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Chapter 8: Problem 28

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 m/s relative to the earth at an angle of 35.0\(^\circ\) above the horizontal. If your mass is 70.0 kg and the rock's mass is 3.00 kg, what is your speed after you throw the rock? (See Discussion Question Q8.7.)

Short Answer

Expert verified
Your speed is approximately 0.44 m/s in the opposite direction to the rock's horizontal motion.

Step by step solution

01

Identify the Conservation of Momentum

In a frictionless environment, momentum is conserved. This means that before you throw the rock, the total momentum is zero (since everything is at rest). After the rock is thrown, the total momentum should still be zero, ensuring that your backward momentum balances the forward momentum of the rock.
02

Break Down the Rock's Momentum Components

The rock's velocity is given as 12.0 m/s at 35.0掳 above horizontal. Break this into horizontal and vertical components:- Horizontal: \( v_{x, ext{rock}} = 12.0 \times \cos(35.0掳) \)- Vertical: \( v_{y, ext{rock}} = 12.0 \times \sin(35.0掳) \)
03

Calculate Rock's Horizontal Momentum

Use the rock's mass and its horizontal velocity component to find the horizontal momentum of the rock:\[ p_{x, ext{rock}} = 3.00 imes v_{x, ext{rock}} = 3.00 \times (12.0 \times \cos(35.0掳)) \]
04

Calculate Your Momentum

Because momentum is conserved and initially zero, your momentum will be equal and opposite to the rock's horizontal momentum:\[ p_{x, ext{you}} = -p_{x, ext{rock}} \]Since your momentum \( p_{x, ext{you}} \) is also equal to \( 70.0 \times v \), where \( v \) is your speed:\[ 70.0 \times v = -3.00 \times (12.0 \times \cos(35.0掳)) \]
05

Solve for Your Speed

Now solve for your speed \( v \) using the equation from Step 4:\[ v = \frac{-3.00 \times (12.0 \times \cos(35.0掳))}{70.0} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictionless Surface
Imagine standing on ice鈥攁 surface so slick there's nearly no resistance. That's the magic of a frictionless surface. It's like a skating rink, just smoother! In physics, when we say a surface is frictionless, we mean there's no force slowing you down. This concept is crucial for understanding momentum. In our ice scenario, a frictionless surface means that once you start moving, you continue at the same speed and direction unless acted upon by another force. Think about it as the ultimate ice rink, where you glide effortlessly after a gentle push. The absence of friction is why the conservation of momentum becomes straightforward鈥攁s there's no energy lost to friction. This lets us focus solely on the initial and final states of motion.
Momentum Components
Momentum is a fancy way of talking about movement. It's the dialogue of mass and velocity. When we break things down into components, we're simplifying complex movements. Imagine if you swung a hammock: part of your motion is forward, and part is upward. For the rock in the ice-throwing exercise, its momentum is divided into horizontal and vertical components. The horizontal component affects how you glide back on the ice, while the vertical component gives the rock a slight lift. We use trigonometry to calculate these components:
  • Horizontal: The angle's cosine multiplied by speed, because sideways movement depends on how direct or side-slanted the launch is.
  • Vertical: The sine of the angle times speed. This bit looks at lifting and dropping.
Understanding these components helps us predict motion and balance forces.
Horizontal and Vertical Motion
In physics, moving objects typically have both horizontal and vertical motion. These two components of motion operate independently of each other. Think of a basketball player grabbing the ball in the air: they might jump vertically and toss the ball horizontally. This motion separation shines in our rock affair. The rock's vertical motion, accentuated by the throw angle, won鈥檛 affect your slide on the ice鈥攂ecause we're dealing with a frictionless surface. Meanwhile, the horizontal motion is directly tied to how you move, due to the conservation of momentum. This division allows us to handle one part of the problem at a time, simplifying our calculations.
Impulse and Momentum
Impulse and momentum are the backbone of motion mechanics. Impulse is the kick that starts movement; it's force applied over time, like when you fling open a door. Momentum, on the other hand, is the raw power of an object on the move鈥攊t鈥檚 the mass dancing with velocity.
  • Impulse changes momentum, a concept seen when you throw the rock. The force you applied to the rock changes not only its momentum but also yours.
  • Momentum is conserved, especially important on our frictionless ice. The total momentum before the rock speaks is zero. Post-throw, it remains zero when your slide equals the rock's rush forward.
Grasping these ideas helps us navigate scenarios where motion, force, and mass interplay intricately.

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Most popular questions from this chapter

Starting at \(t\) = 0, a horizontal net force \(\vec{F} = (0.280 N/s)t\hat{\imath} + (-0.450 N/s^2)t^2\hat{\jmath}\) is applied to a box that has an initial momentum \(\vec{\rho} = (-3.00 kg \cdot m/s)\hat{\imath}+ (4.00 kg \cdot m/s)\hat{\jmath}\). What is the momentum of the box at \(t\) = 2.00 s?

In July 2005, \(NASA\)'s "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the surface at 37,000 km/h. The original speed of the comet at that time was about 40,000 km/h, and its mass was estimated to be in the range (0.10 - 2.5) \(\times\)10\(^{14}\) kg. Use the smallest value of the estimated mass. (a) What change in the comet's velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet's velocity? Would this change be noticeable? (The mass of the earth is 5.97 \(\times\) 10\(^{24}\) kg.)

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: (1) 0.300 kg, (0.200 m, 0.300 m); (2) 0.400 kg, (0.100 m, \(-\)0.400 m); (3) 0.200 kg, (\(-\)0.300 m, 0.600 m). Find the coordinates of the center of mass of the system of three chocolate blocks.

Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of \((20.0 m/s)\hat{\imath} - (4.0 m/s)\hat{\jmath}\). During the 3.00 ms that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 N)\hat{\imath} + (110 N)\hat{\jmath}\). What are the \(x\)- and \(y\)-components (a) of the impulse of the net force applied to the ball; (b) of the final velocity of the ball?

In Section 8.5 we calculated the center of mass by considering objects composed of a \(finite\) number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$x_{cm} = {1\over M}\int x \space dm \space y_{cm} = {1 \over M}\int y \space dm$$ where \(x\) and \(y\) are the coordinates of the small piece of the object that has mass \(dm\). The integration is over the whole of the object. Consider a thin rod of length \(L\), mass \(M\), and cross-sectional area \(A\). Let the origin of the coordinates be at the left end of the rod and the positive \(x\)-axis lie along the rod. (a) If the density \(\rho = M/V\) of the object is uniform, perform the integration described above to show that the \(x\)-coordinate of the center of mass of the rod is at its geometrical center. (b) If the density of the object varies linearly with \(x-\)that is, \(\rho = ax\), where a is a positive constant\(-\)calculate the \(x\)-coordinate of the rod's center of mass.
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