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

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 that’s mentioned, 299 grams, π‘š sub one. And the speed of π‘š sub one before it reaches the second ball, 51 centimeters per second, we’ll call 𝑣 sub one. The second sphere, the one that’s initially at rest, has a mass of 390 grams, which we’ll name π‘š 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. We’ll call that speed 𝑣 sub two.

Let’s start off by sketching out this collision. In this situation, before the collision, mass one is moving at speed 𝑣 one to encounter mass two, which has a speed of zero. It’s stationary. Then after the spheres collide, mass one is stationary, not in motion at all, and mass two is moving at a speed 𝑣 sub two. It’s that speed that we want to solve for. And we’ll solve for 𝑣 sub two by recalling that momentum is conserved.

Mathematically, we can write this out as saying that the initial momentum, 𝑝 sub 𝑖, of a system is equal to its final momentum, 𝑝 sub 𝑓. 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‘š sub one and π‘š sub two. So the initial momentum in the system is equal to π‘š one 𝑣 one plus π‘š two times its initial speed. Since π‘š sub two is initially at rest, the second term is zero. And the initial momentum of our system reduces to π‘š one times 𝑣 one.

When we consider our system after the collision, we see that now π‘š one is not in motion and π‘š two is at a speed 𝑣 sub two. So the term π‘š one times zero is equal to zero. And our final momentum of the system is π‘š two times 𝑣 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‘š one 𝑣 one is equal to π‘š two 𝑣 two. Since 𝑣 two is what we want to solve for, we can rearrange this equation. And we see that 𝑣 sub two is equal to the ratio of mass one to mass two multiplied by the speed of mass one initially, 𝑣 one.

When we plug in the given values for these three terms and calculate 𝑣 sub two, we find it’s 39.1 centimeters per second. That’s the speed of the second ball after the collision.

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