10/14 Energy Practice  Text: Chapter 6 Energy  Lab: “Ballistic Pendulum”  Energy example (Like “Ramp Launch”)  No HW assigned  Exam 2 Thursday, 10/ Wit Wit 114 (only if needed) Please send if other time needed Also let me know if you are a “6-8” person

Your Lab “Ballistic Pendulum” v v v = 0 h Use conservation of momentum for the collision Use conservation of energy for the swing You will find the initial velocity of the ball by measuring the height. (and some math) You will also find the initial velocity of the ball by measuring its velocity with projectile motion ideas.

Your Lab “Ballistic Pendulum” Here is a quick primer on momentum. When two blocks collide, they exert equal and opposite forces on each other. (Third Law) Since the contact time is the same for both we can say that: F A,B  t = -F B,A  t Where the minus sign means direction. We call F A,B  t the “change in momentum” and the changes in momentum for colliding objects are equal and opposite. If we consider the system of both objects, then these third law pairs are “internal forces” and the momentum of the system as a whole does not change. It is conserved.

Your Lab “Ballistic Pendulum” Note that F A,B  t can be seen hiding in the 2nd law, F net = ma. F net = ma = m  v/  t. So F net  t = m  v and for the system, m  v is 0 and mv is constant. (conserved) The symbol for momentum is p and it is a vector. Changes in p are handled just like changes in v.  p = m  v and p = mv Momentum of the system is the vector sum of the individual momenta.

Bullet sticks in block A A v B,i = 100 m/s m B = 0.05 kg m A = 2 kg v A+B,f = ? A = (2.05)v A+B,f v A+B,f = 2.4 m/s  p = m(  v) = F  t p A+B,i = p A+B,f m B v B,i +m A v A,i = (m B + m A )v A+B,f Is energy conserved?No. Mechanical energy was lost in friction between the bullet and block. (All directions to the right) For system, F net = 0

Block up a ramp with friction: A block is held against a compressed spring then released. Find: The velocity at B The height at C The velocity at D Lost energy = -f R,B  x =  k N R,B  x N R,B = W Y = 4 / 5 W E.B = 4 / 5 mg = 4N PE g,B = mgh B = 15J PE and h = 0 PE s,A - lost energy = KE B + PE g,B v B = 8.9m/s B A D C KEPE g PE s KEPE g PE s KEPE g PE s 0.5m 5.0m 3.0m 4.0m PE s,A = 1 / 2 k  x 2 = 40J = 0.25(4)(5) = 5J 40515J ? KE B = 20J = 1 / 2 mv 2 m = 0.5kg k = 320N/m  k = 0.25 g = 10N/kg W E,B WYWY N R,B WXWX f R,B KEPE g PE s 20J ?

m = 0.5kg k = 320N/m  k = 0.25 g = 10N/kg Block up a ramp with friction: A block is held against a compressed spring then released. Find: The velocity at B The height at C The velocity at D PE C = mgh C = 22.2J PE and h = 0 KE B + PE g,B = KE C + PE g,C v C = v X,B = 4 / 5 v B = 7.16m/s Energy conserved KE C = 1 / 2 mv 2 = 12.8J h C = 4.44m KEPE g PE s KEPE g PE s KEPE g PE s KEPE g PE s 0.5m 40J 20J15J v B = 8.9m/s 20J15J12.8J? v C = ? v C = v X,B Projectile motion! B A D C 5.0m 3.0m 4.0m

Block up a ramp with friction: A block is held against a compressed spring then released. Find: The velocity at B The height at C The velocity at D KE D = 1 / 2 mv 2 = 35J PE and h = 0 KE B + PE g,B = KE D v D = 11.8m/s KEPE g PE s KEPE g PE s KEPE g PE s KEPE g PE s 0.5m 20J15J35J v B = 8.9m/s v X,D = 7.16m/s Also note: The angle  = Arccos = 53  40J 20J15J12.8J22.2J 35J B A D C m = 0.5kg k = 320N/m  k = 0.25 g = 10N/kg  5.0m 3.0m 4.0m