Practice test – Chapters 1-4 – PHY1033C

 

Your test will have 20 or 25 questions.  Some will be multiple choice, and some will be short answer.  The questions below are representative of what you will see (but of course will be different).

 

1. A child is on a flatcar moving due east at 8 m/s.  He rolls a ball at a speed of 6 m/s due north relative to the flatcar.  Make a sketch that shows the speed of the ball relative to the ground.  Estimate this speed.

 

Your answer would be to draw a sketch similar to the one shown.  The speed estimate of 10 m/s is from measuring the length of the resultant arrow or by using Pythagorean theorem.

 

2. A ball is tossed upward at a starting velocity of 15 m/s.  Assuming the acceleration of gravity is g = 10 m/s2, the ball will reach a maximum height of ___m.

 

(1) 11.3     (2) 15.0      (3) 18.5      (4) 20.8      (5) 24.5

 

Since the ball loses 10 m/s each second, the peak height is reached in 1.5 s.  As stated in your book, the average velocity going up is thus 7.5 m/s.  The maximum height is thus 7.5 m/s times 1.5 s, or 11.3 m.

 

3. A graph of velocity for a certain object as a function of time is shown below.  The acceleration for this object is ___ m/s2.

 

(1) 4/3    (2) 3/4    (3) -4/3     (4) -3/4     (5) none of these

 

The acceleration is defined as (change in v)/(change in t).  Thus, a = -3/4 m/s2.

 

4.  A 12-N force is applied to a 4-kg mass on a frictionless surface, as shown.  At t = 3 s the acceleration of the mass is ___ m/s2.     

 

(1) 1        (2) 2        (3) 3         (4) 4         (5) 6

 

From Newton’s law F = ma, a = (12 N)/(4 kg) = 3 m/s2.

 

5.  Three seconds after a rock is dropped it hits the ground.  The speed of the rock just before impact is ___ m/s.  Assume g = 10 m/s2, and neglect air resistance.

 

(1) 15        (2) 20        (3) 25        (4) 30         (5) 40

 

The speed increases 10 m/s each second.  After 3 s, the speed must then be 30 m/s.

 

6. The rock in the above problem must have been dropped from a height of ___ meters.

 

(1) 20        (2) 25        (3) 30        (4) 40          (5) 45

 

The rock’s average speed on the way down is 15 m/s.  Height = (15 m/s) x (3 s) = 45 m.

 

7.  A force F = 12 N parallel to a frictionless horizontal surface accelerates two masses connected by a cord, as shown.  The tension in the connecting cord is ___ N.

 

(1) 3.0       (2) 3.6       (3) 4.0       (4) 4.5       (5) 6.0

 

 

The tension in the cord accelerates only the 2-kg mass, whereas the force F = 12 N accelerates all 8 kg.  Since 12 N accelerates 8 kg, only 3 N is needed to accelerate 2 kg by the same amount of acceleration.

 

8. A cart rolls down a frictionless ramp starting from rest.  Between t = 0 and t = 1 s, the cart moves 1 m.  Therefore, between t = 1 and t = 2 s, the cart moves ____ m.

(1) 1     (2) 2      (3) 3      (4)     (5)

 

The total distance traveled by an object undergoing constant acceleration is proportional to the square of the time.  The total distance traveled after t seconds must then be in the ratio 1,4,9, etc.  This tells us that the distance traveled during the 2nd second is 3 m.

 

9. A falling skydiver of mass 100 kg experiences 500 N of air resistance.  The skydiver’s acceleration is

 

(1) 0.2 g   (2) 0.3 g   (3) 0.4 g   (4) 0.5 g   (5) more than 0.5 g

 

The force of gravity on the skydiver is his weight, 1000 N, acting downward.  Air resistance is 500 N upward.  If there were no air resistance, the net force would be 1000 N and his acceleration would be g.   But since the net force on him with air resistance is now 500 N downward, his acceleration is only half as much, or g/2.

 

10. A feather and a coin will have equal accelerations when falling in a vacuum because

 

(1) their velocities are the same  

(2) the force of gravity is the same for each in a vacuum

(3) the force of gravity does not act in a vacuum 

(4) The ratio of each object’s weight to its mass is the same

(5) none of these

 

See pages 62 and 63 in your book!

 

11. A skydiver jumps from a plane.  As her velocity increases her acceleration

 

(1) increases   (2) decreases   (3) remains the same

 

This is the same concept as discussed in question 9.  As her velocity increases the net force on her decreases because of air resistance.  Since the net force decreases, her acceleration decreases.  Eventually, the air resistance exactly balances her weight, and as a result, the net force is zero.  At this point, she has no acceleration, and she falls at terminal velocity.

 

12. According to the special theory of relativity, all laws of nature are the same in reference frames that

 

(1) undergo constant acceleration   (2) move in constant circles   (3) move at constant velocity

(4) all of these   (5) none of these

 

Answer number (3) is one of the two biggies of special relativity.

 

13. A spaceship moving at a velocity of v = 0.8 c relative to an observer on earth fires a missile backwards at a velocity of 0.5 c relative to the spaceship.  What is the velocity of the missile as seen by the earth observer according to classical physics?  What is the velocity of the missile as seen by the earth observer according to special relativity?

 

I discussed in class the equation (page 703) for the addition of velocities (remember the sign error in the denominator).  Let v2 = 0.8 c be the velocity of the ship and v1 = –0.5 c be the velocity of the missile relative to the ship.  In classical physics, the earth observer would see

 

 V = - 0.5 c + 0.8 c = 0.3 c.

 

In relativity, the earth observer would see V = (0.3 c)/(1 – 0.4) = + 0.5 c.