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Problem 1: Kinematics (15 pts) A particle moves along a straight line x. At time t = 0, its position is at x = 0. The velocity, V , of the object changes as a function of time, t, as shown in the figure. V is in m/s, x is in meter, and t is in seconds. (a) What is x at t = 1 sec? (b) What is the acceleration (m/s^2 ) at t = 2 sec? (c) What is x at t = 4 sec? (d) What is the average speed (m/s) between t = 0 and t = 3 sec?
− 6
−
−
2
4
6
1 2 3 4
V (m/s)
t (sec)
Solution:
(a) From the area blow the velocity curve, we find x = 6m at t = 1.
(b) From the slope of the velocity curve at t = 2, we find a = (− 6 − 6)/2 = − 6 m/s^2.
(c) The area below the v = 0-line has a negative contribution to the displacement. We find x = 0 at t = 4.
(d) The area below the v = 0-line has a negative contribution to the total distance traveled. Average speed =(6+3+3)/3 = 4m/s. (Different from average velocity which is (6+3−3)/3 = 2 m/s)
Problem 2: Surveillance Balloon (15 pts) A gun crew observes a remotely controlled balloon launching an instrumented spy package in enemy territory. When first noticed the balloon is at an altitude of 800m and moving vertically upward at a constant velocity of 5m/s. It is 1600m down range. Shells fired from the gun have an initial velocity of 400m/s at a fixed angle θ (sin θ = 3/5 and cos θ = 4/5). The gun crew (using its 8.01 ballistic knowledge) waits and fires so as to destroy the balloon. Assume g = 10m/s^2. Neglect air resistance. (a) What is the flight time of the shell before it strikes the balloon? (b) What is the altitude of the collision? (c) How long did the gun crew wait before they fired?
θ
1600m
400m/s
5m/s
Solution:
(a) The motion in the x-direction is a constant velocity motion. We find the flight time = 1600 m/vx = 1600/(400 cos θ) = 1600/(1600/5) = 5sec. Flight time = 5sec.
(b) From the flight time, the initial velocity in the y-direction and the acceleration in the y-direction, we can calculate the altitude of the shell: h = vyt− 12 gt^2 = 12005 × 5 − 12 × 10 ×25 = 1200 − 125 = 1075m. Altitude = 1075m.
(c) After the waiting time plus the flight time, the balloon should reach the same altitude as the shell. Let tw be the waiting time. We have h = (tw + 5) × 5 + 800 = 1075. tw + 5 = 275/5 = 55sec. So tw = 50sec. The waiting time = 50sec.
Problem 4: Force and Acceleration (25 pts) A particle of mass m = 5kg, is momentarily at rest at x = 0 at t = 0. It is acted upon by two forces F 1 and F 2. F 1 = 70ˆjN. The direction and magnitude of F 2 are unknown. The particle experiences a constant acceleration, a, in the direction as shown. Note: sin θ = 4/5, cos θ = 3/5, and tan θ = 4/3. Neglect gravity.
(a) Find the missing force F 2. Either give magnitude and direction of F 2 or its components. Plot F 2 on the figure. What angle does F 2 make to the x-axis? (b) What is the velocity vector of the particle at t = 10sec? (c) What third force, F 3 , is required to make the acceleration of the particle zero? Either
give magnitude and direction of F 3 or its components. (d) What is the vector sum of the three forces: F 1 + F 2 + F 3 =?
θ
a=10m/s
y
x
2
F (^2)
F =70 j N 1
Solution:
(a) F 2 x = max − F 1 x = 5 × 10 × cosθ − 0 = 30N. F 2 y = may − F 1 y = 5 × 10 × sin θ − 70 = − 30 N.
F^ 2 = (30, −30)N.
(b) v = 10 × a = (60, 80)m/s, or |v| = 100m/s with direction θ relative to x-axis.
(c) The force F 3 cancel the total acceleration. So F 3 = −ma. F^ 2 = (− 30 , −40)N.
(d) F 1 + F 2 + F 3 = (0, 70) + (30, −30) + (− 30 , −40) = (0, 0) = 0.
