On a frictionless air hockey table, a puck of mass 5.0 kg moving at 2:0m=s approaches an identical puck that is stationary. After collision, the �rst puck moves at 30 degrees� above the original line of motion; the second puck moves 60 degrees� below.
(a) What are the speeds of the puck after collision?
(b) Was the collision elastic?
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3.8
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy to determine the speeds of the pucks after the collision and whether the collision was elastic.
(a) Calculating the speeds of the pucks after collision:
1. Before the collision, the momentum of the system is conserved. The momentum of an object is defined as the product of its mass and velocity.
Momentum of the first puck before collision = mass * velocity = 5.0 kg * 2.0 m/s = 10 kg m/s
Momentum of the second puck before collision = mass * velocity = 5.0 kg * 0 m/s = 0 kg m/s (stationary)
2. After the collision, the momentum of the system is still conserved. However, the direction of the momentum changes due to the collisions, so we need to consider both the magnitudes and directions of the velocities.
3. We can split the velocities of the pucks into their horizontal (x) and vertical (y) components. Let's assume the positive x-direction is the initial line of motion of the first puck.
4. After the collision, the first puck moves at 30 degrees above the original line of motion, and the second puck moves 60 degrees below the original line of motion. We can use trigonometry to calculate the x and y components of the velocities.
Velocity of the first puck after collision:
- x-component: v1x = v1 * cos(30 degrees)
- y-component: v1y = v1 * sin(30 degrees)
Velocity of the second puck after collision:
- x-component: v2x = v2 * cos(-60 degrees)
- y-component: v2y = v2 * sin(-60 degrees)
Remember that cos(-60 degrees) = cos(60 degrees) and sin(-60 degrees) = -sin(60 degrees).
5. We can now write down the equation for the conservation of momentum in the x-direction:
m1 * v1x = m2 * v2x (since the initial x-velocity of the second puck is zero)
5.0 kg * v1 * cos(30 degrees) = 5.0 kg * v2 * cos(60 degrees)
We can simplify this equation to:
v1 * cos(30 degrees) = v2 * cos(60 degrees)
6. Next, we can write down the equation for the conservation of momentum in the y-direction:
m1 * v1y = -m2 * v2y (since the initial y-velocity of the second puck is zero, and the opposite sign is due to the opposite direction)
5.0 kg * v1 * sin(30 degrees) = -5.0 kg * v2 * sin(60 degrees)
We can simplify this equation to:
v1 * sin(30 degrees) = -v2 * sin(60 degrees)
7. We now have a system of two equations (one for x and one for y) with two unknowns (v1 and v2). We can solve this system of equations simultaneously to find the values of v1 and v2.
From equation 5: v1 * cos(30 degrees) = v2 * cos(60 degrees)
Divide both sides by cos(30 degrees):
v1 = (v2 * cos(60 degrees)) / cos(30 degrees)
Substitute this result into equation 6: v1 * sin(30 degrees) = -v2 * sin(60 degrees)
Replace v1 with the expression from equation 5:
(v2 * cos(60 degrees)) / cos(30 degrees) * sin(30 degrees) = -v2 * sin(60 degrees)
Simplify the equation:
(v2 * (sqrt(3)/2)) / (sqrt(3)/2) = -v2 * (sqrt(3)/2)
Cancel out the common terms:
v2 = -v2
Since v2 cannot be equal to its negation, this implies v2 = 0.
Now, substitute the value of v2 into the equation for v1 from equation 5:
v1 = (v2 * cos(60 degrees)) / cos(30 degrees)
v1 = (0 * cos(60 degrees)) / cos(30 degrees)
v1 = 0
Therefore, the speed of the first puck after the collision (v1) is 0 m/s, and the speed of the second puck (v2) is also 0 m/s.
(b) Since the speed of both pucks after the collision is 0 m/s, this implies that all kinetic energy has been lost. In an elastic collision, kinetic energy would be conserved. Therefore, the collision in this case is NOT elastic.