# Question Video: Studying the Collision of Two Spheres Where One Is Moving and the Other Is Resting on the Same Line Mathematics

A sphere of mass 299 g was moving horizontally in a straight line at 51 cm/s. It collided with another sphere of mass 390 g that was at rest. As a result of the impact, the first sphere came to rest. Determine the speed of the second sphere after the impact.

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### Video Transcript

A sphere of mass 299 grams was moving horizontally in a straight line at 51 centimeters per second. It collided with another sphere of mass 390 grams that was at rest. As a result of the impact, the first sphere came to rest. Determine the speed of the second sphere after the impact.

We can call the mass of the first sphere thatβs mentioned, 299 grams, π sub one. And the speed of π sub one before it reaches the second ball, 51 centimeters per second, weβll call π£ sub one. The second sphere, the one thatβs initially at rest, has a mass of 390 grams, which weβll name π sub two. Knowing that the first mass comes to a rest after the collision, we want to calculate the speed of the second sphere after the impact. Weβll call that speed π£ sub two.

Letβs start off by sketching out this collision. In this situation, before the collision, mass one is moving at speed π£ one to encounter mass two, which has a speed of zero. Itβs stationary. Then after the spheres collide, mass one is stationary, not in motion at all, and mass two is moving at a speed π£ sub two. Itβs that speed that we want to solve for. And weβll solve for π£ sub two by recalling that momentum is conserved.

Mathematically, we can write this out as saying that the initial momentum, π sub π, of a system is equal to its final momentum, π sub π. Written another way, we can say that the initial mass of a system multiplied by its initial speed is equal to its final mass times its final speed.

In our given system, we have two masses: π sub one and π sub two. So the initial momentum in the system is equal to π one π£ one plus π two times its initial speed. Since π sub two is initially at rest, the second term is zero. And the initial momentum of our system reduces to π one times π£ one.

When we consider our system after the collision, we see that now π one is not in motion and π two is at a speed π£ sub two. So the term π one times zero is equal to zero. And our final momentum of the system is π two times π£ two.

And by the principle of the conservation of momentum, the initial momentum of our system is equal to the final momentum. That is, π one π£ one is equal to π two π£ two. Since π£ two is what we want to solve for, we can rearrange this equation. And we see that π£ sub two is equal to the ratio of mass one to mass two multiplied by the speed of mass one initially, π£ one.

When we plug in the given values for these three terms and calculate π£ sub two, we find itβs 39.1 centimeters per second. Thatβs the speed of the second ball after the collision.