# A ball with a mass of #2 kg# is rolling at #4 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

The velocity of the first ball is

The velocity of the second ball is

In an elastic collision, we have conservation of momentum and conservation of kinetic energy.

and

and

Therefore,

Verificaition

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The post-collision velocities of the balls can be calculated using the conservation of linear momentum. The equation is (m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f), where (m1) and (m2) are the masses, (v1i) and (v2i) are the initial velocities, and (v1f) and (v2f) are the final velocities.

Given: (m1 = 2 \ kg), (v1i = 4 \ m/s), (m2 = 4 \ kg), (v2i = 0 \ m/s) (resting ball)

Substitute values into the conservation of linear momentum equation and solve for (v1f) and (v2f):

(2 * 4 + 4 * 0 = 2 * v1f + 4 * v2f)

(8 = 2 * v1f + 0)

(v1f = 4 \ m/s)

Now, substitute (v1f) into the equation:

(2 * 4 + 4 * 0 = 2 * 4 + 4 * v2f)

(8 = 8 + 0)

(v2f = 0 \ m/s)

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- A ball with a mass of #2 kg# moving at #3 m/s# hits a still ball with a mass of #4 kg#. If the first ball stops moving, how fast is the second ball moving?
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