Question:

A 2 kg block initially at rest is placed on a frictionless horizontal surface. It is then struck by a 1 kg block that is moving with a velocity of 4 m/s to the right. After the collision, the 2 kg block moves to the right with a velocity of 2 m/s. What is the final velocity of the 1 kg block after the collision?

Answer:

The conservation of linear momentum states that the total momentum of an isolated system remains constant before and after a collision.

Let's denote the initial velocity of the 2 kg block as v1_i and the initial velocity of the 1 kg block as v2_i.

Given: Mass of the 2 kg block, m1 = 2 kg Mass of the 1 kg block, m2 = 1 kg Initial velocity of the 2 kg block, v1_i = 0 m/s Initial velocity of the 1 kg block, v2_i = 4 m/s to the right Final velocity of the 2 kg block, v1_f = 2 m/s to the right

According to the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision: Total initial momentum = m1 * v1_i + m2 * v2_i

After the collision: Total final momentum = m1 * v1_f + m2 * v2_f

Since the initial velocity of the 2 kg block is 0 m/s, its initial momentum is 0.

Total initial momentum = m2 * v2_i

Total final momentum = m1 * v1_f + m2 * v2_f

Using the conservation of linear momentum, we can write:

Total initial momentum = Total final momentum m2 * v2_i = m1 * v1_f + m2 * v2_f

Substituting the given values: 1 kg * 4 m/s = 2 kg * 2 m/s + 1 kg * v2_f

Simplifying the equation: 4 kg m/s = 4 kg m/s + 1 kg * v2_f

Rearranging the equation to solve for v2_f: 1 kg * v2_f = 4 kg m/s - 4 kg m/s v2_f = 0 m/s

Therefore, the final velocity of the 1 kg block after the collision is 0 m/s.

Hence, the final velocity of the 1 kg block after the collision is 0 m/s.