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You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 $\mathrm{m} / \mathrm{s}$ relative to the earth at an angle of $35.0^{\circ}$ above the borizontal. If your mass is 70.0 $\mathrm{kg}$ and the rock's mass is 15.0 kg, what is your speed after you throw the rock (see Discussion Question $Q 8.7$ ?

$2.1 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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in this question. You are a person standing in a large sheet of ice holding a rock in order to get off the ice. You throw the rock such that the rock mix an angle. Tita, with the horizontal on the rock moves with Velocity V. R. By throwing the rock to the right, you will move to the left as a consequence of momentum conservation. Then, in this question, we have to evaluate what is your velocity after throwing the rock? So for that we have to use, of course, the momentum conservation. So before throwing the rock, your net momentum, WASP, you and zero. And after throwing the Rock, we haven't at Momentum P. M. So before throwing the rock, there was anything moving there for the net momentum. Zero. After throwing the rock, the rock moves to the right and new moves to the left. Let me choose this reference frame where everything that points to the right is positive, and everything that points the left is negative. In this reference frame, the momentum off the rock is given by the mass off the rock times, the velocity off the rock, and then your momentum is given by minus your mass times your velocity where the minus sign is due to the fact that you are moving to the left. So you have a negative velocity Now don't be naive. Don't be naive because we're right. In the law of conservation, of momentum in the horizontal axis on that velocity of the rock VR do not point in the horizontal direction. So actually, what you will plug in here isn't the velocity off the rock, but on Lee, the X component off the velocity off the rock V X R. Okay, now let's serve this equation for V. By doing that, we'll get the following. Your velocity is given by the mass off the rock divided by your mass times the X component off the velocity off the rock. How can we calculate the X component off the velocity of the rock? For that, we have to do a little geometry. Say the velocity of the rock points in this direction. You have the horizontal right here on. Do you have a new angle, Tita? Now notice the following. This is the components that you want to evaluate. You want the horizontal component Now let me close this so that we get a nice triangle which is a rectangle triangle. Now, in order to figure out what is the value off the X component off the velocity you can do the following This is we are component X. This is the full velocity. As you can see, the full velocity is like a high pot a news so you can use a co sign to evaluate the X component off the velocity. Because the co sign off Tita is given by the judges and side. So we are X divided by the high Potter news The reform What you get is the following the X component off the velocity is given by we are the few velocity times they call Sign off Dita. Now you can plug in this expression into the equation for your velocity. By doing that, you get the following Your velocity is given by the mass off the rock divided by your mass times The full velocity off the rock times the co sign off, Tita, Now you're plugging all values that you have to get. The following the mass of the rock is 15 kg Your mass 70 kg VR the velocity of the rock is 12 m per second and the angle Tita is 35 degrees. So here you have the co sign off 35 and this results in a velocity off approximately 2.1 m per second and this is the answer to this question.

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