Question Video: Finding the Time Taken by a Body on a Smooth Inclined Plane to Move Up and Return given Its Initial Velocity Mathematics

Video Transcript

A body was projected at 16 meters per second up a smooth plane inclined at an angle 𝛼 to the horizontal, where sin 𝛼 is 45 over 49. Determine the time that the body took to return to the point of projection. Take 𝑔 to be equal to 9.8 meters per square second.

To answer a question like this, we’re going to begin by simply sketching a diagram. We have a smooth plane inclined at an angle 𝛼 to the horizontal. Now, the fact that the plane is smooth indicates to us that there will be no frictional forces acting on the body. We’re also told that sin of 𝛼 is equal to 45 over 49. Now, what we’re not going to do is use this equation to work out the value of 𝛼. Instead, we’re going to draw a little right-angle triangle with an included angle of 𝛼. We know that sin of 𝛼 is opposite over hypotenuse, so the opposite side in this triangle must be 45 units and the hypotenuse must be 49 units.

Then, we can use the Pythagorean theorem to find the length of the missing side. This theorem says that the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. So, if we let the side we’re looking for be equal to 𝑎 units, we get 𝑎 squared plus 45 squared equals 49 squared. That’s 𝑎 squared plus 2025 equals 2401. Then, we subtract 2025 from both sides, so 𝑎 squared is 376, meaning 𝑎 is the square root of this value. That’s two times root 94. Now, that’s really useful because we now know the length of the adjacent side. And this will allow us to calculate exact values for cos 𝛼 and, in fact, for tan 𝛼, should we need that too.

But let’s go back to our diagram. The body is being projected up a smooth plane. Now, what this means is the body exerts a downward force on the plane due to its mass. The downward force is mass times gravity. Let’s call that 𝑚𝑔 since we’re not actually given the mass of the body. We are told that 𝑔 is equal to 9.8 meters per square second. But we’ll leave it as 𝑔 for now. There’s only one more force that we’re interested in. And that’s the normal reaction force of the plane on the body. Remember, that acts perpendicular to the plane. So, what next? Well, we’re given an initial velocity. We might call that 𝑢 or 𝑣 sub zero. And that’s equal to 16 meters per second.

We’re trying to find the amount of time that it took for the body to return to the point of projection. So what we’re going to do is actually begin by finding the amount of time it takes before the body comes to rest, in other words, when its final velocity 𝑣 is equal to zero. And in order to do this, we need to calculate the acceleration. And so, we’ll use the formula 𝐹 equals 𝑚𝑎. Force is mass times acceleration. We’re going to carry this calculation out parallel to the plane. And so, we add a right-angle triangle to our weight force. And we need to add this triangle because this force acts vertically downwards, not parallel, not perpendicular to the plane. The component of the weight that acts parallel to the plane is the side I’ve called 𝑥. And we, of course, have this included angle. This angle is 𝛼.

We go back to the convention for labeling our triangles, and we see we’re looking to find the opposite. And we know the hypotenuse or at least we have an expression for the hypotenuse in terms of 𝑚. sin 𝛼 is opposite over hypotenuse, so sin 𝛼 is 𝑥 over 𝑚𝑔. By multiplying both sides of this equation by 𝑚𝑔, we get 𝑥 is equal to 𝑚𝑔 sin 𝛼. sin 𝛼 is 45 over 49, so we get 𝑥 equals 45 over 49 𝑚𝑔. Now, the particle’s moving up the plane, so let’s define a positive direction as being up the plane. The force is acting in the opposite direction, so the resultant force that acts parallel to the plane is negative 45 over 49 𝑚𝑔. That’s equal to mass times acceleration, 𝑚𝑎.

And next, we notice that we can divide through by 𝑚. Now, remember, we can only divide through by a variable if we’re certain it’s not equal to zero. This represents a mass, so it absolutely can’t be. And we’ve now found the acceleration of the body. It’s negative 45 over 49 𝑔 meters per square second. Now, it actually makes a lot of sense that the acceleration is negative. There are no forces helping to move the body up the plane, so we would assume it’s slowing down. Now, we didn’t actually need to calculate the value of cos 𝛼. And so, we can get rid of our right-angle triangle. It’s always sensible to construct these when given values for sin 𝛼, cos 𝛼, or tan 𝛼 because it can create easier calculations later on down the line.

Now, we said we’re going to begin by calculating the amount of time it takes for the particle to reach a speed of zero. So, its initial speed 𝑢 was 16 meters per second, and its final speed 𝑣 is zero. Its acceleration is negative 45 over 49 𝑔. And we’re looking for 𝑡. And so, we’re going to now use one of the equations of constant acceleration. The one that we’re looking for that links 𝑢, 𝑣, 𝑎, and 𝑡 is 𝑣 equals 𝑢 plus 𝑎𝑡. We replace the values we know in this formula, and we get the equation zero equals 16 minus 45 over 49 𝑔𝑡. We solve first by adding 45 over 49 𝑔𝑡 to both sides and then dividing through by 45 over 49 𝑔 or 45 over 49 times 9.8.

And so, in doing so, we see that the time it takes for the body to reach a speed of zero is 16 over nine seconds. So, how does this help us calculate the total time that the body takes to return to its starting point? Well, in a mathematically perfect world — that is, one where, in this case, there’s no friction nor air resistance or anything like that — the time it takes to reach that point where 𝑣 equals zero will be the exact same time it takes to return to the starting point. And so, let’s define the total time as being capital 𝑇. And we know that we’re just going to double 16 over nine. 16 over nine times two is 32 over nine, so we can say that the total time the body takes to return to the point of projection is 32 over nine seconds.