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Problem 1: 2D collision (15 pts) A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-direction with a speed 2v and the second particle is at rest. After the collision, the second particle is moving in the direction 45◦^ below the x-axis and with a speed √ 2 v.
2v
y x Before collision
y x After collision
2v
?
45 o
(a) Find the velocity of the first particle after the collision. (ie find the x- and y-components of the velocity.) (b) Find the total kinetic energy of the two particles before and after the collision. (c) Is the collision elastic or inelastic? Solution: (a) The velocity of the second particle after the collision is (v 2 x, v 2 y) = (v, −v). From mo- mentum conservation in x-direction 2vm = v 1 xm+v 2 xm, we find v 1 x = v. From momentum conservation in y-direction 0 = v 1 ym + v 2 ym, we find v 1 y = v.
(b) Before: K = 12 m(2v)^2 = 2mv^2. After: K = 12 m(v 12 x + v^21 y) + 12 m(v^22 x + v 22 y) = 2mv^2 (c) The collision is elastic.
Problem 2: Dynamics (15 pts) Two balls of mass m and 2m are connected by a rod of length L. The mass of the rod is small and can be treated as zero. The size of the balls can also be neglected. We also assume the center of the rod is fixed, but the rod can rotate about its center in the vertical plane without friction.
L
θ (^) m
2m
(a) (b) (c)
m
2m
(d)
fixed
(a) Find the center of mass of the two balls. That is find the distance between the center of mass and the ball of mass 2m. (b) Find the moment of inertia of the two balls about the rotation axis. (c) Find the gravity induced angular acceleration of the rod when the angle between the rod and the vertical line is θ as shown. (d) If the rod starts its swing from the horizontal position, what is the angular velocity of the rod when it reaches the vertical position? Solution: (a) Let d be the distance between the center of mass and the ball of mass 2m. We have (2m)d = m(L − d). We find d = L/.
(b) Moment of inertia about the axis (the center of the rod): I = 2m(L/2)^2 +m(L/2)^2 = 34 mL^2.
(c) Torque about the axis (the center of the rod): T = 2mg L 2 sin θ − mg L 2 sin θ. Angular acceleration = T /I = (^23) Lg sin θ.
(d) Change in the potential energy ∆U = 2mg(L/2) − mg(L/2). From the energy conserva- tion ∆U = K = Iω^2 /2, we find the angular velocity ω =
√2∆U
I =
√ (^4) g 3 L.
Problem 4: Collision (15 pts) Initially, a rod of mass m and length L moves without rotation on a frictionless surface in a direction perpendicular to the rod. The speed of the rod is v. At time t = 0, one end of the rod collides with (or brushes over) a fixed object. Just after the collision, the rod is still parallel to the rod before the collision and the center of the rod still moves in the same direction as before. But the speed of the center of the rod is reduced to 34 v
v v 3v/
before just after
A
L ?
?
v fixed object
(a) Find the angular velocity ω of the rod just after the collision. (Hint: the total angular momentum about the collision point A is conserved during the collision.) (b) Find the total kinetic energy of the rod after the collision. (c) Find the speeds of the two ends of the rod just after the collision. Solution: (a) Angular momentum about point A: Before collision: mv L 2 After collision: m^34 v^ L 2 + Icω. (After collision, the angular momentum contains both contributions from the orbital motion of the center of mass and the spinning motion about the center of mass.) Ic = 121 mL^2 is the moment of inertia about the center of mass of the rod. From the conservation of the angular momentum mv L 2 = m^34 v^ L 2 + Icω, we find the angular velocity ω = mv^ L^2 − Imc 34 v^ L^2 = (^32) Lv.
(b) Before collision: K = 12 mv^2. After collision: K = 12 m(^34 v )^2 + 12 Icω^2 = 38 mv^2.
(c) Speed of the top end: 34 v + L 2 ω = 32 v. Speed of the bottom end: 34 v − L 2 ω = 0.
