Answer to Question #182187 in Mechanics | Relativity for Te Waiwhakaata Ratu

Question

Two identical balls collide as shown in the figure above. Ball A has an initial speed of 2.0 m s⁻¹ and Ball B is initially stationary.

After the collision, Ball B is moving at a speed of 1.6 m s⁻¹ at an angle of 60° to the direction it was orginally travelling in.

What is the speed of Ball A after the collision? (to 2 s.f)

Answer 1

The momentum is concerved in both x and y projections:

$m v_{1x} +mv_{2x}= mv_{3x}+mv_{4x}$

$m v_{1y} +mv_{2y}= mv_{3y}+mv_{4y}$

We can eliminate m from both equations, so only velocities remain.

$2+0 = 1.6 \cos 60^\circ + v_{4x}$

$v_{4x}=2-0.8=1.2$ m/s

$0 + 0=1.6 \sin 60^\circ + v_{4y}$

$v_{4y}=1.386$ m/s

The total speed of ball A is

$v_4 = \sqrt{1.2^2+1.386^2}=1.833$ m/s

Answer: 1.83 m/s