two identical pucks collide on an air hockey table. one puck was originally at rest. if the incoming puck has a speed of 6 m/s and scatters to an angle of 30 degree, what is the velocity (magnitude and direction) of the second puck if it scatters at 60 degree?
Please I need final answer with magnitude and direction.
Thank you
To solve this problem, we can use the law of conservation of momentum. The law states that the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the velocities of the two pucks before the collision as v1 and v2, and the angles at which they scatter as θ1 and θ2, respectively.
Given:
Initial speed of incoming puck (v1) = 6 m/s
Scattering angle of incoming puck (θ1) = 30 degrees
Scattering angle of second puck (θ2) = 60 degrees
Since the first puck is at rest initially, its initial velocity (v1) can be considered as zero.
Momentum before collision = Momentum after collision
Initial momentum = m1 × v1 + m2 × v2
Final momentum = m1 × v1' + m2 × v2'
Where m1 and m2 are the masses of the two pucks, and v1', v2' are their final velocities. Since the pucks are identical, their masses can be assumed to be the same.
Now, let's break down the given information:
v1 = 0 m/s (Initial velocity of first puck)
v1' = ? (Final velocity of first puck)
θ1 = 30°
θ2 = 60°
From the conservation of momentum equation, we have:
m1 × v1 + m2 × v2 = m1 × v1' + m2 × v2'
Since the first puck is at rest initially (v1 = 0), the equation simplifies to:
m2 × v2 = m1 × v1' + m2 × v2'
Now let's focus on the scattering angles. According to the laws of physics, the scattering angle is measured with respect to the initial direction of the incoming puck.
In our case, the incoming puck scatters at an angle of 30 degrees. Therefore, the scattering angle for the second puck (θ2) should be measured with respect to the direction of the incoming puck before the collision.
Hence, we can say that the scattering angle for the second puck relative to the incoming puck is:
Scattering angle (θ2) = |θ1 - θ2|
θ2 = |30° - 60°| = | -30° | = 30°.
Now, we have:
m2 × v2 = m1 × v1' + m2 × v2'
We know that the scattering angle is the angle between the initial direction of the incoming puck and the final direction of each puck after the collision.
Using the given information: θ1 = 30° and θ2 = 30°
We can conclude that the final velocity of each puck will be the same in magnitude but in opposite direction.
Therefore, the final velocity of the second puck (v2') will be 6 m/s, directed at an angle of 30 degrees in the opposite direction.
So, the magnitude of the velocity of the second puck is 6 m/s, and the direction is 180° - 30° = 150°.
To calculate the velocity of the second puck after the collision, we can use the law of conservation of linear momentum. According to this law, the total momentum before and after the collision should remain the same if no external forces act on the system.
Let's assume that the mass of each puck is m.
Before the collision:
The first puck is at rest, so its initial velocity (v1i) is zero.
The second puck has a speed of 6 m/s, and its initial velocity (v2i) makes an angle of 0 degrees with the positive x-axis.
After the collision:
The first puck scatters at an angle of 30 degrees with the positive x-axis, forming an angle of 30 degrees with the scattered velocity of the second puck.
The second puck scatters at an angle of 60 degrees with the positive x-axis.
Now, let's break down the velocities into their x and y components.
For the first puck, after the collision:
vx1f = v1f * cos(30°)
vy1f = v1f * sin(30°)
For the second puck, after the collision:
vx2f = v2f * cos(60°)
vy2f = v2f * sin(60°)
Since the mass of both pucks is the same (m), the momentum before the collision is equal to the momentum after the collision:
m * v2i = m * v1f + m * v2f
We can ignore the mass (m) on both sides of the equation and substitute the velocities with their components:
v2i = vx1f + vx2f (in the x-direction)
0 = vy1f + vy2f (in the y-direction)
Now, let's plug in the values we have:
v2i = vx1f + vx2f
6 m/s = v1f * cos(30°) + v2f * cos(60°)
0 = vy1f + vy2f
0 = v1f * sin(30°) + v2f * sin(60°)
From the first equation, we can solve for v1f:
v1f = (6 m/s - v2f * cos(60°)) / cos(30°)
Plug this value into the second equation and solve for v2f:
0 = [(6 m/s - v2f * cos(60°)) / cos(30°)] * sin(30°) + v2f * sin(60°)
Now, solve this equation to find v2f.
momentum is conserved
... initial momentum-x = 6 m ... initial momentum-y = 0
let m be the mass of a puck
... a is the moving puck , b is the stationary (2nd) puck
the sum of the final x-momenta is equal to the initial x-momentum
... [va * m * cos(30º)] + [vb * m * cos(60º)] = 6 * m
... [va * cos(30)] + [vb * cos(60º)] = 6
the sum of the final y-momenta is equal to the initial y-momentum
... [va * m * sin(30º)] - [vb * m * sin(60º)] = 0
... [va * sin(30º)] = [vb * sin(60º)]
solve the system of equations for vb (and va)