Video: Impulse Applied to an Object

The π‘₯-component of the force applied on a golf ball of mass 38 g by a golf club is shown in the diagram. What is the π‘₯-component of the impulse applied on the golf ball between 0 ms and 50 ms? What is the π‘₯-component of the impulse applied on the golf ball between 50 ms and 100 ms?

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Video Transcript

The π‘₯-component of the force applied on a golf ball of mass 38 grams by a golf club is shown in the diagram. What is the π‘₯-component of the impulse applied on the golf ball between zero milliseconds and 50 milliseconds? What is the π‘₯-component of the impulse applied on a golf ball between 50 milliseconds and 100 milliseconds?

We can call these π‘₯-components of the impulse 𝐽 sub π‘₯ one and 𝐽 sub π‘₯ two, respectively. When we look at the diagram, we see it’s a graphical representation of the force in the π‘₯-direction applied to the golf ball versus the time in milliseconds over which that force is applied. Over a time span of zero to 100 milliseconds, the force varies from zero to 300 newtons. Since we want to solve for components of impulse, we’ll consider that impulse in general is equal to the integral of the applied force times 𝑑𝑑. We know that this integral represented graphically would equal area under the curve of force versus time. So on our diagram, we can graphically represent what 𝐽 sub π‘₯ one and 𝐽 sub π‘₯ two are.

𝐽 sub π‘₯ one, which is the impulse applied in the π‘₯-direction to the golf ball from zero to 50 milliseconds, is equal to the area under the curve for that time interval. Similarly, 𝐽 sub π‘₯ two, it’s the area under the curve from 50 to 100 milliseconds. We can write then that 𝐽 sub π‘₯ one is equal to the area of the triangle shown, one-half the base of 50 times 10 to the negative third seconds or 50 milliseconds times the height of 300 newtons. When we multiply these three values together, we find a result of 7.5 kilograms-meters per second. That’s the π‘₯-component impulse delivered to the golf ball from zero to 50 milliseconds.

Next, we want to solve for 𝐽 sub π‘₯ two, the area under the curve from 50 to 100 milliseconds. That area is equal to the area of the rectangle with base 100 minus 50, or simply 50 times 10 to the negative third seconds, multiplied by the height of 300 newtons. This results in a value of 15 kilograms-meters per second. That’s the π‘₯-component of the impulse delivered to the golf ball from 50 to 100 milliseconds.

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