Conservation of Linear Momentum

[vladittude0583]

Homework Statement

A particle (m1) w/mass 0.080 kg has an initial velocity of 50m/s in the +x-direction and collides with a particle (m2) w/mass 0.060 kg with an initial velocity of 50m/s in the +y-direction. After the collision, particle (m1) and particle (m2) are stuck together and travel at some unknown velocity with an unknown angle. What is the lost in kinetic energy due to the collision?

Homework Equations

Initial Linear Momentum = Final Linear Momentum
delta K = Kf - Ki

The Attempt at a Solution

What I did was set-up a x- and y-component of the conservation of linear momentum to solve for the final velocities in their respective components. I know that the velocity that results of the two particles being stuck together after the collision is the "final" velocity which would be responsible for the final kinetic energy right? Beyond this, I do not know how to solve for the lost in kinetic energy? Could someone please tell me how to solve for this? Thanks.

 

[Snazzy]

You can calculate the kinetic energy each car had before the collision and what the kinetic energy of the two masses stuck together was after the collision, and you can calculate the loss in kinetic energy from that. Remember that energy is a scalar quantity, not a vector.

 

[Moetasim]

I would approach this problem by first understanding the concept of conservation of linear momentum. This law states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the system consists of two particles (m1 and m2) colliding with each other and sticking together.

To solve for the lost kinetic energy, we can use the formula delta K = Kf - Ki, where delta K is the change in kinetic energy, Kf is the final kinetic energy, and Ki is the initial kinetic energy.

We can calculate the initial kinetic energy of each particle using the formula K = 1/2 * m * v^2, where m is the mass and v is the velocity. For particle m1, K1 = 1/2 * 0.080 kg * (50 m/s)^2 = 100 J. For particle m2, K2 = 1/2 * 0.060 kg * (50 m/s)^2 = 75 J.

After the collision, the two particles are stuck together and travel at some unknown velocity with an unknown angle. We can use the conservation of linear momentum to solve for this final velocity. Since momentum is a vector quantity, we need to consider the x- and y-components separately.

In the x-direction, the initial momentum is m1 * v1x = 0.080 kg * 50 m/s = 4 kg*m/s. The final momentum is (m1 + m2) * vf * cos(theta), where vf is the final velocity and theta is the angle between the final velocity and the x-axis. Similarly, in the y-direction, the initial momentum is m2 * v2y = 0.060 kg * 50 m/s = 3 kg*m/s. The final momentum is (m1 + m2) * vf * sin(theta).

Since the particles stick together after the collision, we can set the initial and final momenta equal to each other. This gives us two equations:

m1 * v1x = (m1 + m2) * vf * cos(theta)
m2 * v2y = (m1 + m2) * vf * sin(theta)

Solving for vf in both equations and setting them equal to each other, we get:

vf = ( m1 * v1x / cos(theta) + m2 * v2y / sin(theta) )