Help pleass URGENT If a 0.16 kg 8-ball at rest is hit by the 0.17 kg cue ball that is moving at a speed of 2 m/s, what is the speed of the 8-ball if the cue ball is completely stopped after the collision?

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

When a collision occurs, the velocities change to v' and the final momentum is:

In a system of two masses, the law of conservation of linear momentum is simplified to:

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

The speed of the 8-ball is 2.125 m/s after the collision.

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b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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