The momentum of a system before a collision is 2.4 × 103 kilogram meters/second in the x-direction and 3.5 × 103 kilogram meters/second in the y-direction. What is the magnitude of the resultant momentum after the collision if the collision is inelastic?

A1.7 × 103 kilogram meters/second
B2.1 × 103 kilogram meters/second
C3.4 × 103 kilogram meters/second
D4.2 × 103 kilogram meters/second
E5.7 × 103 kilogram meters/second ...?

If inelastic, then the total amount of system momentum before the collision (and after) can be determined by using the Pythagorean theorem. It is expressed as:

c^2 = a^2 + b^2
c^2 = (2.4x10^3)^2 + (3.5x10^3)^2
c = 4.2 x10^3 kilogram meters/second -------> OPTION C

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tardymanchester tardymanchester · Answer 2 · 2019-03-11T02:28:17-04:00

Answer:

Option D - \(4.2\times 10^3\) kilogram meters/second.

Step-by-step explanation:

Given : The momentum of a system before a collision is \(2.4 \times 10^3\) kilogram meters/second in the x-direction and \(3.5 \times 10^3\) kilogram meters/second in the y-direction.

To find : What is the magnitude of the resultant momentum after the collision if the collision is inelastic?

Solution :

In inelastic,

The magnitude of the resultant momentum after the collision (and before) can be determined by using the Pythagorean theorem.

\(c^2 = a^2 + b^2\)

Where, a is the momentum before collision \(a=2.4 \times 10^3\)

b is the momentum after collision \(b=3.5 \times 10^3\)

c is the total amount of momentum before and after collision.

Substitute the value,

\(c^2 = (2.4\times10^3)^2 + (3.5\times10^3)^2\)

\(c^2 =5760000+ 12250000\)

\(c=\sqrt{18010000}\)

\(c=4243.819\)

\(c=4.2\times 10^3\) kg m/s

Therefore, Option D is correct.

The magnitude of the resultant momentum after the collision if the collision is inelastic is \(4.2\times 10^3\) kilogram meters/second.