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IRINA_888 [86]
2 years ago
12

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?
Physics
1 answer:
lisov135 [29]2 years ago
7 0

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

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initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

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As we have to find the time taken, the formula can be altered as,

t = \frac{v-u}{a}

where, t - time taken to reach a final speed

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a - acceleration.

Substituting all the given values,

t =\frac{781000 - 226000} {4* 1014}

= 1.3875 × 10⁻⁹ seconds.

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