Question

Cassy shoots a large marble (Marble A, mass: 0.06kg) at a smaller marble (Marble B, mass: 0.03kg) that is sitting still. Marble A was initially moving at a velocity of 0.7m/s, but after the collision it has a velocity of -.02m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work.

Answer 1

The question can be solved using conservation of linear momentum.

`M_(a)` = 0.06kg and
`M_(b)` = 0.03kg

Let the initial velocity of Marble A be ,
`V_(a1)` = 0.7m/s

Let the initial velocity of Marble B be,
`V_(b1)` = 0m/s

Let the velocity of Marble A after collisiong ,
`V_(a2)`= -0,02m/s

Let the velocity of Marble B after collision be
`V_(b2)`

From the conservation of linear momentum equation. We get,

`M_(a)V_(a1)+M_(b)V_(b1)=M_(a)V_(a2)+M_(b)V_(b2)`

Substituting the values we get,

(0.06)(0.7) + 0 = (0.06)(-0.02) + (0.03)
`V_(b2)`

we get,
`V_(b2)` = 1.44m/s