Question

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

Answer 1

Answer:

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 = $\frac{(m_2-m_1)u_2}{ m_2 +m_1} + \frac{2m_1u_1}{m_1+m_2}$

= $\frac{(M-4M)\times 0}{ M +4M} + \frac{2\times 4M\times 5v_0}{4M+M }$

= 8v₀ .

Answer 2

The change in the velocity in a collision can be determined by its initial and final velocities.

After the collision, the velocity of block Y will be 8v0.

Velocity After Collision

Given that Block X has an initial velocity of 5v0, the mass of 4M and has a final velocity after the collision is 3v0.

Block Y has an initial velocity of 0v0 as it is at rest before the collision and has a mass of M.

Let us consider that the initial velocity of block X can be represented by u_x and its final velocity by v_x. The initial velocity of block Y can be represented by u_y and its final velocity by v_y.

Then after the collision, the final velocity of Block Y can be given below.

$v_y = \dfrac {(M - 4M)u_y} {M+4M} + \dfrac {2\times 4M \times u_x}{M+4M}$

$v_y = \dfrac {(-3M)\times 0}{5M} + \dfrac {8M\times 5}{5M}$

$v_y = 0 + 8$

$v_y = 8 v_0$

Hence we can conclude that, after the collision, the velocity of block Y will be 8v0.

To know more about the velocity, follow the link given below.

https://brainly.com/question/14620367.