A hockey puck B rests on frictionless, level ice and is struck by a second puck A, which was
originally traveling at 40.0 m/s and which is deflected 30.0° from its original direction. Puck B
acquires a velocity at a 45.0° angle to the original direction of A. The pucks have the same mass.
Determine the speed of each puck after the collision.
I'm kind of stuck on this question. I know that its an elastic collision, and that B is initially at rest. However, I have no idea what to do from here.
call the original direction the x-direction
all of the original momentum is in the x-direction ... carried by A
after the collision, there are momenta in the ± y-direction
... they are equal and opposite (cancel each other)
the sum of the two x-momenta is equal to the original momentum
40.0 m/s = vA cos(30º) + vB cos(45º)
vA sin(30º) = vB sin(45º)
Oh okay, I think you helped me figure it out.
Since the y-momenta cancel, piy = pif.
initially, piy = 0
pif = vA sin 30 + vB sin 45
0 = vA sin 30 + vB sin 45
vA = vB sin45 / sin 30
substitute that into your equation to find VB and then solve the rest
thanks alot!
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's denote the initial velocity of puck A as vA and the final velocity as v'A, and the final velocity of puck B as vB.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
mAvA = mA'v'A + mBvB
Since puck B is initially at rest (vB = 0):
mAvA = mA'v'A
2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
0.5mA(vA^2) = 0.5mA'(v'A^2) + 0.5mB(vB^2)
Since puck B is initially at rest (vB = 0):
0.5mA(vA^2) = 0.5mA'(v'A^2)
Now, we need to find a way to solve these two equations simultaneously to determine the final velocities.
From the given information, we can determine the angles at which puck A and puck B move after the collision.
The angle of puck A, θA = 30.0° (deflected from its original direction).
The angle of puck B, θB = 45.0°.
Now, we can calculate the components of the velocities:
vA_x = vA * cos(θA)
vA_y = vA * sin(θA)
vB_x = vB * cos(θB)
vB_y = vB * sin(θB)
Since puck B is initially at rest, its velocities are zero:
vB_x = 0
vB_y = 0
Now, we can rewrite the conservation of momentum equation in terms of velocity components:
mA * vA_x = mA' * v'A_x + mB * vB_x
mA * vA_y = mA' * v'A_y + mB * vB_y
Since puck B is initially at rest, vB_x = 0 and vB_y = 0:
mA * vA_x = mA' * v'A_x
mA * vA_y = mA' * v'A_y
Finally, we can solve these two equations simultaneously to find the final velocities v'A_x and v'A_y.
Let me do the calculations for you.
To solve this problem, we can use the principles of conservation of linear momentum and conservation of kinetic energy.
First, let's identify the known information:
- Initial speed of puck A: 40.0 m/s
- Angle of deflection for puck A: 30.0°
- Angle of the velocity of puck B after collision: 45.0°
- Mass of puck A = mass of puck B (let's represent it as m)
Now, let's break down the problem into components:
- Before the collision, the momentum is only in the x-direction because puck A is traveling in the x-direction and puck B is initially at rest.
- After the collision, both pucks will have momentum components in both the x and y directions.
Step 1: Calculate the initial momentum of puck A in the x-direction.
- Momentum = mass × velocity
- Initial momentum (puck A) in the x-direction = mass × (initial speed of A) × cos(angle of deflection of A)
Step 2: Calculate the initial momentum of puck A in the y-direction.
- The initial momentum in the y-direction is given by the equation: momentum (y-dir) = mass × (initial speed of A) × sin(angle of deflection of A)
- Since no other forces act in the y-direction before and after the collision, the y-component of momentum is conserved.
Step 3: Calculate the initial momentum of puck B in the x-direction.
- Initially at rest, puck B has no momentum in the x-direction.
Step 4: Calculate the initial momentum of puck B in the y-direction.
- Initially at rest, puck B has no momentum in the y-direction.
Step 5: Apply the principle of conservation of linear momentum in the x-direction.
- After the collision, the total momentum in the x-direction is conserved.
- Initial momentum (puck A in x-dir) = Final momentum (puck A in x-dir) + Final momentum (puck B in x-dir)
Step 6: Apply the principle of conservation of linear momentum in the y-direction.
- After the collision, the total momentum in the y-direction is conserved.
- Initial momentum (puck A in y-dir) = Final momentum (puck A in y-dir) + Final momentum (puck B in y-dir)
Step 7: Find the final velocities of puck A and puck B.
- The final velocity of B can be found using the final momentum of B in the x-direction and y-direction.
- The final velocity of A can be found using the final momentum of A in the x-direction and y-direction.
This step-by-step approach will help you tackle this problem systematically. Let me know if you need help with the calculations or any further assistance.