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