A 12.0 g wad of sticky clay is hurled horizontally at a 100...

Name: Problem 25, Chapter 9: Physics for Scientists and Engineers with Modern Physics
Uploaded: 2021-01-07T09:36:38-08:00
Duration: 5 min 8 s
Channel: Vishal Gupta
Description: A 12.0 g wad of sticky clay is hurled horizontally at a 100 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650 , what was the speed of the clay immediately before impact?

A 12.0 g wad of sticky clay is hurled horizontally at a 100 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650 , what was the speed of the clay immediately before impact?

Key Concepts

Conservation of Momentum

In collisions, the total momentum of the system is conserved provided there is no external impulse. In a completely inelastic collision, even though kinetic energy is not conserved, the momentum before impact equals the momentum immediately after impact, which lets us relate the velocity of the clay prior to collision to the combined mass and velocity of the clay-block system after collision.

Work-Energy Principle

After the collision, the kinetic energy of the combined system is gradually dissipated by work done against friction. By equating the initial kinetic energy of the system to the work done by friction over the distance before the system stops, one can solve for the velocity of the system immediately after the impact.

Friction Force

The friction force, calculated as the product of the coefficient of friction and the normal force, plays a critical role in the energy dissipation process. By knowing the friction force and the sliding distance, we can determine the amount of work done by friction, which is then used in analyzing the post-collision motion through the work-energy principle.

Transcript

00:01 Hi, the given problem on the horizontal rough surface.

00:05 Suppose this is the wooden block of mass capital m and here this is the clay of mass small m.

00:14 Suppose initial speed of the clay is u then the speed achieved by this wooden block after the impact with clay is v i and its final speed after traveling through 7 .5 meters means vf comes to be zero.

00:34 The weight of the combined block with clay will be acting vertically downward m plus m g so the reaction offered by this surface horizontal surface to this block will be n which is given by n is equal to m plus m into g and force of friction as the block will be moving rightward a force of friction will be acting on it towards left which is given as mu into n or we can say this is mu into m plus m into g then as per newton's second of motion this is also given as the product of mass of this block with the clay and the acceleration produced in this block along with the clay so canceling this m plus m acceleration in the motion of this block or we can say retardation as the block finally comes to rest.

01:34 This comes this will be given as negative of mu into g as this mu is given as 0 .650 then multiplied with 9 .8 meter per second is square.

01:50 Retardation in the motion of this block comes out to be minus v.

01:58 We will keep it as it is, we will calculate it later on.

02:01 Now this is the retardation in the motion of the block.

02:04 Now for the motion of the block, its initial speed is vi.

02:16 This speed is after the impact with the clay.

02:19 And that is missing, we don't know this...