Kinetic Energy and XY velocity components

Question

Initially, a 0.200-kg ball moves in the +x direction with a speed equal to 8.00 m/s. This ball

collides elastically with an identical 0.200-kg ball that initially is at rest. After the collision, one

ball moves at an angle of −20.0° with respect to the +x direction with a speed equal to 7.52 m/s.

After the collision, what are the x- and y-components of the velocity, 𝒗⃗ , of the other ball, and the

kinetic energies, K, of each ball?

Answer 1

By Law of Conservation of Momentum: mv = mv₁ + mv₂, or v = v₁ + v₂;

We project this equation, on x-axis 8 = 7.52cos20° + v_2x;

on y-axis 0 = - 7.52sin20° + v_2y.

From here v_2x = 8 - 7.52cos20° = 0.93 m/s

v_2y = 7.52sin20° = 2.57 m/s

K₁ =mv₁²/2 = 0.200kg·7.52m²/s² = 5.66 J

K₂ = m(v_2x² + v_2y²)/2 = 0.2kg(0.93² + 2.57²)m²/s²/2 = 0.747 J