A puck (uniform disk) of mass m and radius r is travelingat a speed v on frictionless ice. It strikes a second identical puck a glancing blow (at point P) and sticks to it (using
VelcroTM). After the collision, the two-puck-object rotates with angular velocity ω about its center of mass, and its center of mass moves with speed V . In parts (a) to (d),express your answers in terms of m, r, and/or v.
a) Before the collision, what is the angular momentum of the system about point P?
b) Using the parallel axis theorem, determine the moment of inertia of the two-puck-object about its center of mass.
c) Using conservation of total angular momentum, determine the angular velocity, ω.
d) Using conservation of total linear momentum, determine the speed, V .
e) What is the change in kinetic energy of the system? Express your answer as a numerical fraction of the original kinetic energy in the system.
a) To determine the angular momentum of the system about point P before the collision, we need to consider the angular momentum of each puck separately and then sum them up.
The angular momentum of an object is given by the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since the first puck is traveling at a speed v, its angular momentum about point P is given by L1 = mvr, where m is the mass of the puck and r is its radius.
Since the second puck is initially at rest, its angular momentum about point P is zero.
Therefore, the total angular momentum of the system about point P before the collision is L = L1 + L2 = mvr.
b) The moment of inertia of the two-puck-object about its center of mass can be determined using the parallel axis theorem. According to the theorem, the moment of inertia about any axis parallel to and a distance d away from an axis through the center of mass is given by I = Icm + Md^2, where Icm is the moment of inertia about the center of mass, M is the total mass of the system, and d is the distance between the two axes.
The moment of inertia of a single puck about its center of mass can be calculated using the formula Icm = 1/2mr^2.
Since we have two identical pucks stuck together, the total mass of the system is 2m.
Applying the parallel axis theorem, the moment of inertia of the two-puck-object about its center of mass is I = 1/2mr^2 + 2m(r)^2 = 1/2mr^2 + 2mr^2 = 5/2mr^2.
c) Conservation of total angular momentum states that the total angular momentum before the collision is equal to the total angular momentum after the collision. Using this principle, we can set up an equation to solve for the angular velocity ω.
Before the collision, we found that the angular momentum about point P is mvr.
After the collision, the two-puck-object rotates about its center of mass with angular velocity ω. The moment of inertia about the center of mass is 5/2mr^2 (as found in part (b)). Therefore, the angular momentum about the center of mass is given by Lcm = Icm * ω = (5/2mr^2) * ω.
Since the pucks stick together after the collision, the total angular momentum of the system after the collision is the same as before, which is mvr.
Setting up the equation: mvr = (5/2mr^2) * ω.
Solving for ω, we get ω = 2vr/5r^2 = 2v/5r.
d) Conservation of total linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. Using this principle, we can set up an equation to solve for the speed V.
Before the collision, the first puck has momentum mv, and the second puck has zero momentum.
After the collision, the two-puck-object moves with speed V. Since the pucks stick together, the total mass of the system in this case is 2m.
Setting up the equation: mv = 2mV.
Solving for V, we get V = v/2.
e) To determine the change in kinetic energy of the system, we need to calculate the initial and final kinetic energies and then find their difference.
The initial kinetic energy of the system is given by KE_initial = 1/2mv^2 + 0 (since the second puck is at rest initially).
The final kinetic energy of the system is given by KE_final = 1/2(2m)V^2 + (1/2)(5/2mr^2)ω^2, where V is the speed of the system and ω is the angular velocity of the system.
Using the values we found in parts (c) and (d), we can calculate the final kinetic energy.
After simplifying the expression, we get KE_final = 1/2mv^2 + (1/2)(25/20)m(v/r)^2.
To find the change in kinetic energy, we subtract the initial kinetic energy from the final kinetic energy: ΔKE = KE_final - KE_initial = [1/2mv^2 + (1/2)(25/20)m(v/r)^2] - [1/2mv^2 + 0].
After further simplification, we get ΔKE = (1/2)(5/8)m(v/r)^2.
Therefore, the change in kinetic energy of the system is (1/2)(5/8)m(v/r)^2.