∙A ball with a mass of 0.600 kg is initially at rest. It is struck by a second ball having a mas

∙A ball with a mass of 0.600 kg is initially at rest. It is...

$\bullet$ A ball with a mass of 0.600 $\mathrm{kg}$ is initially at rest. It is struck by a second ball having a mass of 0.400 $\mathrm{kg}$ , initially moving with a velocity of 0.250 $\mathrm{m} / \mathrm{s}$ toward the right along the $x$ axis. After the collision, the 0.400 $\mathrm{kg}$ ball has a velocity of 0.200 $\mathrm{m} / \mathrm{s}$ at an angle of $36.9^{\circ}$ above the $x$ axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the 0.600 kg ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

$\bullet$ A ball with a mass of 0.600 $\mathrm{kg}$ is initially at rest. It is struck by a second ball having a mass of 0.400 $\mathrm{kg}$ , initially moving with a velocity of 0.250 $\mathrm{m} / \mathrm{s}$ toward the right along the $x$ axis. After the collision, the 0.400 $\mathrm{kg}$ ball has a velocity of 0.200 $\mathrm{m} / \mathrm{s}$ at an angle of $36.9^{\circ}$ above the $x$ axis in the first
quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the 0.600 kg ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

Key Concepts

Conservation of Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant during a collision. It is used to set up equations for the sum of the momenta before and after the collision in both the horizontal and vertical directions independently, ensuring that the momentum vector is conserved throughout the event.

Vector Resolution

In many collision problems, especially when motion is not confined to a single straight line, it is essential to break velocity vectors into their horizontal and vertical components. This process, typically using trigonometric functions, allows for the correct application of conservation laws along each axis separately to solve for unknowns such as speeds and angles after the collision.

Kinetic Energy Analysis

Analyzing the kinetic energy before and after a collision helps determine how the energy is redistributed during the event. While momentum is always conserved in isolated systems, kinetic energy may not be, particularly in inelastic collisions. Evaluating the change in kinetic energy provides insights into the efficiency of the collision and whether some energy was transformed into other forms, such as heat or sound.

Transcript

00:01 So initially, a 0 .40 kilogram object is moving with a velocity of 0 .250 meter per second.

00:11 And it hits a stationary object of mass 0 .6 .0 kilograms.

00:17 Note that all numbers have three significant digits.

00:20 From here on, i'll just write down one significant, i'll write it down without any significant digits, but they all have three significant digits.

00:30 And after the collision, the 0 .4 kilogram mass is moving at an angle of of 36 .9 degrees with a velocity 0 .2 meters per second.

00:46 Let's say the other object, the 0 .6 kilogram, is moving at an angle theta below the x -axis with a velocity v.

01:00 Now we want to find out the magnitude of this velocity and the angle it makes with the x -axis for part a.

01:08 Now, we simply use conservation of momentum in both the directions.

01:12 Let's say our hard horizontal direction is x and vertical direction is y.

01:18 So let's go with first y direction.

01:20 Initially, this object was moving horizontally and this object was not moving, which means initial momentum in y was zero.

01:32 But finally, let's call this object a and let's call this object b.

01:40 Object a has y momentum mass, which is 0 .4 times velocity, which is 0 .2 times sine theta, which is sign 36 .9 degrees.

01:57 And for object b, the y momentum would be mass 0 .6 kilogram times velocity v times sine theta...

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