newton·seconds or kilogram·meters/second.
8 . Momentum is conserved in all collisions. Kinetic energy is conserved only in elastic collisions.
9 . Using the reasoning from question #1, if weight is mg , then m = W /g .10 . The acceleration of a projectile is always g ; i.e., 10 m/s^2 , downward. Even though the velocity is
instantaneously zero, the velocity is still changing, so the acceleration is not zero. (By the way, the
answer “−10 m/s^2 ” is wrong unless you have clearly and specifically defined the down direction as
negative for this problem.)
11 . The magnitude of the resultant force is found by placing the component vectors tip-to-tail. This gives
a right triangle, so the magnitude is given by the Pythagorean theorem, 50 N. The angle of the resultant
force is found by taking the inverse tangent of the vertical component over the horizontal component,
tan−1 (40/30). This gives the angle measured from the horizontal.
12 .
friction force divided by normal force. μ has no units.13 . Acceleration is the slope of a velocity-time graph.
14 . Displacement is the area under a velocity-time graph (i.e., the area between the graph and the
horizontal axis).
15 . Velocity is the slope of a position-time graph. If the position-time graph is curved, then instantaneous
velocity is the slope of the tangent line to the graph.
16 . Because acceleration is not zero, the object cannot be moving with constant speed. If the signs of
acceleration and velocity are the same (here, if velocity is positive), the object is speeding up. If the
signs of acceleration and velocity are different (here, if velocity is negative), the object is slowing
down.
17 . An object always moves in the direction indicated by the velocity.
18 . Near the surface of a planet, mg gives the gravitational force. Newton’s law of gravitation, Gm 1 m (^2)
/r 2 , is valid everywhere in the universe. (It turns out that g can be found by calculating GM (^) planet /R
planet^
(^2) , where R
planet^ is the planet’s radius.)
19 . An object in uniform circular motion experiences a centripetal , meaning “center seeking,” force.
This force must be directed to the center of the circle.
20 . This and all three kinematics equations are valid only when acceleration is constant. So, for example,
this equation can NOT be used to find the distance travelled by a mass attached to a spring. The spring
force changes as the mass moves; thus, the acceleration of the mass is changing, and kinematics
equations are not valid. (On a problem where kinematics equations aren’t valid, conservation of
energy usually is what you need.)
Answers to Electricity and Magnetism Quiz
1 . F = qE .