Two identical pucks are on an air table. Puck A has an initial velocity of 2.0m/s in the +x-direction. Puck B is at rest. Puck A collides elastically with Puck B and A moves off at 1.0m/s at an angle of 60 degrees above the x-axis. What is the speed and direction of puck B after the collision?
You have two unknowns: the final x and y components of the velocity of Puck B: Vbx and Vby. The two momentum equations (x and y) will let you solve for each. You should not have to use the energy conservation equation at all; they have provided more information than you need. Mathematically speakking, the problem is overdetermined
M*Vo = M*Va*cos60 + M*Vbx
0 = M*Va*sin60 -M*Vby
2 = 1/2 + Vbx
Vbx = 3/2
Vby = sqrt3/2
Vbx^2 + Vby^2 = 9/4 + 3/4 = 3
After collision
Vb^2 + Va^2 = 4 = Vo^2
so kinetic energy really is conserved.
What kind of drug is this guy on. How would anyone understand that answer
You literally explained nothing. Your numbers just spill onto the page with no explanation of where they came from
Given:
Ma = M kg, Va = 2 m/s.
Mb = M kg, Vb = 0.
V3 = 1m/s[60o] CCW = Velocity of Ma after collision.
V4 = ? = Velocity of Mb after collision.
Momentum before = Momentum after.
Ma*Va + Mb*Vb = Ma*V3 + Mb*V4.
M*2 + M*0 = M*1[60] + M*V4,
2M = M[60] + M*V4,
Divide by M:
2 = 1[60] + V4,
2 = 0.5 + 0.866i + V4,
V4 = 1.5 - 0.866i = 1.732[-30o] = 30o S. of W. = 1.732m/s[210o] CCW. = Velocity of Mb after collision.
=
To solve this question, we can use the principle of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
In an elastic collision, the total momentum before and after the collision remains the same.
Let's denote the mass of each puck as m.
Initially:
Puck A's momentum (pA) = m * 2.0 m/s (since it moves in the +x-direction)
Puck B's momentum (pB) = 0 (as it is at rest)
After the collision:
Puck A's momentum (pA') = m * 1.0 m/s (at an angle of 60 degrees above the x-axis)
Using the conservation of momentum:
pA + pB = pA' + pB'
m * 2.0 + 0 = m * 1.0 * cos(60) + pB'
Simplifying:
2.0 = 0.5 + pB'
So, pB' = 2.0 - 0.5 = 1.5 m/s
2. Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before and after the collision also remains the same.
Initially:
Puck A's kinetic energy (KEA) = 0.5 * m * (2.0)^2 = 2m J
Puck B's kinetic energy (KEB) = 0
After the collision:
Puck A's kinetic energy (KEA') = 0.5 * m * (1.0)^2 = 0.5m J
Puck B's kinetic energy (KEB') = 0.5 * m * (vB')^2
Using the conservation of kinetic energy:
KEA + KEB = KEA' + KEB'
2m + 0 = 0.5m + 0.5 * m * (vB')^2
Simplifying:
2m = m + 0.5 * m * (vB')^2
So, (vB')^2 = (2m - m) / (0.5m)
(vB')^2 = 2 / 0.5
(vB')^2 = 4
vB' = 2 m/s
Therefore, the speed of Puck B after the collision is 2.0 m/s, and its direction is opposite to the initial direction of Puck A (i.e., in the -x-direction).