Momentum And Impulse

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| By Kevin Shaw
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Kevin Shaw
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Quizzes Created: 7 | Total Attempts: 5,161
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  • 1/13 Questions
    The diagram below shows a 4.0-kilogram cart moving to the right and a 6.0-kilogram cart moving to the left on a horizontal frictionless surface. When the two carts collide they lock together. The magnitude of the total momentum of the two-cart system after the collision is   - ProProfs

    The diagram below shows a 4.0-kilogram cart moving to the right and a 6.0-kilogram cart moving to the left on a horizontal frictionless surface. When the two carts collide they lock together. The magnitude of the total momentum of the two-cart system after the collision is  

    • 0.0 Ns
    • 15 Ns
    • 6.0 Ns
    • 30.0 Ns
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About This Quiz

Explore the principles of momentum and impulse with this interactive quiz. Engage with scenarios involving collisions, recoil, and force calculations to deepen your understanding of physics fundamentals, enhancing both academic learning and practical application skills.

Momentum And Impulse - Quiz

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  • 2. 

    A 0.050-kilogram bullet is fired from a 4.0 kilogram rifle that is initially at rest. If the bullet leaves the rifle with momentum having a magnitude of 20.0 Ns, the rifle will recoil with a momentum having a magnitude of

    • 1,600 Ns

    • 20.0 Ns

    • 80.0 Ns

    • 0.25 Ns

    Correct Answer
    A. 20.0 Ns
    Explanation
    When the bullet is fired from the rifle, according to the law of conservation of momentum, the total momentum before firing should be equal to the total momentum after firing. Initially, the rifle is at rest, so its momentum is zero. The bullet has a momentum of 20.0 Ns. Therefore, the total momentum after firing is also 20.0 Ns. Since the rifle is the only other object involved, its momentum must also be 20.0 Ns for the total momentum to be conserved. Therefore, the rifle will recoil with a momentum having a magnitude of 20.0 Ns.

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  • 3. 

    What impulse must be applied to a 25.0 kg cart to cause a velocity change of 12.0 m/s?

    • 0.48 Ns

    • 2.1 Ns

    • 13.0 Ns

    • 300.0 Ns

    Correct Answer
    A. 300.0 Ns
    Explanation
    To calculate the impulse, we can use the formula impulse = mass × velocity change. In this case, the mass of the cart is given as 25.0 kg and the velocity change is given as 12.0 m/s. Plugging these values into the formula, we get impulse = 25.0 kg × 12.0 m/s = 300.0 Ns. Therefore, the correct answer is 300.0 Ns.

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  • 4. 

    A 6.0 × 10-3 kg insect is flying horizontally with a momentum of 4.8 × 10-2 Ns headwind. What is its resulting velocity relative to the ground?

    • 8.0 m/s

    • 6.0 m/s

    • 0.80 m/s

    • 0.60 m/s

    Correct Answer
    A. 8.0 m/s
    Explanation
    The momentum of an object is equal to its mass multiplied by its velocity. In this question, the momentum of the insect is given as 4.8 × 10-2 Ns. The mass of the insect is given as 6.0 × 10-3 kg. To find the velocity, we can rearrange the formula for momentum to solve for velocity. The velocity is equal to the momentum divided by the mass. Plugging in the given values, we get 4.8 × 10-2 Ns divided by 6.0 × 10-3 kg, which equals 8.0 m/s. Therefore, the resulting velocity of the insect relative to the ground is 8.0 m/s.

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  • 5. 

    A ball is hit with a bat. A student determines that the momentum of the ball is 1.0 Ns.  What is the mass of the ball if it has a velocity of 2.0 meters per second?  

    • 0 kg

    • 0.50 kg

    • 1.0 kg

    • 2.0 kg

    Correct Answer
    A. 0.50 kg
    Explanation
    The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the momentum of the ball is 1.0 Ns and its velocity is 2.0 meters per second. To find the mass, we can rearrange the equation to solve for mass: mass = momentum / velocity. Substituting the given values, we get mass = 1.0 Ns / 2.0 m/s = 0.50 kg. Therefore, the mass of the ball is 0.50 kg.

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  • 6. 

    A 0.40-kg flashlight is constructed to withstand a specific force of impact that protects the bulb. A standard 0.10 second is used as time-of-impact in calculating a flashlight’s specifications.  What is the impact force on the 0.40 kg flashlight if it falls 2.5 meters and hits a concrete floor at 7.0 m/s, stopping in 0.10 seconds?

    • 1.0 N

    • 2.8 N

    • 28 N

    • 98 N

    Correct Answer
    A. 28 N
    Explanation
    The impact force can be calculated using the equation F = m * Δv / Δt, where F is the force, m is the mass, Δv is the change in velocity, and Δt is the time taken. In this case, the mass of the flashlight is 0.40 kg, the change in velocity is 7.0 m/s (since it stops), and the time taken is 0.10 seconds. Plugging these values into the equation, we get F = 0.40 kg * 7.0 m/s / 0.10 s = 2.8 N. Therefore, the impact force on the flashlight is 2.8 N.

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  • 7. 

    A 60.0 kg student on ice skates stands at rest on a frictionless frozen pond holding a 10.-kg brick. He throws the brick east with a speed of 18 m/s.  What is the resulting velocity of the student?

    • 3.0 m/s west

    • 3.0 m/s east

    • 18 m/s west

    • 18 m/s east

    Correct Answer
    A. 3.0 m/s west
    Explanation
    When the student throws the brick east with a speed of 18 m/s, according to the law of conservation of momentum, the student will experience an equal and opposite force in the opposite direction. This means that the student will move in the opposite direction with a velocity of 3.0 m/s west.

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  • 8. 

    A train car with a mass of 3.0 E 4 kg traveling north at 1.5 m/s collides and couples with a 3.2 E 4 kg train car going south at 0.80 m/s. What is the velocity of the coupled cars after the collision?

    • 0.31 m/s north

    • 0.31 m/s south

    • 0.97 m/s north

    • 0.97 m/s south

    Correct Answer
    A. 0.31 m/s north
    Explanation
    The velocity of the coupled cars after the collision is 0.31 m/s north. This can be determined using the principle of conservation of momentum. The total momentum before the collision is the sum of the individual momenta of the two train cars. After the collision, the total momentum remains the same. Since the two train cars are moving in opposite directions, their momenta have opposite signs. By applying the conservation of momentum equation, the final velocity of the coupled cars can be calculated as the ratio of the total momentum to the total mass of the cars.

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  • 9. 

    A 1,000-kg cannon fires a 10 kg projectile horizontally at a velocity of 300 m/s. What is the recoil velocity of the cannon?

    • 0.3 m/s

    • 3 m/s

    • 30 m/s

    • 300 m/s

    Correct Answer
    A. 3 m/s
    Explanation
    When the cannon fires the projectile, according to Newton's third law of motion, the cannon experiences an equal and opposite force. This force causes the cannon to move in the opposite direction with a certain velocity, known as the recoil velocity. The mass of the cannon is 1,000 kg, and the mass of the projectile is 10 kg. Since the velocity of the projectile is 300 m/s, the momentum of the projectile is 10 kg * 300 m/s = 3000 kg*m/s. To conserve momentum, the cannon must have an equal and opposite momentum, so its recoil velocity can be calculated as 3000 kg*m/s divided by the mass of the cannon (1,000 kg), which equals 3 m/s.

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  • 10. 

    A student on roller skates throws a basketball forward. How does the momentum of the student on skates compare to the momentum of the basketball?  

    • The velocity of the student is equal in magnitude but opposite in direction to the velocity of the basketball.

    • The velocity of the student is equal in magnitude and in the same direction as the velocity of the basketball.

    • The product of the mass times the velocity of the student is equal in magnitude and in the same direction as the product of the mass times the velocity of the basketball.

    • The product of the mass times the velocity of the student is equal in magnitude but opposite in direction to the product of the mass times the velocity of the basketball.

    Correct Answer
    A. The product of the mass times the velocity of the student is equal in magnitude but opposite in direction to the product of the mass times the velocity of the basketball.
    Explanation
    When the student throws the basketball forward, both the student and the basketball experience a change in momentum. According to the law of conservation of momentum, the total momentum before and after the throw must be the same. Since the basketball is thrown forward, its momentum is in the positive direction. To maintain the conservation of momentum, the momentum of the student on skates must be equal in magnitude but opposite in direction to the momentum of the basketball. This means that the product of the mass times the velocity of the student is equal in magnitude but opposite in direction to the product of the mass times the velocity of the basketball.

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  • 11. 

    During batting practice, a 0.30 kg baseball is hit with a bat that exerts a force of 350 N on the ball. The ball left the bat at 80. 0 m/s. If the incoming pitch was traveling at 60.0 m/s, how long did the ball stay in contact with the bat?

    • 0.12 s

    • 0.50 s

    • 0.85 s

    • 1.4 s

    Correct Answer
    A. 0.12 s
    Explanation
    The time the ball stays in contact with the bat can be determined using the impulse-momentum principle. The impulse experienced by an object is equal to the change in momentum it undergoes. In this case, the impulse exerted on the ball by the bat is equal to the force applied multiplied by the time of contact. The change in momentum of the ball is equal to the mass of the ball multiplied by the difference in velocities before and after contact. By rearranging the equation, we can solve for the time of contact, which is 0.12 s.

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  • 12. 

    Which is an acceptable unit for impulse?

    • N•m

    • J•s

    • J/s

    • kg•m/s

    Correct Answer
    A. kg•m/s
    Explanation
    Impulse is defined as the change in momentum of an object, and momentum is the product of mass and velocity. Therefore, the unit for impulse should also involve mass and velocity. The unit kg•m/s represents the product of mass (kg) and velocity (m/s), making it an acceptable unit for impulse.

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  • 13. 
    A 3.0 kilogram steel block is at rest on a frictionless horizontal surface. A 1.0 kilogram lump of clay is propelled horizontally at 6.0 meters per second toward the block as shown in the diagram below.   Upon collision, the clay and steel block stick together and move to the right with a speed of - ProProfs

    A 3.0 kilogram steel block is at rest on a frictionless horizontal surface. A 1.0 kilogram lump of clay is propelled horizontally at 6.0 meters per second toward the block as shown in the diagram below.   Upon collision, the clay and steel block stick together and move to the right with a speed of

    • 1.5 m/s

    • 3.0 m/s

    • 2.0 m/s

    • 6.0 m/s

    Correct Answer
    A. 1.5 m/s
    Explanation
    When the clay lump collides with the steel block, momentum is conserved. Since the system is isolated and there is no external force acting on it, the total momentum before the collision is equal to the total momentum after the collision. Initially, the clay lump has a momentum of (1.0 kg) * (6.0 m/s) = 6.0 kg·m/s to the right, and the steel block has a momentum of zero since it is at rest. After the collision, the clay and steel block stick together and move to the right with a common velocity. Let's call this final velocity V. The total mass of the system is (3.0 kg + 1.0 kg) = 4.0 kg. Therefore, the total momentum after the collision is (4.0 kg) * V. Setting the initial and final momenta equal, we have 6.0 kg·m/s = (4.0 kg) * V. Solving for V gives V = 1.5 m/s.

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  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 16, 2010
    Quiz Created by
    Kevin Shaw

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