Physics 225 – Dr. Shapiro

Midterm Exam #1 (Section 2)

 

Closed book.  One 3”´5” note card, the handout and a calculator may be used.  Please do not write on the test or the handout.  Both must be turned in with your test answers.  Note: This is one hour exam.

 

Partial credit will be given.  Please show all your work.

 

 

1) A person is driving a car at 50 km/hr on a straight road, when she notices a stop sign that is exactly 50 m in front of her.  It takes her 0.3 s to apply the brakes.  She applies the brakes carefully so that her deceleration is constant, and she comes to a stop just as she gets to the stop sign.  (a) How far did she travel during the 0.3 s it took her to apply the brakes?  (b) What was her deceleration?  (c) How long did it take her to stop after she applied the brakes?

 

a) In 0.3 s the car travels a distance = vt.  Since , this distance is 4.167m.

b) Using the equations of motion for constant acceleration in one dimension, we have

and

where t’ is the stopping time, and a is the acceleration (actually a deceleration in this case).  Solving the first equation for t’, and substituting the result in the second equation, we have .  Solving for a, we find a = 2.10 m/s2.

c) Substituting this in the first equation of motion, we find t’ = 6.61s.

 

 

 

2) A and B are two position vectors locating points P1 and P2, respectively.

 

     A = (5i – 7j + 2k)m

 

     B = (4i + 3j + 3k)m

 

(a) What is the displacement between points P1 and P2?  (b) What is the shortest distance between the two points?  (c) What is the angle between vector A and the x-axis?

 

a) The displacement vector is C = B-A = (-1i + 10j +1k)m.

b) The shortest distance between the two points is the length of the displacement vector |C| =((-1)2 + 102 + 12)1/2m = 10.1m.

c) There is more than one way to find the angle.  One approach is to note that A·i = |A|cos(theta), so , which implies that q = 55.5 deg. Another way to solve for the angle q is to note that the vector A consists of a component parallel to the x-axis, 5i, and a component perpendicular to the x-axis, -7j + 2k.  The tan(theta) is just the length of the perpendicular component divided by the parallel component.  This gives the same result.

 

3) A ball rolls horizontally off the top of a stairway with a speed of 1.52 m/s.  The steps are 20.3 cm high and 20.3 cm wide.  Which step does the ball hit first?  (See the diagram below.)

 

 

 

 

 

 

 

 

 

 





Probably the easiest way to solve this problem is to compute the time required for the ball to drop 20.3 cm, and then compare how far the ball moves horizontally in that time.  Using we find it takes 0.2035s.  In this time the ball has moved horizontally a distance of 0.3094m, so it already is over the second step.  The time it takes to fall to the level of the second step is 0.2878s using the same equation of motion in the vertical direction.  But in that time, the ball has moved 43.75 cm in the horizontal direction, so it’s over the third step.  It reaches the level of the third step in 0.3525s.  The horizontal distance traveled in that time is 53.58cm, so it hits on the third step.

 

4) In the diagram shown below the incline plane is frictionless.  The incline plane makes an angle of 30 degrees with the horizontal.  The two blocks are being pushed up the plane at constant velocity.  (a) What is the uphill force on the 5 kg block?  (b) What is the uphill force on the 10 kg block?  (c) If the uphill force is suddenly removed, what is the downhill force on the 10 kg block?

 

 

 

 

 

 

 

 

 

 


(a) The net force on the 5 kg block must be zero because there is no acceleration.  The downhill force on the 5kg block is 5kg·9.8m/s2·sin(30 deg) = 24.5N, so the uphill force must be equal to this.

(b) The uphill force on the 10kg block must balance the downhill force on it.  The downhill force from the block’s weight is 10kg·9.8m/s2·sin(30 deg)= 49N, but there also is a force from the 5 kg block on the 10 kg block (the reaction force to the uphill force on the 5 kg block) of 24.5N, so the net downhill force is 73.5N.  The uphill force must be equal to this in magnitude.

(c) If the uphill force is suddenly removed the downhill force on the 10kg block is just the downhill component of its weight = 49N, since both blocks have the same acceleration.