# a uniform rod of length l rests on a frictionless horizontal surface. the rod pivots about a fixed frictionless axis at one end. the rod is initially at rest. a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. the mass of the bullet is one-fourth the mass of the rod.

### James

Guys, does anyone know the answer?

get a uniform rod of length l rests on a frictionless horizontal surface. the rod pivots about a fixed frictionless axis at one end. the rod is initially at rest. a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. the mass of the bullet is one-fourth the mass of the rod. from EN Bilgi.

## A uniform rod of length L rests on a frictionless horizontal

Find step-by-step Physics solutions and your answer to the following textbook question: A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?.

### Question

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

### Explanation

Verified

**a)**To solve the first part of this problem, we have to assume that since external forces which are acting on the givens system do not produce any torque, the Law of conservation of angular momentum can be applied.

The Law of conservation of angular momentum states that initial and final angular momentum will be equal, total angular momentum will remain constant, whenever there is no torque produced by the external forces. Therefore, it can be expressed by the following equation:

L_{1} = L_{2} L 1 =L 2 Where: L_{1} L 1

- initial angular momentum

L_{2} L 2

- final angular momentum

Also, we will be applying an equation that determines angular momentum:

L = I \omega L=Iω Where: I

I - moment of inertia

\omega ω - angular speed

## Create a free account to see explanations

Continue with Google

Continue with Facebook

Already have an account?

### Related questions

PHYSICS

The left-hand end of a uniform rod of length L and mass m is attached to a vertical wall by a frictionless hinge. The rod is held at an angle

\theta

θ above the horizontal by a horizontal wire that tuns between the wall and the right-hand end of the rod. (a) If the tension in the wire is T, what is the magnitude of the angle

\theta

θ that the rod makes with the horizontal? (b) The wire breaks and the rod rotates about the hinge. What is the angular speed of the rod as the rod passes through a horizontal position?

PHYSICS

A rod of mass m and radius R rests on two parallel rails that are a distance d apart and have a length L. The rod carries a current I in the direction shown and rolls along the rails without slipping. A uniform magnetic field B is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

PHYSICS

A long, uniform rod of length L and mass M is pivoted about a frictionless, horizontal pin through one end. The rod is nudged from rest in a vertical position. At the instant the rod is horizontal, find (a) its angular speed, (b) the magnitude of its angular acceleration, (c) the x and y components of the acceleration of its center of mass, and (d) the components of the reaction force at the pivot.

PHYSICS

A thin rod of length h and mass M is held vertically with its lower end resting on a frictionless, horizontal surface. The rod is then released to fall freely. (a) Determine the speed of its center of mass just before it hits the horizontal surface. (b) What If? Now suppose the rod has a fixed pivot at its lower end. Determine the speed of the rod's center of mass just before it hits the surface.

## Solved A uniform rod of length L rests on a frictionless

Answer to Solved A uniform rod of length L rests on a frictionless

© 2003-2022 Chegg Inc. All rights reserved.

## A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one

Click here👆to get an answer to your question ✍️ A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one - sixth the mass of the rod. What is the final angular velocity of the rod ?

Question

## A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. What is the final angular velocity of the rod ?

**A**ω=

9L v

**B**ω=

9L 2v

**C**ω=

9L 3v

**D**ω=

9L 5v Hard Open in App Solution Verified by Toppr

Correct option is B)

Angular momentum about O remains constant just before and just after collision∴L i =L f or ( 6 m v 2 L )=Iω =[ 3 mL 2 + 6 m 4 L 2 ]ω Solving, we get 12 mvL = 24 9mL 2 ω= 9L 2v

Video Explanation

Was this answer helpful?

2 0

Guys, does anyone know the answer?