Question

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 direction of the cue ball and 2.0 N·s perpendicular to that direction. What is the momentum of the cue ball after the collision?

Answer 1

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 .

$P = \sqrt{x^2+y^2}$

$P = \sqrt{7^2+(-2)^2}$

      P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s