AP Physics 1 Exam Question:

A 0.2 kg ball is initially at rest on a frictionless surface. Another 0.4 kg ball approaches it with an initial velocity of 4 m/s to the right. Upon collision, the two balls stick together and move as a single unit. Calculate the final velocity of the balls after the collision.

Solution:

To solve this problem, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before a collision is equal to the total momentum after the collision, provided no external forces are acting on the system.

The momentum, p, of an object is given by its mass, m, multiplied by its velocity, v: p = mv.

Let's denote the initial velocity of the 0.2 kg ball as v1i and the initial velocity of the 0.4 kg ball as v2i. Since the 0.2 kg ball is initially at rest, v1i = 0 and v2i = 4 m/s.

After the collision, the two balls stick together and move with a common final velocity, vf.

Applying the conservation of linear momentum, we have:

(0.2 kg)(0 m/s) + (0.4 kg)(4 m/s) = (0.2 kg + 0.4 kg)vf

0 + 1.6 kg·m/s = 0.6 kg·vf

Simplifying the equation, we have:

1.6 kg·m/s = 0.6 kg·vf

Dividing both sides by 0.6 kg, we get:

vf = (1.6 kg·m/s) / (0.6 kg) ≈ 2.67 m/s

Therefore, the final velocity of the balls after the collision is approximately 2.67 m/s to the right.

NOTE: It is important to note that since the question states the two balls stick together after the collision, we assume that there are no external forces acting on the system of the two balls. This allows us to apply the conservation of linear momentum principle.