a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.060 kg and is initially at rest. The collision is not head-on. After the collision, pluck A goes in 65degree above + Xaxis, Pluck B is 37 degree below +X axis. what's final speed of pluck A and pluck B?
you need to work the conservation of momentum in two directions, x, and y.
To find the final speeds of puck A and puck B after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's break down the problem into two components: the x-component and the y-component.
In the x-component, we know that puck B is initially at rest, so its initial momentum is zero (mB * vBx = 0). Puck A has a mass of 0.025 kg and is moving along the x-axis with a velocity of +5.5 m/s. So, the initial momentum of puck A in the x-component is mA * vAx = 0.025 kg * 5.5 m/s = 0.1375 kg * m/s.
After the collision, puck A goes in the 65-degree direction above the positive x-axis, while puck B goes in the 37-degree direction below the positive x-axis. Let's find the final velocities of both pucks in the x-component.
Using the conservation of momentum, we have:
mAx * vAx + mBx * vBx = mAx * vAfx + mBx * vBfx
Since mBx is initially zero, the equation simplifies to:
mAx * vAx = mAx * vAfx
Substituting the values, we get:
0.025 kg * 5.5 m/s = 0.025 kg * vAfx
Solving for vAfx, we find:
vAfx = 5.5 m/s
So, the final velocity of puck A in the x-component is 5.5 m/s.
Now, let's consider the y-component. Here, we need to find the final velocities of both pucks in the y-direction.
Using the conservation of kinetic energy, we have:
(1/2) * mAy * vAy^2 + (1/2) * mBy * vBy^2 = (1/2) * mAy * vAfAy^2 + (1/2) * mBy * vBfy^2
Since both pucks are initially at rest in the y-direction, the equation simplifies to:
0 + 0 = (1/2) * mAy * vAfAy^2 + (1/2) * mBy * vBfy^2
Substituting the values, we get:
(1/2) * 0.025 kg * (0)^2 + (1/2) * 0.06 kg * (0)^2 = (1/2) * 0.025 kg * vAfAy^2 + (1/2) * 0.06 kg * vBfy^2
0 = (1/2) * 0.025 kg * vAfAy^2 + (1/2) * 0.06 kg * vBfy^2
This equation tells us that the sum of the kinetic energies before the collision (which is zero) equals the sum of the kinetic energies after the collision.
Since the pucks separate in a direction perpendicular to the x-axis, their final velocities in the y-component can be determined from the given angles.
Using sine and cosine, we can find the y-components of the final velocities:
vAfAy = vAfy * sin(65 degrees)
vBfy = vBfy * sin(-37 degrees)
Now, substituting these equations into the conservation of kinetic energy equation, we can solve for vAfy and vBfy.
0 = (1/2) * 0.025 kg * (vAfy * sin(65 degrees))^2 + (1/2) * 0.06 kg * (vBfy * sin(-37 degrees))^2
Simplifying the equation, we get:
0 = 0.000364 * (vAfy)^2 + 0.001055 * (vBfy)^2
Since the equation must equal zero, we can conclude that both terms must be zero:
0.000364 * (vAfy)^2 = 0
0.001055 * (vBfy)^2 = 0
Therefore, both vAfy and vBfy must be zero.
In summary, after the collision, the final speed of puck A in the x-component is 5.5 m/s, and both pucks come to rest in the y-component.