A 0.30-kg puck, initially at rest on a frictionless...

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of u 5 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of u 5 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg
puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

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Key Concepts

Conservation of Momentum

This principle states that in an isolated system with no external forces, the total momentum remains constant. In collision problems, it is applied in each independent direction (typically x and y) to relate the velocities before and after the collision, making it possible to solve for unknown variables like post-collision speeds and directions.

Vector Decomposition

When dealing with collisions that involve angles, it is essential to break down velocities and momentum into their horizontal and vertical components. This process enables the separate application of conservation laws along each coordinate axis, ensuring that the directional aspects of the motion are accurately accounted for during analysis.

Kinetic Energy Analysis (Elastic vs. Inelastic Collisions)

While conservation of momentum always holds in collisions, the conservation of kinetic energy depends on the type of collision. In perfectly elastic collisions, kinetic energy is conserved, whereas in inelastic collisions some kinetic energy is converted into other forms of energy. Calculating the kinetic energies before and after the collision helps in determining the energy lost and the nature of the collision.

Transcript

00:01 In this question, we have a 2d collision between two parks.

00:05 One is 0 .2 kg, the other one is 0 .3 kg.

00:10 Okay.

00:13 So initially, m1, which is 0 .2 kg, is moving to the right, 2 meters per second.

00:22 And then it collides with a stationary park, m2, which is 0 .3 kg.

00:29 And then after that m1 moves at 1 meters per second at an angle of 53 degrees to the horizontal axis and then puck two will just move down okay with some angle 5 from the horizontal okay so in part a we want to find a velocity of the 0 .3 kg puck and in part b we want to find a fraction of kinetic energy loss in the collision okay so this question is a 2d collision problem.

01:06 Okay, and then clearly we'll be using conservation of momentum.

01:12 Okay, and then in this 2d collision, i'll be using vector notation to write my answer.

01:21 Okay, and after that, i can use pythagoras theorem to find a many tube, okay, and then use the tangent to find the direction.

01:35 So the unit vector i has to the right, j has to up.

01:40 Okay.

01:42 So the conservation of momentum equation will be with an s -m -1 -v -1i plus m2v -2i and is equal to m1v plus m2 v2f.

01:57 Okay, then we just collect the terms.

02:00 M1 is 0 .2 .2.

02:02 Kjj.

02:02 M2 is 0 .3 kg.

02:08 V1i is 2i is 2i, and then 2i is 2i and then v2i is 0...