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

Knight Calc 5th Edition

Ch 11: Impulse and Momentum

Problem 25a

Chapter 11, Problem 25a

A 50 g ball of clay traveling at speed v₀ hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface. What is the speed of the brick after the collision?

Verified step by step guidance

Step 1: Identify the type of collision. Since the ball of clay sticks to the brick after the collision, this is an inelastic collision. In such collisions, momentum is conserved but kinetic energy is not.

Step 2: Write the equation for conservation of momentum. The total momentum before the collision equals the total momentum after the collision. Use the formula:

m_{1} v_{0} + m_{2} v_{2} = (m_{1} + m_{2}) v_{f}

, where

m_{1}

and

v_{0}

are the mass and velocity of the clay,

m_{2}

and

v_{2}

are the mass and velocity of the brick, and

v_{f}

is the final velocity of the combined system.

Step 3: Substitute the given values into the momentum equation. The mass of the clay is 50 g (convert to kg: 0.05 kg), the mass of the brick is 1.0 kg, the initial velocity of the brick is 0 m/s (since it is at rest), and the initial velocity of the clay is

v_{0}

. The equation becomes:

0.05 v_{0} + 1.0 (0) = (0.05 + 1.0) v_{f}

Step 4: Simplify the equation to isolate

v_{f}

. After simplifying, the equation becomes:

v_{f} = (0.05 v_{0})/(1.05)

Step 5: Conclude that the final velocity of the brick and clay system depends on the initial velocity of the clay,

v_{0}

. To find the numerical value, substitute the value of

v_{0}

into the equation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event. In this scenario, the clay ball collides with the brick, and since no external forces are acting on the system, the momentum of the clay ball and brick combined must remain constant throughout the collision.

Elastic vs. Inelastic Collisions

Collisions can be classified as elastic or inelastic based on whether kinetic energy is conserved. In this case, the clay ball sticks to the brick after the collision, indicating an inelastic collision where kinetic energy is not conserved, but momentum is. This distinction is crucial for calculating the final speed of the combined mass after the collision.

Mass and Velocity Relationship

The relationship between mass and velocity is fundamental in determining the outcome of collisions. In this problem, the mass of the clay ball (50 g) and the brick (1.0 kg) will influence the final velocity of the combined system after the collision. Understanding how mass affects momentum allows for the calculation of the final speed using the conservation of momentum principle.

Recommended video:

Guided course

20:32

Mass Spectrometers

Related Practice

Textbook Question

A 70.00 kg football player is gliding across very smooth ice at 2.00 m/s. He throws a 0.450 kg football straight forward. What is the player's speed afterward if the ball is thrown at 15.0 m/s relative to the ground?

views

Textbook Question

A proton is traveling to the right at 2.0 x 10⁷ m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

views

Textbook Question

Dan is gliding on his skateboard at 4.0 m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.0 m/s. How fast is Dan going as his feet hit the ground? Dan's mass is 50 kg and the skateboard's mass is 5.0 kg.

Textbook Question

A 50 g marble moving at 2.0 m/s strikes a 20 g marble at rest. What is the speed of each marble immediately after the collision?

views

Textbook Question

A package of mass m is released from rest at a warehouse loading dock and slides down the 3.0-m-high, frictionless chute of FIGURE EX11.24 to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

views

Textbook Question

Two objects collide and bounce apart. FIGURE EX11.31 shows the initial momenta of both and the final momentum of object 2. What is the final momentum of object 1? Write your answer using unit vectors.

views

A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface. What is the speed of the brick after the collision?

Verified video answer for a similar problem:

Key Concepts

Conservation of Momentum

Elastic vs. Inelastic Collisions

Mass and Velocity Relationship

A 50 g ball of clay traveling at speed v₀ hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface. What is the speed of the brick after the collision?