Block X of mass 4M travels at a speed 5v0 toward block Y of mass M, which is initially at rest. After block X elastically collides with block Y, block X has a speed of 3v0 in the same direction as before the collision. What is the speed of block Y immediately after the collision

Explanation:

Initial velocity of block X

u₁ = 5 v₀ ,

Initial velocity of block Y

u₂ = 0 ,

Final velocity of block X

v₁ = 3 v₀

Final velocity of block Y

v₂ = ?

Mass of block X

m₁ = 4 M

Mass of block Y

m₂ = M .

For elastic collision , velocity for Y is given by the following expression

v₂ =

=

= 8v₀ .

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s