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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.
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What is the momentum of the ball after it has been in flight for 0.36 seconds?
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We can call the ballโs momentum after itโs been in flight for this amount of time ๐ and begin on our solution by drawing a diagram of this situation.
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Our sketch represents that weโve thrown this ball at an angle, weโve called ๐, of 40 degrees above the horizontal at an initial speed, weโve called ๐ฃ sub ๐, of 18 meters per second.
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After a time in flight, weโve called ฮ๐ก, of 0.36 seconds, we want to know the momentum of the ball.
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If we let upward motion be motion in the positive ๐ฆ-direction and motion to the right be in the positive ๐ฅ and if we recall that momentum ๐ is equal to an objectโs mass times its velocity, then we can write that momentum ๐ is equal to our ballโs mass ๐ multiplied by the square root of ๐ฃ sub ๐ฅ squared plus ๐ฃ sub ๐ฆ squared.
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Here, ๐ฃ sub ๐ฅ is the ballโs speed in the ๐ฅ-direction at a time value of 0.36 seconds.
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And likewise, ๐ฃ sub ๐ฆ is the ballโs vertical speed at that same time.
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Since weโve been told the ballโs mass ๐, that means if we can solve for ๐ฃ sub ๐ฅ and ๐ฃ sub ๐ฆ, weโll be able to solve for momentum ๐.
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Starting with ๐ฃ sub ๐ฅ, when we look at our diagram, we see that we can represent the speed of the ball in the ๐ฅ-direction in terms of ๐ฃ sub ๐ and ๐.
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๐ฃ sub ๐ฅ which, recall, is the ballโs ๐ฅ-velocity component at time ๐ก equals 0.36 seconds is equal to the ballโs initial launch speed ๐ฃ sub ๐ times the cosine of the angle ๐.
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Plugging in for these values, when we calculate ๐ฃ sub ๐ฅ, we find itโs approximately 13.79 meters per second.
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This is the ballโs speed in the ๐ฅ-direction and is constant throughout the ballโs flight.
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Now that we know ๐ฃ sub ๐ฅ, we move on to solving for ๐ฃ sub ๐ฆ, the ballโs vertical speed component at time ๐ก equal 0.36 seconds.
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Unlike the ballโs ๐ฅ-component speed, its ๐ฆ-component speed is not constant in time, but constantly changes under the influence of gravity.
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As a projectile, the ballโs motion can be described by the kinematic equations.
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In particular, we recall the relationship final speed is equal to initial speed plus acceleration times time.
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When we write this equation in terms of our variables, we can write that ๐ฃ sub ๐ฆ is equal to the initial speed of the ball in the ๐ฆ-direction minus the acceleration due to gravity ๐ times ฮ๐ก.
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Weโll treat gravitational acceleration ๐ as exactly 9.8 meters per second squared.
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Looking again at our diagram, we see that we can write the expression for ๐ฃ sub ๐ฆ๐, the initial speed of the ball in the vertical direction, as the product of ๐ฃ sub ๐, the ballโs initial speed, times the sin of the angle ๐.
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Since weโre given ๐ฃ sub ๐, ๐, and ฮ๐ก, and ๐ is a constant, weโre ready to plug in and solve for ๐ฃ sub ๐ฆ.
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When we do and enter this expression on our calculator, we find that the ๐ฃ sub ๐ฆ is approximately 8.042 meters per second.
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Now that we know ๐ฃ sub ๐ฅ and ๐ฃ sub ๐ฆ, weโre ready to plug in and solve for momentum ๐.
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When we do plug in, weโre careful to convert our mass into units of kilograms.
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And we find that ๐ is equal to 2.4 kilograms meters per second.
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Thatโs the overall momentum of this ball 0.36 seconds after its been launched.