Video Transcript
In the reality television show
“Amazing Race,” a contestant is firing 12-kilogram watermelons from a slingshot to
hit targets down the field. The slingshot is pulled back 1.5
meters and the watermelon is at ground level. The launch point is 0.3 meters
above ground. And the target is 10 meters away
horizontally from the launch point. Calculate the spring constant of
the slingshot.
We can call the spring constant of
the slingshot 𝑘. And we’ll start out by drawing a
diagram of the situation. In this scenario, we have our
watermelons of mass we’ve called 𝑚, 12 kilograms, being loaded into an elastic
slingshot. The slingshot has been pulled back
a total distance Δ𝑥 of 1.5 meters. And when the slingshot is released,
the watermelon is launched from an initial height we’ve called ℎ of 0.3 meters. If it’s aimed on target, the
watermelon then travels a horizontal range we’ve called 𝑅 of 10 meters before it
hits the target. Based on this information, we want
to solve for the spring constant of this elastic slingshot.
We know that when it’s released,
the slingshot will give the watermelon kinetic energy. And because it raises the
watermelon a height ℎ, it also gives that watermelon potential energy due to
gravity. When we recall that elastic
potential energy equals one-half 𝑘, the spring constant, times the displacement,
Δ𝑥, squared, that gravitational potential energy equals mass times 𝑔 times ℎ,
where we’ll treat 𝑔, the acceleration due to gravity, as exactly 9.8 meters per
second squared.
And we recall that an object’s
kinetic energy is half its mass times its speed squared. Then we can write that one-half the
spring constant, 𝑘, times the displacement, Δ𝑥, squared is equal to 𝑚 times 𝑔
times ℎ, where 𝑚 is the mass of the watermelon and ℎ is its launch height, plus
one-half the watermelon’s mass times its launch speed, 𝑣, squared.
If we rearrange this equation to
solve for 𝑘, we see that we know the mass of the watermelon 𝑚, we know the
displacement of the slingshot Δ𝑥, we know the acceleration due to gravity 𝑔, the
height ℎ. But the speed 𝑣 of the watermelon
when it’s launched is the one thing we don’t yet know. That means if we can solve for 𝑣,
then we can solve for 𝑘. So let’s figure out what that
launch speed 𝑣 of the watermelon is.
To do that, let’s start by dividing
the motion of our watermelon projectile into vertical, 𝑦, and horizontal, 𝑥,
directions. If we call the total time that our
watermelon is in flight before it reaches our target 𝑡, then we can write that 𝑣
sub 𝑥, the horizontal speed of our projectile, is equal to the range, 𝑅, over the
time it travels, 𝑡.
We’ve called 𝑣 the launch speed of
our watermelon at its initial launch point. And if we look at the right
triangle formed by the slingshot, the ground, and the support post, we can see that
the horizontal speed of our object is equal to its overall initial speed, 𝑣,
multiplied by the cosine of the angle that we’ve called 𝜃, the angle between the
slingshot and the ground.
Looking at this equation, we can
solve for 𝜃 because we know the height ℎ as well as Δ𝑥. And we’re given the range 𝑅. But we don’t know the total time in
flight 𝑡, which will let us solve then for 𝑣, the initial launch speed of the
watermelon. So let’s now move to considering
motion in the 𝑦 or vertical direction to see if that can help us solve for the
overall time, 𝑡.
In the vertical direction, our
projectile experiences a steady acceleration due to gravity. And therefore, its motion is
described by the kinematic equations. Say we pick two points along the
watermelon’s trajectory. The first point is at its launch
point. And the second is at the zenith of
its trajectory, where its altitude is highest. And we’ll call these points 𝑖 and
𝑓 for initial and final.
There is a kinematic equation which
says that final speed is equal to initial speed plus acceleration times time
passed. If we apply that particular
equation to our chosen initial and final points, we can write that 𝑣 sub 𝑓 is
equal to 𝑣 sub 𝑖 minus 𝑔 times one-half the total time in flight 𝑡.
Recalling that we’re only
considering vertical motion at the moment, we know that the vertical speed of our
projectile at point 𝑓 is zero. It’s neither going up nor down. So 𝑣 sub 𝑓 in our equation is
zero. And we can write that 𝑣 sub 𝑖 is
equal to 𝑔 times 𝑡 over two. 𝑣 sub 𝑖 is the initial speed of
our projectile in the 𝑦 or vertical direction.
Based on our diagram, we can write
𝑣 sub 𝑖 as the overall launch speed, 𝑣, multiplied by the sin of the angle
𝜃. When we rearrange this equation to
solve for 𝑡, we find it’s equal to two times 𝑣 times the sin of 𝜃 all over
𝑔.
What we’ll do now is take this
expression for the overall time in flight 𝑡 and substitute it in for 𝑡 in our
equation for 𝑣 times the cos of 𝜃. When we do, we find that 𝑣 cos 𝜃
equals 𝑅𝑔 over two 𝑣 sin 𝜃. And when we multiply both sides by
the denominator, we have an expression on the left-hand side of our equation that we
can simplify through a trigonometric identity. Two times the cosine of an angle
times the sine of that same angle is equal to the sine of twice that angle. So we can write that 𝑣 squared
times the sin of two 𝜃 is equal to 𝑅 times 𝑔. Or solving for 𝑣, it’s equal to
the square root of 𝑅 times 𝑔 over the sin of two 𝜃.
We now have an expression for 𝑣
that we can plug into our original energy balance equation to help us solve for the
spring constant 𝑘. When we do plug in this term, we
see we can factor out 𝑔 from this equation. And looking at this simplified
equation, we see we have values for 𝑚, 𝑔, Δ𝑥, ℎ, and 𝑅.
The last thing to solve for before
calculating 𝑘 is 𝜃. Looking at the right triangle,
where we’ve defined 𝜃, we see that the sine of this angle is equal to ℎ over Δ𝑥,
both of which are known. If we take the arcsine of both
sides of the equation, we find that 𝜃 is equal to the inverse sin of ℎ, 0.3 meters,
divided by Δ𝑥, 1.5 meters. Plugging this in on our calculator,
we find that 𝜃 is approximately 11.54 degrees.
Now we’re ready to plug in for our
variables and solve for 𝑘. When we enter this expression on
our calculator, we find that, to two significant figures, 𝑘 equals 1400 newtons per
meter. That’s the value of the spring
constant of the slingshot.
