A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball.

Question

A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. As a result,
the incident ball is deflected 30.00
from its initial path (remember the 900
rule for same masses).
a. Draw a vector diagram showing the momentum of each ball after the collision.
b. What is the velocity of the incident ball after the collision? (ans: 0.35 m/s, 300
to incident
direction)
c. What is the velocity of the struck ball after the collision? (ans: 0.20 m/s, 600
to incident ball
direction)

Answer 1

Use conservation of linear momentum and the 90° rule.

 

Horizontal direction:

 

v = initial speed of moving ball

u₁ = final speed of moving ball

u₂ = final speed of stationary ball

 

mv = mu₁cos30° + mu₂cos60°

 

(i)   0.40 = 0.866u₁ + 0.5u₂

 

Vertical direction:

 

0 = mu₁sin30° - mu₂sin60°

 

(ii)  0.5u₁ - 0.866u₂ = 0

 

Solve (i) and (ii) simultaneously for u₁ and u₂ by substitution, elimination, or Cramer's rule and get

 

u₁ ≅ 0.346 m/s at 30° above incidence

u₂ ≅ 0.200 m/s at 60° below incidence