A particle of mass m kinetic energy K momentum p collides head on elastically with another particle of mass 2m at rest

A) Momentum of first particle ?
B) momentum of second particle?
C) kinetic energy of first particle?
D) kinetic energy of second particle?

dont copy kid go do it youre self

To solve this problem, we can use the conservation laws of momentum and kinetic energy in an elastic collision. Let's begin by finding the momenta of the two particles before and after the collision.

A) Momentum of the first particle:

The momentum of a particle is given by the product of its mass and velocity.
Let the velocity of the first particle before the collision be v1 and after the collision be v1'.

The momentum before the collision is given by:
Momentum_before = mass * velocity_before

Since the first particle is colliding head-on with the second particle, its velocity before the collision is positive (moving to the right). Therefore:
Momentum_before = m * v1

During an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. As the second particle is initially at rest, its momentum before the collision is zero.

Therefore, we can write the equation:
m * v1 = m * v1' + 2m * v2'

Now let's move on to finding the momentum of the second particle.

B) Momentum of the second particle:

Since the second particle is initially at rest, its momentum before the collision is zero. After the collision, the momentum of the second particle is given by:
Momentum_second = mass * velocity_second

Therefore, the momentum of the second particle after the collision is:
Momentum_second = 2m * v2'

Now, let's move on to finding the kinetic energy of each particle.

C) Kinetic energy of the first particle:

The kinetic energy of a particle is given by one-half times its mass times the square of its velocity.
Kinetic_energy_first = (1/2) * mass * velocity_first^2

Therefore, the kinetic energy of the first particle before the collision is:
Kinetic_energy_first_before = (1/2) * m * v1^2

For an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Hence, we can write the equation:
(1/2) * m * v1^2 = (1/2) * m * v1'^2 + (1/2) * 2m * v2'^2

Finally, let's find the kinetic energy of the second particle.

D) Kinetic energy of the second particle:

Similar to the first particle, the kinetic energy of the second particle after the collision is given by:
Kinetic_energy_second = (1/2) * mass * velocity_second^2

Therefore, the kinetic energy of the second particle after the collision is:
Kinetic_energy_second = (1/2) * 2m * v2'^2

To find v1', v2', we need to solve the system of equations obtained from the conservation of momentum and kinetic energy equations.

small letters for particle 1

capitals for particle 2

before
Particle 1
p = m v
K = .5 m v^2
Particle 2
p = 2 m* 0 = 0
K = (1/2)(2m) 0^2 = 0

after: va is particle 1 after, V is particle 2
p still = m v
but now m va + 2m V
K still .5 m v^2
but now .5m va^2 + .5 (2m)V^2
so
m v = m va + 2m V or v = va + 2V
so v^2 = va^2 + 4 va V + 4 V^2
and
v^2 = va^2 + 2 V^2
so
4 va V +4V^2 = 2 V^2
2va V = -V^2
V = -2va
va = -V/2
then
v= -V/2+ 2V
v = 3V/2
V = (2/3)v
va = -(1/3)v
take it from there