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Ch 11: Impulse and Momentum
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 11: Impulse and MomentumProblem 38
Chapter 11, Problem 38

A 550 g cart is released from rest on a frictionless, 30° ramp, 120 cm from the bottom of the ramp. It rolls down, bounces off a rubber block at the bottom, and then rolls 80 cm back up the ramp. A high-speed video shows that the cart was in contact with the rubber block for 25 ms. What was the average force exerted on the cart by the block?

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Step 1: Calculate the initial velocity of the cart just before it hits the rubber block. Use the conservation of energy principle to find the velocity at the bottom of the ramp. The potential energy at the top of the ramp is converted into kinetic energy at the bottom. The formula is: \( m g h = \frac{1}{2} m v^2 \), where \( h \) is the height of the ramp, \( m \) is the mass of the cart, \( g \) is the acceleration due to gravity, and \( v \) is the velocity.
Step 2: Determine the height \( h \) of the ramp using trigonometry. The ramp length is given as 120 cm, and the angle is 30°. Use \( h = L \sin(\theta) \), where \( L \) is the ramp length and \( \theta \) is the angle of inclination.
Step 3: After the cart bounces off the rubber block, calculate the velocity of the cart as it moves back up the ramp. Use the conservation of energy principle again, but this time for the motion after the bounce. The cart rolls back up to a height \( h' \), which can be calculated using \( h' = L' \sin(\theta) \), where \( L' \) is the distance the cart rolls back up the ramp (80 cm). The velocity after the bounce can be found using \( \frac{1}{2} m v'^2 = m g h' \).
Step 4: Calculate the change in velocity \( \Delta v \) of the cart due to the collision with the rubber block. The change in velocity is \( \Delta v = v - v' \), where \( v \) is the velocity before the collision and \( v' \) is the velocity after the collision.
Step 5: Use the impulse-momentum theorem to find the average force exerted on the cart by the block. The impulse-momentum theorem states \( F_{avg} \Delta t = m \Delta v \), where \( F_{avg} \) is the average force, \( \Delta t \) is the contact time (25 ms), \( m \) is the mass of the cart, and \( \Delta v \) is the change in velocity. Rearrange the formula to solve for \( F_{avg} \): \( F_{avg} = \frac{m \Delta v}{\Delta t} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). This principle is crucial for understanding how forces affect the motion of the cart as it interacts with the rubber block, allowing us to calculate the average force exerted during the collision.
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Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the potential energy of the cart at the top of the ramp is converted into kinetic energy as it rolls down, and this energy transformation is essential for analyzing the cart's motion and the effects of the collision.
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Impulse and Momentum

Impulse is defined as the change in momentum of an object when a force is applied over a period of time. The relationship between impulse and momentum is key to determining the average force exerted on the cart by the rubber block, as it involves calculating the change in momentum during the brief contact time of 25 ms.
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Related Practice
Textbook Question

A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. FIGURE P11.40 shows the force of the wall on the ball during the collision. What is the value of Fmax , the maximum value of the contact force during the collision?

Textbook Question

A tennis player swings her 1000g racket with a speed of 10 m/s. She hits a 60g tennis ball that was approaching her at a speed of 20 m/s. The ball rebounds at 40 m/s. If the tennis ball and racket are in contact for 10 ms, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

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Textbook Question

An object at rest explodes into three fragments. FIGURE EX11.32 shows the momentum vectors of two of the fragments. What is the momentum of the third fragment? Write your answer using unit vectors.

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Textbook Question

A 20 g ball of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay? Give your answer as an angle north of east.

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Textbook Question

At the center of a 50-m-diameter circular ice rink, a 75 kg skater traveling north at 2.5 m/s collides with and holds on to a 60 kg skater who had been heading west at 3.5 m/s. Where will they reach it? Give your answer as an angle north of west.

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Textbook Question

A particle of mass m is at rest at t = 0. Its momentum for t > 0 is given by px = 6t² kg m/s, where t is in s. Find an expression for Fx(t), the force exerted on the particle as a function of time.

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