A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original
1 answer:
Answer:
P = 7.28 N.s
Explanation:
given,
initial momentum of cue ball in x- direction,P₁ = 9 N.s
momentum of nine ball in x- direction, P₂ = 2 N.s
momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s
momentum of the cue after collision = ?
using conservation of momentum
in x- direction
P₁ + p = x + P₂
p is the initial momentum of the nine balls which is equal to zero.
9 + 0 = x + 2
x = 7 N.s
momentum in x-direction.
equating along y-direction
P'₁ + p = y + P'₂
0 + 0 = y + 2
y = -2 N.s
the momentum of the cue ball after collision is equal to resultant of the momentum .
P = 7.28 N.s
the momentum of the cue ball after collision is equal to P = 7.28 N.s
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We want to find y in terms of d, φ and θ.
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Therefore
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From (1), obtain the required expression for y.
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