Physics 225 – Dr. Shapiro

Midterm Exam #2 (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 a one hour exam.

 

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

 

 

1) A 10 kg crate is on an incline plane that makes an angle of 20 degrees with the horizontal.  When the block is set in motion, it is observed to slide downhill at constant speed.  When the block reaches the bottom of the incline plane, it is projected back up the plane with an initial velocity of 1.2 m/s.  (a) What is the coefficient of kinetic friction between the block and the incline plane?  (b) When the block is projected back up the plane, how far along the plane does it travel before stopping?

 

 

 

 

 

 

 

 

 


(a) Since the crate is not accelerating, the downhill gravitational force must be balanced by the uphill frictional force.  So , which implies that

(b) When the block is projected uphill, it starts with a kinetic energy of .  The block will stop when all the kinetic energy has been converted into gravitational potential energy and thermal energy.  It is convenient to think in terms of the distance moved along the incline, so .  Note that represents the vertical rise of the block, when the block slides a distance d along the incline plane.  Solving for d we get

 

2) A steel ball with a mass of 0.25 kg is attached to a 0.25 m cord with negligible mass.  The steel ball is being rotated in a vertical circle (see the diagram) at a constant tangential speed of 1.6 m/s  (a) What is the tension in the cord when the ball is at the top of the circle?  (b) What is the tension in the cord when the ball is at the bottom of the circle?

 

 

 

 

 

 

 

 


(a) At the top of the circle there are two downward forces acting on the ball, T and mg.  These together provide the necessary centripetal force to keep the ball moving in a circle, so  or

(b) At the bottom of the circle T and mg are in opposite directions, so , or

 

3) A 10 kg block is placed gently on a vertical spring.  The spring is an ideal Hooke’s Law spring.  It compresses 12.5 cm.  The block then is pushed down an additional 25 cm and released.  (a) What is the spring constant?  (b) How much elastic energy is stored in the spring when it is compressed the full 37.5 cm?  (c) What is the maximum height that the block reaches above the release point?

 

(a) At a compression of 12.5cm the spring force is just balancing the weight of the block, so  and

(b)

(c) When the block is released it will rise until all the elastic energy stored in the spring has been converted into gravitational potential energy of the block, so , or

 

 

4) A 10 kg body is traveling at 2.0 m/s with no external force acting on it.  At a certain instant an internal explosion occurs splitting the body into two equal chunks (5 kg each).  The explosion gives the chunks an additional 15 J of kinetic energy.  Neither chunk leaves the line of original motion.  (a) Determine the momentum of each of the chunks after the explosion.  (b) Determine the kinetic energy of each of the chunks after the explosion.

 

Initially the momentum is 20kg·m/s and the kinetic energy is 20J.  After the explosion the total momentum is unchanged (all forces are internal), and the total energy is now 35J, so  and .  These equations can be solved for v1 and v2 (requires use of quadratic formula).  It turns out that there are two solutions, but only one satisfies the energy equation.  The results are v1 = 0.268m/s and v2 = 3.732m/s.

 

A second approach is to work in a coordinate system that moves with the center of mass.  Since before the collision the system consists of only a single mass, it’s easy to see that the velocity of the center of mass is just 2.0m/s.  In this coordinate system the initial momentum is zero, and the initial kinetic energy is also zero.  After the explosion, the total momentum in this coordinate system must still be zero, but the total kinetic energy is now 15J.  This yields the following two equations:  and .  Here v1’ and v2’ refer to velocities in the center of mass reference frame.  From the first of these equations v1’= - v2’, then from the second equation v1’= 1.7321m/s and v2’=-1.7321m/s.  Then in the “lab” system v1= v1’+2.0m/s = 3.732m/s and v2= v2’+2.0m/s =0.268m/s as before.

 

(a) The momenta then are mv1 = 5kg·0.268m/s = 1.34kg·m/s and mv2 = 5kg·3.732m/s = 18.66kg·m/s.

 

(b) KE = 0.5mv2 = p2/2m.  So KE1 = 0.1796J and KE2 = 34.8196J.