Question

A 3.0-kg ball moving eastward at 4.0 m/s suddenly collides with and sticks to a 5.0-kg ball moving northward at 3.0 m/s. what is the magnitude of the momentum of this system just after the collision?

Answer 1

Magnitude of the momentum of this system just after the collision = 19.2Kgm/s

What is law of conservation of momentum?

Law of conservation of momentum states that the total momentum after the collision must be equal to the total momentum before the collision.

The momentum of each ball is given by:

p = mv

where m is the mass of the ball and v its velocity.

The momentum of ball 1 is:

p = mv

= (3.0 kg)(4.0 m/s)

= 12.0 kg m/s in the eastward direction

The momentum of ball 2 is:

p = mv

= (5.0 kg)(3.0 m/s)

= 15.0 kg m/s in the northward direction

The two momenta are in perpendicular directions, so the magnitude of the total momentum can be found as:

$p = \sqrt{p1^{2} +p2^{2} }$

$p = \sqrt{12^{2} +15^{2} }\\p = \sqrt{144 + 225} \\p = \sqrt{369}\\ p = 19.2 kgm/s$

and due to the law of conservation of the momentum, this is also equal to the total momentum after the collision.

To learn more about law of conservation of momentum from the given link

brainly.com/question/12773641

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