Two pucks collide on a level air-hockey table. Puck A has a mass of 0.0250 kg and is moving initially due eastward at a speed of 5.50m/s. Puck B has a mass of 0.0500kg and is initially at rest. Right after collision, puck A is moving in a direction of 65 degrees north of east; and puck B is moving in a direction of 37 degress south of east. Find the speed of each puck right after the collision.
Well, you know momentum is conserved in NS, and EW directions. So write the momentum conservation equations in NS, and EW directions.
And, you have KE conservation.Three equations, three unknowns (velocity A, velocity B).
Start with the NS equation.
To find the speed of each puck right after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's start by finding the initial momentum of each puck:
Momentum of puck A before collision = mass of puck A * velocity of puck A
= 0.0250 kg * 5.50 m/s = 0.1375 kg·m/s
Momentum of puck B before collision = mass of puck B * velocity of puck B
= 0.0500 kg * 0 m/s = 0 kg·m/s (since puck B is initially at rest)
Since there are no external forces acting on the system, the total momentum before and after the collision must be the same. Therefore, the total momentum after the collision is also 0.1375 kg·m/s.
Now, let's find the velocity components of each puck after the collision:
Velocity component of puck A after collision in the y-direction = speed of puck A * sin(angle of direction)
= speed of puck A * sin(65 degrees)
Velocity component of puck B after collision in the y-direction = speed of puck B * sin(angle of direction)
= speed of puck B * sin(37 degrees)
Using the principle of conservation of momentum, we can write the equation:
Momentum of puck A after collision + Momentum of puck B after collision = 0
(mass of puck A * velocity component of puck A after collision) + (mass of puck B * velocity component of puck B after collision) = 0
(0.0250 kg * speed of puck A * sin(65 degrees)) + (0.0500 kg * speed of puck B * sin(37 degrees)) = 0
Substituting the given values into the equation, we get:
(0.0250 kg * speed of puck A * sin(65 degrees)) + (0.0500 kg * speed of puck B * sin(37 degrees)) = 0
(0.0250 kg * speed of puck A * 0.9063) + (0.0500 kg * speed of puck B * -0.6018) = 0
(0.0226 kg * speed of puck A) - (0.0301 kg * speed of puck B) = 0
0.0226 kg * speed of puck A = 0.0301 kg * speed of puck B
Simplifying the equation further, we get:
speed of puck A = (0.0301 kg * speed of puck B) / 0.0226 kg
Now, using the fact that the total momentum after the collision is 0.1375 kg·m/s, we can write another equation:
Momentum of puck A after collision + Momentum of puck B after collision = 0.1375 kg·m/s
(mass of puck A * velocity component of puck A after collision) + (mass of puck B * velocity component of puck B after collision) = 0.1375 kg·m/s
(0.0250 kg * speed of puck A * cos(65 degrees)) + (0.0500 kg * speed of puck B * cos(37 degrees)) = 0.1375 kg·m/s
(0.0250 kg * speed of puck A * 0.4226) + (0.0500 kg * speed of puck B * 0.7986) = 0.1375 kg·m/s
(0.0106 kg * speed of puck A) + (0.0399 kg * speed of puck B) = 0.1375 kg·m/s
Substituting the value of speed of puck A from the previous equation, we get:
(0.0106 kg * [(0.0301 kg * speed of puck B) / 0.0226 kg]) + (0.0399 kg * speed of puck B) = 0.1375 kg·m/s
Now, solving this equation will give us the value of speed of puck B. Once we have the value of speed of puck B, we can substitute it back into speed of puck A equation to find the value of speed of puck A.