A 0.300-kg puck, initially at rest on a horizontal,...

A $0.300-\mathrm{kg}$ puck, initially at rest on a horizontal, frictionless surface, is struck by a $0.200-\mathrm{kg}$ puck moving initially along the $x$ axis with a speed of $2.00 \mathrm{~m} / \mathrm{s}$. After the collision, the $0.200-\mathrm{kg}$ puck has a speed of $1.00 \mathrm{~m} / \mathrm{s}$ at an angle of $\theta=53.0^{\circ}$ to the positive $x$ axis (see Fig. 9.14). (a) Determine the velocity of the $0.300-\mathrm{kg}$ puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

Key Concepts

Conservation of Momentum

In a collision, the total momentum of a closed system is conserved. This means that the vector sum of the momenta before the collision equals the vector sum after the collision, even if energy is not fully conserved. This principle is especially useful when analyzing collisions in two dimensions, where momentum must be conserved independently along each axis.

Vector Components Analysis

Breaking down the momentum vectors into their x and y components allows one to solve collision problems in two dimensions. By resolving each velocity into its horizontal and vertical components, the conservation of momentum can be applied separately along the x-axis and y-axis, simplifying the process of finding unknown quantities after the collision.

Kinetic Energy in Collisions

Kinetic energy, which depends on the mass and the square of the velocity of an object, may not be conserved in collisions even though momentum is. By comparing the total kinetic energy before and after a collision, one can determine how much kinetic energy has been lost, which is essential for analyzing inelastic collisions.

Elastic vs Inelastic Collisions

Collisions are classified as elastic or inelastic based on energy conservation. In elastic collisions, both momentum and kinetic energy are conserved. In inelastic collisions, momentum is conserved but some kinetic energy is lost to factors like deformation or heat. Determining the fraction of kinetic energy lost helps in identifying the nature of the collision.

Transcript

00:01 Welcome today we'll be talking about collisions and momentum so first let's draw a small diagram of the problem so we can see what's happening and we can see here that after colliding with a stationary puck we'll call m2 the first puck goes off at a 53 degree angle above the horizontal at one meter per second and we don't know the second the final velocity of the second so so this is what we're looking for in the problem and like always let's start by bringing it down into x and y components so we'll have that in the x direction we have the mass of the first time its velocity and the second one has no velocity as it's stationary so that'll be zero so we'll just put equal to m1 v1 final along x plus m2 v final along x so from here we'll isolate for v2 final we'll get v2 final along x is equal to m1 v1 minus m1 v1 final multiplied by cos of 53 and remember this is because of trigonometry we know that the x component is the hypotenuse which is the v1 final times cost 53 and then we'll divide that by the mass of the second so we'll get v2 final along x is equal to 0 .2 times 2 minus 0 .2 times 2 times 1 times cost of 53 divided by the mass of the second which is 0 .3 and here we will get a value approximately 0 .93 meters per second.

03:02 This will be approximately 0 .932 meters per second.

03:12 So for the y, we'll do the same thing.

03:16 Just this first velocity does not have any y component.

03:23 So instead of having m1 v1, we'll have 0 is equal to m1 v1.

03:32 Plus m2 v final of 2 along y this time we'll get v2 final along y is equal to m1 v1 final along y this is negative because it got subtracted divided by m2 and if we look at our diagram it makes sense because we know that the ball is hit the first ball goes off in the positive direction so most likely the second one had to go at some sort of negative direction below the horizontal so this makes sense and now substituting in for our values we'll get the b2 final y is equal to 0 .2 times the velocity final along y which is 1 times sine of 53 divided by the mass of the second, which is 0 .3.

04:52 And here, we'll get a value of 0 .53 meters per second, 532 meters per second.

05:12 From there, we know that v2 final is equal to the square root when this is negative.

05:22 Can't forget that, is equal to v2 final along x plus v2 final along y, squared and square root so we'll get v2 final is equal to the square root of 0 .932 squared plus negative 0 .532 squared and we'll get an answer for v2 final of approximately 1 .15 meters per second and now we just need to find the angle so we'll call this angle theta so we'll have theta is equal to 10 inverse y over x for the velocities and we know that theta will be equal at 10 inverse we know y is 0 .532 divided by 0 .9 32 and because we can draw a small diagram of the components in the x and y so we know that this x component will be a positive x and we know that this y component will be negative y and we just draw a small plane this is positive x this is positive y this is along negative y so we know that this angle will be below the horizontal so we don't have to add the negative.

07:25 But if we add the negative, it'll tell us the same thing.

07:27 So either way, we get the value for this angle to be approximately 30 degrees...

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