Video Transcript
Two spheres of mass two 𝑚 and five 𝑚 were moving in the same direction along the same straight line on a smooth horizontal table. The lighter sphere was in front and was moving at four meters per second and the heavier sphere was behind and was moving at six meters per second. Given that when the two spheres collided, the lighter sphere’s speed increased to six meters per second, find the speed of the heavy one rounded to the nearest hundredth.
In order to answer this question, we will use the conservation of momentum, which states that the momentum before is equal to the momentum after. In a collision’s problem of this type, we use the formula 𝑚 one 𝑢 one plus 𝑚 two 𝑢 two is equal to 𝑚 one 𝑣 one plus 𝑚 two 𝑣 two, where 𝑚 one and 𝑚 two are the masses of the objects that collide. 𝑢 one and 𝑢 two are their velocities before the collision. And 𝑣 one and 𝑣 two are the velocities after. We will begin by sketching before and after diagrams.
We are told that the two spheres have masses two 𝑚 and five 𝑚 and that the lighter sphere is in front. This lighter sphere is moving at 4 meters per second. And the heavier sphere is moving at six meters per second. We are told that after the collision, the lighter sphere’s speed increases to six meters per second. We are trying to calculate the speed of the heavier one, which we will call 𝑣 meters per second.
The momentum before is equal to five 𝑚 multiplied by six plus two 𝑚 multiplied by four. This is equal to five 𝑚 multiplied by 𝑣 plus two 𝑚 multiplied by six. Since the mass cannot equal zero, we can divide through by 𝑚. Five multiplied by six is 30, and two multiplied by four is eight. Therefore, the left-hand side simplifies to 38.
The right-hand side is equal to five 𝑣 plus 12. We can then subtract 12 from both sides such that five 𝑣 is equal to 26. Dividing through by five, we have 𝑣 is equal to 5.2.
As we are asked to give our answer to the nearest hundredth, the speed of the heavier sphere after the collision is 5.20 meters per second.
