Consider a glancing collision between two identical spheres of 0.016 kg, with one of the spheres initially at rest. Initially the incoming projectile has a velocity of 1.50 m/s to the right and after the collision we observe that this sphere leaves the collision region with a velocity of 0.80 m/s at an angle of 25 to its initial direction. (Assume that the +x-axis is to the right and the +y-axis is up along the page. Angles are measured counterclockwise from the +x-axis. Do not assume this collision is elastic.)

A)Find the x-component of the velocity of the second sphere.

B)Find the y-component of the velocity of the second sphere.

(a) The x-component of the second sphere's velocity is 0.775 m/s

(b) The y-component of the second sphere's velocity is -0.338 m/s

Explanation:

Conservation of Linear Momentum

The principle of the conservation of linear momentum states that the total momentum of a closed system of particles is conserved while no external forces act on any part of it, regardless of the interaction between the particles. The momentum is a vector, thus we must analyze both axes in the calculations.

The momentum can be computed as

where m is the mass of the particle, and is its velocity. In a system of two particles, the total initial momentum is

And the final momentum is

Since the total momentum is conserved

According to the conditions of the problem, both masses are identical and sphere 2 is initially at rest (v2=0), thus

Simplifying by m

We know that the first sphere has a velocity of 1.5 m/s to the right. This means that the vertical component of v1 is 0:

We also know after the collision this same sphere travels at 0.8 m/s at an angle of 25°. The components of this velocity are

(a) Solving the equation [1] for v2

The x-component of the second sphere's velocity is 0.775 m/s

(b) The y-component of the second sphere's velocity is -0.338 m/s