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.
