A 0.120 kg billiard ball with velocity v = (2.50 m/s)i strikes a stationary billiard ball of the same mass. After the collision, the first billiard ball has velocity v₁ = (0.70 m/s) i + (-0.30 m/s)j. Ignore friction.

a) What is the final velocity of the second ball?

b) Is the collision elastic? Show your work.

a) This is a collision, in two dimensions, with no net external forces (no friction!), so we can apply conservation of momentum (note that we cannot also assume conservation of energy, since we are not told that the collision was elastic). Thus we have:

p = mv = mv₁ + mv₂

where m is the mass of either ball, and where

v = (2.50 m/s)i    is given,

v₁ = (0.70 m/s) i + (-0.30 m/s)j    is also given,

and the unknown is v₂, the final velocity of the second ball.

We can make life a bit easier by dividing through by the mass, m

v = v₁ + v₂

This is a vector equation, so break it into x and y components:

x:    v = v_1x + v_2x

y:    0 = v_1y + v_2y

or

x:    2.5 m/s = 0.7 m/s + v_2x

y:    0 = -0.3 m/s + v_2y

Each of these is easy to solve:

v_2x = 1.8 m/s

v_2y =0.3 m/s

Thus    v₂ = (1.8 m/s i + 0.3 m/s j)

b) If the collision is elastic, then mechanical energy is conserved, i.e.

½mv² = ½mv₁² + ½mv₂²

again, we can make life simpler by dividing through by ½m, so all we need to check is whether or not

v² = v₁² + v₂²   ?

The left hand side is

v² = (2.50 m/s)² = 6.25 m²/s²

while the right hand side is

(0.7 m/s)² + (-0.30 m/s)² + (1.8 m/s)² + (0.3 m/s)² = 3.91 m²/s²

which is only a bit more than half of the right hand side; thus the mechanical energy is not conserved, and the collision is inelastic. Perhaps lots of energy was lost to sound, heat, cracking one of the balls, etc...

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