50:750:203 General Physics July 1999

Test 1.



Attempt all questions. You must show some working. Credit will not be given for an answer without working.



1. A chimpanzee sitting against his favorite tree gets up and walks 37 m due east and 29 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound? (b) What angle does the shortest distance make with respect to due east?



Shortest distance = length of hypotenuse of triangle

= √{372 + 292} = 47.0 m



The direction is south of east by an angle , where

tan(θ) = 29/37, θ = 38.1o





2. A football player runs the pattern given in the drawing by the three displacement vectors A, B, and C. The magnitudes of these vectors are A = 4.00m, B = 12.0m, and C = 15.0m. Using the component method, find the magnitude and direction of the resultant vector A + B + C.

Choose the x-axis to point to the right and the y-axis to point upwards on the picture.

Vector x-component y-component

A 0 4.00

B 12.00 0

C 15cos(35)=12.29 -15sin(35)=-8.60

- --------------- ---------------

A+B+C 24.29 -4.60

magnitude of resultant = √{(24.29)2 + (-4.60)2} = 24.72 m

tan(θ) = 4.60/24.29, θ = 10.72o



3.(Conceptual question) The average velocity for a trip has a positive value. Is it possible for the instantaneous velocity at any point during the trip to have a negative value? Justify your answer.

Yes, it is possible. The average velocity is the total displacement divided by the total time. The displacement might decrease in some part of the trip, if you back up for some reason. As long as the final value for the total displacement is positive, the average velocity is positive.



4. Starting from rest, a speedboat reaches a speed of 3.9 m/s in 2.0s. What is the boat's speed after an additional 1.0 s has elapsed, assuming the boat's acceleration remains the same?

In general, v = v0 + at

Starting from rest, v = at

If v = 3.9 m/s after 2s,

a = v/t = 3.9m/s / 2.0 s = 1.95 m/s2

After a total time of 3.0 s,

v = (1.95 m/s2)(3.0s) = 5.85 m/s2



5. A soccer player kicks the ball toward a goal that is 28.0 m in front of him. The ball leaves his foot at a speed of 19.0 m/s and an angle of 32.0o above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: the answer is not 19.0 m/s.)

The horizontal component of the velocity is

vx = (19.0m/s)cos(32o) = 16.11 m/s

It remains constant at this value. The goal is 28.0 m from the point where the ball is kicked, so the time taken for the ball to reach the goal is

t = (28.0 m) / (16.11 m/s) = 1.738 s.

The vertical component of the velocity varies as

vy = v0y - gt

where the initial value is

voy = (19.0m/s)sin(32o) = 10.07 m/s

After 1.738s,

vy = 10.07 - (9.80)(2.738) = -6.98 m/s

(The value is negative because the ball is falling as it approaches the goal.)

The speed is

√{(16.11)2 + (-6.96)2} = 17.55 m/s





6. A swimmer, capable of swimming at a speed of 1.5 m/s in still water (i.e. the swimmer can swim with a speed of 1.5 m/s relative to the water), starts to swim directly across a 3.0-km-wide river. However, the current is 1.0 m/s, and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

The long way to do this is to find the net velocity of the swimmer diagonally across the river, and then to find the distance the swimmer has to go in this direction. The short way is to recognize that the component of the swimmer's velocity across the river, and the distance to be traveled directly across the river, are not affected by the motion of the river. Therefore

(a) time taken = width of river/component of velocity across river = 3.0km/1.5m/s = 2000s = 33min 20s

(b) distance the swimmer is carried downstream in this time is

speed of current times time taken

= (1.0m/s)(2000s) = 2000m = 2.0 km



7. An airplane has a mass of 33,000 kg and takes off under the influence of a constant force of 37,000 N. What is the net force that acts on the plane's 85-kg pilot, assuming that he has the same acceleration as the plane?



For the airplane, ∑F = ma

a = ∑F/m = (37000N)/(33000kg) = 1.1212 m/s2

The pilot has the same acceleration, so

∑Fpilot = mpilota = (85kg)(1.1212m/s2) = 95.30N



8. A 0.600-kg kite is being flown at the end of a string. The string is straight and makes an angle of 60.0o above the horizontal. The kite is stationary and the tension in the string is 40.0 N. Determine the force (both magnitude and direction) that the wind exerts on the kite. Specify the angle relative to the horizontal.



One approach is to represent the force by its horizontal and vertical components. Label them as H and V







Force x-component y-component

V 0 V

H H 0

W 0 -mg

T -Tcos(60) -Tsin(60)

- ---------- ---------

∑F H-Tcos(60) V-mg-Tsin(60)

In equilibrium, each component is zero. The x-component is zero if H = Tcos(60) = (40N)(0.50) = 20N

The vertical component is zero if

V = mg+Tsin(60) = (0.6)(9.80)+(40)(0.866) = 40.52 N

The magnitude of the force is

√{H2+V2} = 45.19N

and the angle above the horizontal is

tan-1(V/H) = 63.73o.