Video Transcript
A tennis ball of mass 150 grams is
thrown at an angle of 40 degrees above the horizontal and a speed of 18 meters per
second. What is the momentum of the ball
after it has been in flight for 0.36 seconds?
We can call the ball’s momentum
after it’s been in flight for this amount of time 𝑝 and begin on our solution by
drawing a diagram of this situation. Our sketch represents that we’ve
thrown this ball at an angle, we’ve called 𝜃, of 40 degrees above the horizontal at
an initial speed, we’ve called 𝑣 sub 𝑖, of 18 meters per second. After a time in flight, we’ve
called Δ𝑡, of 0.36 seconds, we want to know the momentum of the ball.
If we let upward motion be motion
in the positive 𝑦-direction and motion to the right be in the positive 𝑥 and if we
recall that momentum 𝑝 is equal to an object’s mass times its velocity, then we can
write that momentum 𝑝 is equal to our ball’s mass 𝑚 multiplied by the square root
of 𝑣 sub 𝑥 squared plus 𝑣 sub 𝑦 squared. Here, 𝑣 sub 𝑥 is the ball’s speed
in the 𝑥-direction at a time value of 0.36 seconds. And likewise, 𝑣 sub 𝑦 is the
ball’s vertical speed at that same time.
Since we’ve been told the ball’s
mass 𝑚, that means if we can solve for 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we’ll be able to
solve for momentum 𝑝. Starting with 𝑣 sub 𝑥, when we
look at our diagram, we see that we can represent the speed of the ball in the
𝑥-direction in terms of 𝑣 sub 𝑖 and 𝜃. 𝑣 sub 𝑥 which, recall, is the
ball’s 𝑥-velocity component at time 𝑡 equals 0.36 seconds is equal to the ball’s
initial launch speed 𝑣 sub 𝑖 times the cosine of the angle 𝜃.
Plugging in for these values, when
we calculate 𝑣 sub 𝑥, we find it’s approximately 13.79 meters per second. This is the ball’s speed in the
𝑥-direction and is constant throughout the ball’s flight. Now that we know 𝑣 sub 𝑥, we move
on to solving for 𝑣 sub 𝑦, the ball’s vertical speed component at time 𝑡 equal
0.36 seconds.
Unlike the ball’s 𝑥-component
speed, its 𝑦-component speed is not constant in time, but constantly changes under
the influence of gravity. As a projectile, the ball’s motion
can be described by the kinematic equations. In particular, we recall the
relationship final speed is equal to initial speed plus acceleration times time. When we write this equation in
terms of our variables, we can write that 𝑣 sub 𝑦 is equal to the initial speed of
the ball in the 𝑦-direction minus the acceleration due to gravity 𝑔 times Δ𝑡.
We’ll treat gravitational
acceleration 𝑔 as exactly 9.8 meters per second squared. Looking again at our diagram, we
see that we can write the expression for 𝑣 sub 𝑦𝑖, the initial speed of the ball
in the vertical direction, as the product of 𝑣 sub 𝑖, the ball’s initial speed,
times the sin of the angle 𝜃. Since we’re given 𝑣 sub 𝑖, 𝜃,
and Δ𝑡, and 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑦.
When we do and enter this
expression on our calculator, we find that the 𝑣 sub 𝑦 is approximately 8.042
meters per second. Now that we know 𝑣 sub 𝑥 and 𝑣
sub 𝑦, we’re ready to plug in and solve for momentum 𝑝. When we do plug in, we’re careful
to convert our mass into units of kilograms. And we find that 𝑝 is equal to 2.4
kilograms meters per second. That’s the overall momentum of this
ball 0.36 seconds after its been launched.
