Question

10. A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30 Degree to the original path, determine a. the speed of the first ball after the collision. b. the speed and direction of the second ball after the collision.

Answer 1

V_1ix = initial velocity of first ball in X-direction = 2 m/s

V_1iy = initial velocity of first ball in Y-direction = 0 m/s

V_1fx = final velocity of first ball in X-direction = V₁ Cos30

V_1fy = final velocity of first ball in Y-direction = V₁ Sin30

V_2ix = initial velocity of second ball in X-direction = 0 m/s

V_2iy = initial velocity of second ball in Y-direction = 0 m/s

V_2fx = final velocity of second ball in X-direction = V₂ Cos $\theta$

V_2fy = final velocity of first ball in Y-direction = - V₂ Sin $\theta$

Using conservation of momentum Along X-direction :

m₁ V_1ix + m₂ V_2ix = m₁ V_1fx + m₂ V_2fx

m₁ (3) + m₂ (0) = m₁ V₁ Cos30 + m₂ V₂ Cos $\theta$

V₁ Cos30 + V₂ Cos $\theta$ = 3                            

V₂ Cos $\theta$ = 3 - V₁ Cos30                             Eq-1

Using conservation of momentum Along Y-direction :

m₁ V_1iy + m₂ V_2iy = m₁ V_1fy + m₂ V_2fy

2 (0) + 2 (0) = 2 (V₁ Sin30) + 2 (- V₂ Sin $\theta$ )

V₂ Sin $\theta$ = V₁ Sin30                            Eq-2

Squaring Eq-1 and Eq-2 and adding

V²₂ Cos² $\theta$ + V²₂ Sin² $\theta$ = (3 - V₁ Cos30 )² + V²₁ Sin²30     

V²₂ = 9 + V²₁ Cos²30 - 6 V₁ Cos30 + V²₁ Sin²30  

V²₂ = 9 + V²₁ - 6 V₁ Cos30        Eq-3

since the collision is elastic

(0.5) m₁ (3)² + (0.5) m₂ (0)² = (0.5) m₁ (V₁)² + (0.5) m₂ (V₂)²

9 = (V₁)² + (V₂)²

V₂² = 9 - V₁²                               Eq-4

Using eq-3 and Eq-4

9 - V₁² = 9 + V²₁ - 6 V₁ Cos30

V₁= 2.6 m/s

from Eq-4

V₂² = 9 - V₁²     = 9 - (2.6)² = 2.24

V₂ = 1.5 m/s            

Using Eq-2

V₂ Sin $\theta$ = V₁ Sin30    

1.5 Sin $\theta$ = (2.6) (0.5)

$\theta$ = 60.5