Physics 101 Exam 1
8 October 2004 Prof L. Weinstein


There are 19 problems. Please give a short explanation for all multiple choice questions. Show your work for all numerical answers.

  1. A net force of 18 N acts on a block for 5 s. During that time, the block has an acceleration of 3 m/s$^2$. What is the mass of the block?

    F = ma, therefore m = F/a = 18 N / 3 m/s^2 = 6 kg (since 1 N = 1 kg m/s^2)



  2. A 1000-kg car travelling at 38 m/s East is seen entering a tunnel. Five seconds later it drives out of the tunnel travelling 27 m/s West.

    a) What was the change in velocity of the car during that time?

    b) What was the acceleration of the car during that time (give magnitude and direction)?


    a) Let's take East as + and West as -.  Then the velocity went from +38 m/s to -27 m/s.
    delta v = v_final - v_initial = -27 m/s - (+38 m/s) = -65 m/s = 65 m/s West.

    b) a = delta v / t = 65 m/s West / 5 s = 13 m/s^2 West

    If you got part (a) wrong, I used that as the starting point for part (b).


  3. You are travelling in a golfcart down the road at a speed of 10 m/s. You want to throw your full water bottle to your friend, who is standing still. You throw the water bottle as you pass your friend. Neglect air resistance. To make sure that she can catch it, you should aim your throw:

  4. = 1.4in \epsffile{golfcart.eps}

    1. at point A (behind your friend)
    2. at point B (at your friend)
    3. at point C (ahead of your friend)
    4. need more information

    The bottle keeps its horizontal velocity.  Therefore you have to aim it behind your friend at point A.  (The layout of this question was confusing to many people so I made it extra credit.)




  5. You drive from Norfolk to Richmond. You travel 10 miles in the first hour, 60 miles in the second hour, and 50 miles in the third hour. What was your average speed?

    v_average = total distance / total time = (10+60+50 miles) / 3 hours = 40 miles/hour



  6. You are driving your car around a curve in a parking lot at about 15 mph. When you reach point A, you hit a large patch of oil and there is suddenly no friction between the tires and the road. Draw a line indicating your motion in the next second.

    \epsffile{curve.eps}figure shows answer to question

    Because there are no forces acting on your car, it goes in a straight line at constant speed.  I took off 5 points for a straight line in the wrong direction.

  7. A 4 kg pumpkin is dropped from the 10th floor of a building and hits the ground with a certain speed. Neglect air resistance. If you want the pumpkin to hit the ground with half the speed, you should drop it from a height

    1. 1/4 as high
    2. 1/3 as high
    3. 1/2 as high
    4. twice as high
    5. four times as high
    6. need more information

    Oops.  This is really an energy question.  I deleted it from the test.


  8. A 0.5 kg apple falls 20 m from a tree. How much time did it take to fall?

    d = (1/2)gt^2
    2d/g = t^2
    sqrt(2d/g) = t
    sqrt((2*20 m)/(10 m/s^2)) = t
    sqrt(4 s^2) = t
    t = 2 s.
    Note that you get the number 2 by the incorrect method where you divide the distance (20 m) by the acceleration (10 m/s^2) to get 2. This gives the wrong units (s^2).

  9. A 1000-kg car is driving down the highway in a straight line at a constant speed of 30 m/s (65 mph). The only horizontal forces on the car are the force due to the engine, the force of friction, and the force of air resistance. The engine of the car applies a force of 1500 N in the direction of travel. What is the total force of air resistance plus friction on the car?

    1. 0 N
    2. between 0 and 1500 N
    3. 1500 N
    4. more than 1500 N
    5. need more information

    The car is moving with constant speed in a straight line, therefore velocity is constant, acceleration is zero and the net force on the car is zero.
    0 = F_net = F_engine + F_air_and_friction
    Therefore F_air_and_friction must be equal and opposite to F_engine, or 1500 N.
    This is NOT a 3rd law problem.  We did this in class on a quiz question.

  10. I throw a 0.5 kg ball straight down from a 1000-m high balloon at a speed of 20 m/s. Two seconds later, what are (a) the velocity and (b) the acceleration of the ball?

    The height and mass of the ball are irrelevent.  The velocity is 20 m/s down.  After one second it has increased by 10 m/s (since g = 10 m/s^2) so it is 30 m/s.  After 2 s, it is 40 m/s.

    The acceleration is 10 m/s^2 (gravity is gravity).

    We did this in class on a quiz question.

  11. On Earth, a 1000-kg Honda Civic can go from 0 to 60 mph in 8 sec. Suppose that the Civic is moved to a planet where the gravitational field is twice as strong (so that $g_{planet} = 20$ m/s$^2$). Assume that the engine can exert the same force on this planet as on Earth. How much time will it take to accelerate from 0 to 60 mph?

    1. 2 sec
    2. 4 sec
    3. 8 sec
    4. 16 sec
    5. 32 sec
    6. Not enough information

    The force is unchanged.  The mass is unchanged.  Therefore the acceleration is unchanged.  Therefore the time to accelerate is unchanged.
    Gravity exerts a vertical force on the car.  This does not affect the horizontal forces and accelerations.  We did this in class and on the homework.

  12. A 1000-kg Civic going 45 miles per hour on dry roads applies the brakes and stops in a period of time $t$. If instead the road is wet so that the brakes only exert half as much force, how much time would it take to stop?
    1. one-fourth as much
    2. half as much
    3. the same
    4. twice as much
    5. four times as much
    6. Not enough information

    In both cases, it has to change speed from 45 mph to zero.  When the roads are dry, it has a_dry = F/m.  When the roads are wet, a_wet = F_wet/m = (F/2)/m = (1/2)a_dry.  Therefore the acceleration is half as large.  Since a = v/t, we know that t = v/a.  If a is half as large, we will need twice as much time to change the velocity by the same amount.
    Alternatively, we have to apply an impulse Ft to change the momentum from (1000 kg)(45 mph) to 0.  If F is half as large, t must double to compensate. 
    I was looking for arguments as to a) why t needed to be larger and b) why it needed to be two times larger, not four times.

  13. (No explanation needed) You turn a fan and it blows air at you. If the fan pushing on the air is the action force, what is the reaction force?

    1. gravity pulling on the air
    2. the air pushing on you
    3. the air pushing on the fan
    4. gravity pulling on the fan
    5. gravity pulling on you
    6. Other
    7. Not enough information

  14. A small 50-kg (110 lb) hockey fan climbs onto the ice and is hit by a large 100-kg (220 lb) ice hockey player travelling at 12 m/s (25 mph) to the right. Assume that the hockey fan was not moving before he was hit and that the fan and the player are all tangled together after they hit.

    Which exerts more force on the other during the collision?

    1. The big hockey player exerts more force on the small hockey fan
    2. The small hockey fan exerts more force on the big hockey player
    3. Both exert the same force on each other
    4. Not enough information

    Newton's third law. (That's all the explanation I needed.)  The fan and the player interact.  Therefore the force exerted on the player equals the force exerted on the fan.  This is just like in class where we looked at lots of different collisions and interactions.
    Another acceptable explanation is: 'Same force, different results'



  15. In the previous problem, what is the speed of the two tangled people immediately after the collision ? Ignore friction.

    1. more than 12 m/s
    2. 12 m/s
    3. less than 12 m/s but not zero
    4. zero
    5. Not enough information

    Conservation of momentum.  Before we have 100 kg moving at 12 m/s (and 50 kg not moving), now we have more mass so the speed must decrease.

  16. In the previous problem, which person has the greater change in velocity?

    1. The big player
    2. The small fan
    3. Both the same
    4. Not enough information

    Same force, different mass.  a = F/m.  The person with the smaller mass will have the larger acceleration.  You could also calculate it exactly.  The speed before was 12 m/s and afterwards it was 8 m/s (in the same direction).  Therefore the small fan had the larger change in  velocity.


  17. In the previous problem, which person has the greater change in momentum?

    1. The big player
    2. The small fan
    3. Both the same
    4. Not enough information

    Same force of collision, same time of collision, therefore they each had the same impulse and thus the same momentum change.  Another way to look at it is that momentum was conserved.  Therefore, if the momentum of one person increased, the other had to decrease by the exact same amount to compensate.  We did this entire collision problem in class (with two cars colliding) and also on the homework (also with two cars colliding).


  18. A 0.15 kg baseball travelling at 30 m/s toward the batter is hit by a baseball bat and then has a velocity of 25 m/s away from the batter. What was the force applied to the baseball by the baseball bat (in Newtons)?

    1. 0.75
    2. 1.5
    3. 5
    4. 8.25
    5. 55
    6. other
    7. Not enough information

    To get the force, we need to know the time.  Ft = delta(p).  We know the change in momentum; we don't know the time.
    Alternatively, we can use F = ma = m delta(v)/t.  We know m and the change in velocity; we don't know the time.  We did a problem just like this in class (with the rocket sled).



  19. You throw a 1-kg rock straight upward at a speed of 20 m/s. What is the net force on the rock when it is half way to the top of its path? Ignore air resistance.

    gravity is gravity.  The only force on the rock is gravity.  F = mg = 1 kg * 10 m/s^2 = 10 N.



  20. Consider a person standing in an elevator that is accelerating upward. The upward normal force $N$ exerted by the elevator floor on the person is

    1. more than the person's weight
    2. equal to the person's weight
    3. less than the person's weight
    4. Not enough information

    If the elevator was standing still (or moving at constant velocity), then you would not be accelerating.  Thus the total force on you would be zero so that the normal force (from the elevator) would equal the force exerted on you by gravity (your weight).
    However, the elevator is accelerating upward.  To make this happen, the total force on you must be upwards so that F_normal must be greater than F_gravity.  This was a homework problem (about jumping).
    So many of you misunderstood this problem that I made it extra credit.