Two masses m amd 2m approach each other along a part right angles to eachother. After collision, they stick together and move off at 2m/s at an angle 37¤ to the original direction of m....what is the initial speeds of the particles???
initial x momentum = m u
initial y momentum = 2 m v
final x momentum = 3 m (2) cos 37
final y momentum = 3 m (2) sin 37
so
u = 6 cos 37
v = 3 sin 37
add up the momentum vectors. Assuming the particles are moving in the +x and +y directions before the collision, with speeds u and v,
<mu,0> + <0,2mv> = <3m*2sin37°,3m*2cos37°>
now just solve for u and v.
To solve this problem, we can use the principles of conservation of momentum and conservation of energy.
Let's break down the information given:
Mass of one particle: m
Mass of the other particle: 2m
Velocity of the particles after collision: 2m/s
Angle of the particles after collision with the original direction of m: 37 degrees
We need to find the initial velocities of the particles before the collision.
First, let's consider the conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision. The momentum formula is defined as the product of an object's mass and its velocity.
Before collision:
Momentum of particle m = m * velocity_m
Momentum of particle 2m = (2m) * velocity_2m
After collision:
Total momentum = (m + 2m) * 2m/s = 3m * 2m/s = 6m^2/s
According to the conservation of momentum, the total momentum before the collision is the same as the total momentum after the collision:
m * velocity_m + (2m) * velocity_2m = 6m^2/s
Secondly, let's consider the conservation of energy. The total kinetic energy before the collision should be equal to the total kinetic energy after the collision. The kinetic energy formula is defined as one-half times the mass times the square of the velocity.
Before collision:
Kinetic energy of particle m = (1/2) * m * (velocity_m)^2
Kinetic energy of particle 2m = (1/2) * (2m) * (velocity_2m)^2
After collision:
Total kinetic energy = (1/2) * (3m) * (2m/s)^2 = 6m^2
According to the conservation of energy, the total kinetic energy before the collision is the same as the total kinetic energy after the collision:
(1/2) * m * (velocity_m)^2 + (1/2) * (2m) * (velocity_2m)^2 = 6m^2
Now we have two equations:
m * velocity_m + (2m) * velocity_2m = 6m^2/s -- (Equation 1)
(1/2) * m * (velocity_m)^2 + (1/2) * (2m) * (velocity_2m)^2 = 6m^2 -- (Equation 2)
Simplifying Equation 2:
(1/2) * m * (velocity_m)^2 + m * (velocity_2m)^2 = 6m^2
Dividing both sides by m:
(1/2) * (velocity_m)^2 + (velocity_2m)^2 = 6m
Let's substitute (velocity_m)^2 = 2h and (velocity_2m)^2 = 8h into Equation 1:
m * 2h + (2m) * 8h = 6m^2/s
2mh + 16mh = 6m^2/s
18mh = 6m^2/s
18h =6m/s
h = (6m/s) / 18 = (1/3)m/s
Therefore:
(velocity_m)^2 = 2h = 2 * (1/3)m/s = (2/3)m/s
(velocity_2m)^2 = 8h = 8 * (1/3)m/s = (8/3)m/s
Finally, taking the square root of the velocities:
velocity_m = sqrt((2/3)m/s) = (√2 / √3) * sqrt(m/s) ≈ 0.816 * sqrt(m/s)
velocity_2m = sqrt((8/3)m/s) = (2 * √2 / √3) * sqrt(m/s) ≈ 1.633 * sqrt(m/s)
Thus, the initial speeds of the particles are approximately 0.816 * sqrt(m/s) and 1.633 * sqrt(m/s) respectively.