Unit: Linear Momentum

Multi-Step Problem Preview

Part 1 of 6 — sign up to solve the full problem!

Start free trial

Part 1

A cue ball (Ball A) of mass mm is moving across a frictionless pool table with an initial velocity v0v_0 in the +x+x direction. It strikes a stationary 8-ball (Ball B) of identical mass mm in a glancing collision. The collision is perfectly elastic.

Compiling TikZ diagram...

Which of the following correctly identifies the conserved quantities for the two-ball system immediately before and after the collision?

Correct!

Solution:

For the system consisting of the cue ball (Ball A) and the 8-ball (Ball B), we analyze the external forces. Since the pool table is frictionless, there are no external dissipative forces like friction acting on the system in the horizontal plane. The vertical forces (gravity and normal force) balance each other. Therefore, the net external force on the system is zero. According to the conservation of momentum principle:

Pi=Pf \vec{P}_i = \vec{P}_f

Thus, total linear momentum is conserved.

The problem explicitly states that the collision is perfectly elastic. By definition, an elastic collision is one where the internal non-conservative work is zero, meaning the total kinetic energy of the system is conserved:

Ki=KfK_i = K_f 12mv02=12mvA2+12mvB2\frac{1}{2} m v_0^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2

Therefore, both momentum and kinetic energy are conserved.

Want to solve all 6 parts?

Sign up for a free account to work through the complete multi-step problem with instant feedback!

Ready to Start Learning?

Sign up now to access the full 8-Ball Pool Glancing Collision lesson and our entire curriculum!