`-100-150=-200+p_(B2)`

`-250=-200+p_(B2)`

Adding 200 to both sides.

`200-250=-200+p_(B2)+200`

`-50=p_(B2)`

∴
`p_(B2)=-50\ kg\ ms^(-1)`

Momentum of block B after collision =
`-50\ kg\ ms^(-1)`