Problem 9.42 Part A One billiard ball is shot east at 1.7 m/s. A second identica

Question

Problem 9.42 Part A One billiard ball is shot east at 1.7 m/s. A second identical billiard ball is shot west at 0.90 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.41 m/s What is the speed of the first ball after the collision? Express your answer to two significant figures and include the appropriate units 1 Value Units Submit My Answers Give Up Part B What is the direction of the first ball after the collision? Give the direction as an angle south of east. Express your answer to two significant figures and include the appropriate units Value Units Submit My Answers Give Up

Explanation / Answer

P(initial) = m(1.7 - 0.90) = 0.8 m N-s[East]

Let the final momentum of the first ball is P at A* from E to W

By the law of momentum conservation in East direction:-

PcosA = P(initial) = 0.8 m

m * v * cosA = 0.8 m

vcosA = 0.8 --------------(i)

On N - W axis:-

PsinA = m * 1.41

m * v * sinA = m * 1.41

vsinA = 1.41 ---------(ii)

By (i)^2 + (ii)^2 :-

v^2 = (0.8)^2 + (1.41)^2

v = sqrt2.6281 = 1.621 m/s

By (ii)/(i) :-

tanA = 1.41

A = 60.43 degree south ofe east

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