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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 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° above horizontal. Break this into horizontal and vertical components:- Horizontal: \( v_{x, ext{rock}} = 12.0 \times \cos(35.0°) \)- Vertical: \( v_{y, ext{rock}} = 12.0 \times \sin(35.0°) \)
03

Calculate Rock's Horizontal Momentum

Use the rock's mass and its horizontal velocity component to find the horizontal momentum of the rock:\[ p_{x, ext{rock}} = 3.00 imes v_{x, ext{rock}} = 3.00 \times (12.0 \times \cos(35.0°)) \]
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:\[ 70.0 \times v = -3.00 \times (12.0 \times \cos(35.0°)) \]
05

Solve for Your Speed

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

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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—a 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—as 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’t affect your slide on the ice—because 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—it’s 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

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.600-kg ball that is traveling horizontally at 10.0 m/s. Your mass is 70.0 kg. (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 m/s in the opposite direction, what is your speed after the collision?

Two ice skaters, Daniel (mass 65.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 53.1\(^\circ\) from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. (a) What are the magnitude and direction of Daniel's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s \((\textbf{Fig. P8.87})\). Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37\(^\circ\) from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

To keep the calculations fairly simple but still reasonable, we model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and are uniform. For a 70.0-kg person, the mass of the upper leg is 8.60 kg, while that of the lower leg (including the foot) is 5.25 kg. Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.

An 8.00-kg block of wood sits at the edge of a frictionless table, 2.20 m above the floor. A 0.500-kg blob of clay slides along the length of the table with a speed of 24.0 m/s, strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

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