AP QUIZ #5 2D MOTION
WWT08: SPEEDBOATS CHANGING VELOCITIESACCELERATION
Two speedboats are racing on a lake. In 10 seconds, Boat A goes from traveling east at 15 m/s to traveling north at
20 m/s. In the same time interval, Boat B goes from 20 m/s east to 2
A cannon, located 60.0m from the base of a vertical 25.0m tall cliff, shoots a 15kg shell at 43.0
degress above the horizontal toward the cliff.
1). What must the minimum muzzle velocity be for the shell to clear the top of the cliff?
2) The ground at the
Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0
Diagram (a) (b) (c)
Position Negative Negative Positive
Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .
9.1. Model: Model the car and the baseball as particles.
9.2. Model: Model the bicycle and its rider as a particl
Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.
Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:
Written in component form, Newton's first law is
( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 N
T1 x = - T1
T1y = 0 N Using Newton's first l
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
13.1. Model: The crankshaft is a rotating rigid body.
Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)
Fs on e =
Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -
Fm on e =
GMm Me (6.67 10 -1
11.1. Visualize:r Please refer to Figure Ex11.1. r
Solve: (b) (c)
(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.
11.2. Visualize:r Please refer to Figure Ex11.2. r
10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).
For the bullet,
For the bowling ball,
1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle
consists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the a
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):
1.013 10 5 Pa N p = 2.69 10 25 m -3 = = V kB T (1.38 10 -23 J K )(273 K )
18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-1
17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =
Kmicro. Solve: The number of atoms is
M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kg
Because helium atoms have an atomic mass number A
16.1. Solve: The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have the
same mass its volume must be
mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 3
16.2. Solve: The volume of the uranium nucleus is
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
T= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz
14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
15.1. Solve: The density of the liquid is
m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 10 -3 10 -3 m 3
Assess: The liquid's density is more than that of water (1000 kg/m3) and is a reasonable number.
15.2. Solve: The volume of the helium
3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the