# Question Video: Finding the Speed of a Ball Moving on a Horizontal Plane after Collision with a Vertical Barrier given Its Impulse Mathematics

A sphere of mass 125 g was moving along a section of horizontal ground at 165 cm/s when it hit a vertical wall. Given that the wall applied an impulse of magnitude 0.214 N ⋅ s to the sphere, find the speed of the sphere as it bounced off the wall.

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### Video Transcript

A sphere of mass 125 grams was moving along a section of horizontal ground at 165 centimeters per second when it hit a vertical wall. Given that the wall applied an impulse of magnitude 0.214 newton seconds to the sphere, find the speed of the sphere as it bounced off the wall.

So we have a sphere moving along a section of horizontal ground. And we’re told that it bounces off a vertical wall. And we’re being asked to calculate the speed of the sphere as it bounced off the wall. So we want to find out how fast the sphere is travelling after it’s collided with the wall. The question tells us the mass of the sphere is 125 grams. And we’re also told that the initial speed of the sphere as it travels towards the wall is 165 centimeters per second. And in addition to this, we’re told that the wall applies an impulse of magnitude 0.214 newton seconds to the sphere.

At this point, we can recall that an impulse can be equated to a change in momentum. This means that when an object experiences an impulse of a certain size, its momentum will change by that exact amount. So when the question tells us that the wall applies an impulse of magnitude 0.214 newton seconds to the sphere, what this means is that the sphere’s momentum changes by 0.214 newton seconds during the collision with the wall. So since this question tells us information about the change in momentum that occurs during the collision, let’s think about the momentum of the sphere before the collision with the wall and after the collision with the wall.

Let’s recall that the momentum of an object 𝑝 is given by that object’s mass 𝑚 multiplied by its velocity 𝑣. This means that the momentum of the sphere before the collision, we could call this its initial momentum 𝑝 𝑖, is equal to its mass, we’ll call this 𝑚 for now, multiplied by its velocity, we’ll call this 𝑣 𝑖 for initial velocity. Both of these quantities are given in the question. We’re told that the mass of the sphere is 125 grams and the speed of the sphere as it approaches the wall is 165 centimeters per second.

Now one thing we need to be careful of here is that momentum, velocity, and even impulse are vector quantities. This means that they can take positive or negative values depending on which way that they’re acting. So we need to decide which way is our positive direction and which way is our negative direction. For this question, let’s say that any vectors acting to the right are positive and any vectors acting to the left are negative.

Because the initial velocity of the sphere is going to the right, this means that 𝑣 𝑖 is pointing in the positive direction, so it takes a positive value. So 𝑣 𝑖 is equal to positive 165 centimeters per second, which expressed in standard units is 1.65 meters per second. We also know that the mass 𝑚 is equal to 125 grams, which expressed in the standard units for mass is 0.125 kilograms.

Using these values, we can see that the initial momentum of the sphere is equal to 0.125 kilograms multiplied by 1.65 meters per second, which gives us a value of 0.20625. Since we used standard units in our calculation and we’re calculating momentum, this value takes the standard units for momentum. Often, we express momentum in kilogram meters per second. But we could equivalently use newton seconds, which are the units used for impulse in the question.

Now that we’ve calculated the initial value of the sphere’s momentum, we can use the information we’ve been given about the impulse it receives to calculate its momentum after the collision. Let’s call this final momentum 𝑝 𝑓. To see exactly how we can do this, we can rewrite this equation: Impulse is equal to a change in momentum. Using symbols, we could write this as 𝐼 equals 𝛥𝑝. Now 𝛥𝑝, the change in momentum, is equal to the difference between the final and initial values of momentum, which we could write as 𝑝 𝑓 minus 𝑝 𝑖.

So if we rearrange this equation by adding 𝑝 𝑖 to both sides, we can see that the final momentum of an object 𝑝 𝑓 is equal to its initial momentum 𝑝 𝑖 plus the impulse 𝐼 that it receives. We can use this equation to calculate the final momentum of the sphere. We just need to substitute in the value of initial momentum, which we’ve already calculated, and a value for the impulse 𝐼.

Now, at this point, we need to be really careful to remember that impulse is a vector quantity and that the question only tells us the magnitude of this quantity. The value that we use for 𝐼 might be positive or negative depending on the direction that the impulse vector acts. Now because the sphere in this question bounces off the wall, we know that its momentum changes direction. This is only possible if it experiences an impulse that opposes its initial momentum. This means that the impulse acts toward the left, which we’ve defined as the negative direction. So in this case, the impulse 𝐼 takes a value of negative 0.214 newton seconds.

Substituting both of these values into this equation gives us 0.20625 minus 0.214, which is equal to negative 0.00775 newton seconds. And the fact that this value’s negative signifies that the sphere is indeed moving to the left. Now all that’s left is to calculate the velocity based on this momentum because momentum is equal to mass times velocity, the final momentum 𝑝 𝑓 must be equal to mass times the final velocity 𝑣 𝑓.

We can rearrange this equation to make 𝑣 𝑓 the subject by dividing both sides by 𝑚. This gives us 𝑣 𝑓 equals 𝑝 𝑓 over 𝑚. Substituting in our calculated value of final momentum and the mass expressed in kilograms gives us negative 0.00775 divided by 0.125, which is equal to negative 0.062. The units here are the standard units of velocity meters per second. Now because the question asks for the speed of the sphere, our answer is given by the magnitude of this velocity, so we can ignore the negative sign.

And the very final step is to convert this speed from meters per second into the units of speed that are used in the question, in this case, centimeters per second. And 0.062 meters per second is 6.2 centimeters per second. So there is our final answer. If a sphere of mass 125 grams was moving along a section of horizontal ground at 165 centimeters per second then bounces off a wall which applies an impulse of 0.214 newton seconds, the speed of the sphere as it bounces off the wall is 6.2 centimeters per second.