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# Simple collision problem?

A billiard ball is shot west at 0.4 m/s. A second, identical ball is shot east at 1.9 m/s. The balls have a glancing collision, which deflects the first ball at 90 deg. and sending it north at 0.4 m/s. What is the speed of the other ball right after the collision?

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- Steve4PhysicsLv 71 year agoFavorite Answer
Only 'simple' when you know how to do it!

Take east as +x and north as +y.

To conserve momentum in the y direction, ball 2 must have a final y component of velocity= -0.4m/s.

The ball 1 has a final x component of momentum = 0 (as it moves in the y direction).

Applying conservation of momentum in the x direction:

m(-0.4) + m(1.9) = m.0 + mv

v = 1.5m/s (x component of ball 2's final velocity)

So ball 2's speed = √(0.4² + 1.5²) = 1.55m/s

(=1.6m/s to 2 sig. figs.)

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