class 051026

1.  A ball of mass .1 kg is released from rest and rolls down a ramp of length 30 cm, with one end of the ramp 10 cm higher than the other.  It rolls onto a horizontal ramp, then rolls off the end of that ramp and falls to the floor.  From the distance of fall and the horizontal range we find that the ball was moving at 60 cm / sec when it left the ramp.

• How much PE did it lose between release and leaving the horizontal ramp?

• The ball will be the system.  `dPE is work done by system against gravity.  System exerts upward grav. force of .1 kg * 9.8 m/s^2 = .98 N against gravity (grav force is downward so force against grav. is upward).  Displacement is 10 cm downward.  Using upward as positive we get

`dPE = .98 N * (-.10 m) = -.098 Joule.

• What was its KE as it left the horizontal ramp?
• At 60 cm/s the ball's KE is 1/2 m v^2 = 1/2 ( .1 kg) ( .6 m/s)^2 = .018 Joules.
• How much work was done on the ball by nonconservative forces between release and leaving the horizontal ramp?
• `dW_noncons_ON = `dPE + `dKE = -.098 J + .018 J = -.080 J.
• This is consistent with our knowledge that the frictional force exerted ON the system is in the direction opposite motion, which would result in negative work being done.  If friction is our only nonconservative force, the we conclude that friction does -.080 J of work on the system.
• How much work was by the ball against nonconservative forces between release and leaving the horizontal ramp?
• `dPE + `dKE + `dW_noncons_BY = 0 so

  -.098 J + .018 J + `dW_noncons_BY = 0 and

  -.080 J + `dW_noncons_BY = 0.  We conclude that

  `dW_noncons_BY = +.080 J.  The system does +.080 J of work against nonconservative forces.

How does all of this fit into the two statement of energy conservation

• `dW_noncons_ON = `dPE + `dKE and
• `dW_noncons_BY + `dPE + `dKE = 0

2.  A net force of 10 N acts on a 5 kg ball for 3 seconds. 

• What is the impulse of this force?
• The impulse of a force is F `dt.  So the impulse of the net force is F_net `dt = 10 N * 3 seconds = 30 N s = 30 kg m/s^2 * s = 30 kg m/s.
• By how much does the momentum of the ball change as a result?
• Momentum is m v, and by the impulse-momentum theorem we know that the impulse is equal to the change in momentum.  So change in momentum is 30 kg m / s.
• If the initial momentum of the ball is 40 kg m/s then what will be its final momentum?
• The final momentum is initial momentum + change in momentum = 40 kg m/s + 30 kg m/s = 70 kg m/s.
• What therefore will be its final velocity?
• Momentum is the product p = m v, so we have

  m v = 70 kg m/s with m = 5 kg.  We get

  v = 70 kg m/s / (5 kg) = 14 m/s.

Using these results:

• By how much will its KE have changed?
• We know the final velocity to be 14 m/s, so we can calculate final KE.  We need the initial velocity to get the initial KE.

   

  The initial momentum was 40 kg m/s, implying an initial velocity v0 = 40 kg m/s / ( 5 kg ) = 8 m/s.

  Thus initial and final KE are

KE0 = 1/2 * 5 kg * (8 m/s)^2 = 160 Joules

KEf = 1/2 * 5 kg * (14 m/s)^2 = 490 Joules.

Change in KE is therefore 490 J - 160 J = 330 J.

• What is the average rate of change of the ball's KE with respect to clock time during this 3 second interval?
• The ave rate of change of KE with respect to clock time is (change in KE) / (change in clock time) = +330 J / (3 sec) = 110 J / s.
• This is the average rate at which work is done by the net force.
• The rate at which work is done with respect to clock time is power, and its unit is the J / s, more commonly called the watt.
• So the average rate at which work is being done is 110 watts.
• Do you think the rate of change of KE with respect to clock time is constant during this interval?

Preliminary Experiment:

Roll a ball down an incline and onto a horizontal incline, from which it will collide immediately after leaving that incline with a small marble supported by a section of straw.  The plane normal to the contact surfaces between the two marbles should be vertical and perpendicular to the direction of the first ball's motion. 

• See how far each ball travels after collision before reaching the floor, and use this information to determine the velocity of each ball after the collision.
• Let the first ball roll down the inclines and off the edge without hitting the second ball.  Use your observation of its horizontal range to determine how fast this ball was traveling before the collision.
• Determine `dv_2, the change in velocity of the second ball in the collision.
• Determine `dv_1, the change in velocity of the first ball in the collision.
• Using the fact that m2 `dv2 = -m1 `dv1, find the ratio m2 / m1 of the mass of the second ball to the mass of the first ball.