Video: Total Momentum of Two-Dimensional Motion

A tennis ball of mass 150 g is thrown at an angle of 40ยฐ above the horizontal and a speed of 18 m/s. What is the momentum of the ball after it has been in flight for 0.36 s?

03:52

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 ballโ€™s momentum after itโ€™s been in flight for this amount of time ๐‘ and begin on our solution by drawing a diagram of this situation. Our sketch represents that weโ€™ve thrown this ball at an angle, weโ€™ve called ๐œƒ, of 40 degrees above the horizontal at an initial speed, weโ€™ve called ๐‘ฃ sub ๐‘–, of 18 meters per second. After a time in flight, weโ€™ve called ฮ”๐‘ก, of 0.36 seconds, we want to know the momentum of the ball.

If we let upward motion be motion in the positive ๐‘ฆ-direction and motion to the right be in the positive ๐‘ฅ and if we recall that momentum ๐‘ is equal to an objectโ€™s mass times its velocity, then we can write that momentum ๐‘ is equal to our ballโ€™s mass ๐‘š multiplied by the square root of ๐‘ฃ sub ๐‘ฅ squared plus ๐‘ฃ sub ๐‘ฆ squared. Here, ๐‘ฃ sub ๐‘ฅ is the ballโ€™s speed in the ๐‘ฅ-direction at a time value of 0.36 seconds. And likewise, ๐‘ฃ sub ๐‘ฆ is the ballโ€™s vertical speed at that same time.

Since weโ€™ve been told the ballโ€™s mass ๐‘š, that means if we can solve for ๐‘ฃ sub ๐‘ฅ and ๐‘ฃ sub ๐‘ฆ, weโ€™ll be able to solve for momentum ๐‘. Starting with ๐‘ฃ sub ๐‘ฅ, when we look at our diagram, we see that we can represent the speed of the ball in the ๐‘ฅ-direction in terms of ๐‘ฃ sub ๐‘– and ๐œƒ. ๐‘ฃ sub ๐‘ฅ which, recall, is the ballโ€™s ๐‘ฅ-velocity component at time ๐‘ก equals 0.36 seconds is equal to the ballโ€™s initial launch speed ๐‘ฃ sub ๐‘– times the cosine of the angle ๐œƒ.

Plugging in for these values, when we calculate ๐‘ฃ sub ๐‘ฅ, we find itโ€™s approximately 13.79 meters per second. This is the ballโ€™s speed in the ๐‘ฅ-direction and is constant throughout the ballโ€™s flight. Now that we know ๐‘ฃ sub ๐‘ฅ, we move on to solving for ๐‘ฃ sub ๐‘ฆ, the ballโ€™s vertical speed component at time ๐‘ก equal 0.36 seconds.

Unlike the ballโ€™s ๐‘ฅ-component speed, its ๐‘ฆ-component speed is not constant in time, but constantly changes under the influence of gravity. As a projectile, the ballโ€™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 ๐‘ฃ sub ๐‘ฆ is equal to the initial speed of the ball in the ๐‘ฆ-direction minus the acceleration due to gravity ๐‘” times ฮ”๐‘ก.

Weโ€™ll treat gravitational acceleration ๐‘” as exactly 9.8 meters per second squared. Looking again at our diagram, we see that we can write the expression for ๐‘ฃ sub ๐‘ฆ๐‘–, the initial speed of the ball in the vertical direction, as the product of ๐‘ฃ sub ๐‘–, the ballโ€™s initial speed, times the sin of the angle ๐œƒ. Since weโ€™re given ๐‘ฃ sub ๐‘–, ๐œƒ, and ฮ”๐‘ก, and ๐‘” is a constant, weโ€™re ready to plug in and solve for ๐‘ฃ sub ๐‘ฆ.

When we do and enter this expression on our calculator, we find that the ๐‘ฃ sub ๐‘ฆ is approximately 8.042 meters per second. Now that we know ๐‘ฃ sub ๐‘ฅ and ๐‘ฃ sub ๐‘ฆ, weโ€™re ready to plug in and solve for momentum ๐‘. When we do plug in, weโ€™re careful to convert our mass into units of kilograms. And we find that ๐‘ is equal to 2.4 kilograms meters per second. Thatโ€™s the overall momentum of this ball 0.36 seconds after its been launched.

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