Conservation of Momentum: The total vector momentum of an isolated system remains constant. pbefore = pafter

Conservation of Momentum Examples Answers 1. A cannonball of mass 5.0 kg is placed in a cannon of mass 425 kg. After firing, the cannonball is moving to the east at 85 m/s. Before 85 m/s east 5.0 kg After 425 kg mcannon = mc = 425 kg mball = mb = 5.0 kg vball = vb = + 85 m/s

Before vb = 85 m/s east mb = 5.0 kg After mc = 425 kg (a) What is the momentum of the cannon-cannonball system before firing? p = m v Before firing, neither the cannon nor the cannonball is moving. pbefore = 0

Before vb = 85 m/s east mb = 5.0 kg After mc = 425 kg (b) What is the momentum of the cannonball after firing? pb = mb vb = ( 5.0 kg )( + 85 m/s ) pb = + 425 kg m/s OR 425 kg m/s east

Before vb = 85 m/s east mb = 5.0 kg After pb = + 425 kg m/s mc = 425 kg (c) What is the momentum of the cannon after firing? pbefore = pafter pbefore = 0 part (a) pafter = pball + pcannon = pb + pc pb = + 425 kg m/s part (b)

Before vb = 85 m/s east mb = 5.0 kg After pb = + 425 kg m/s mc = 425 kg (c) What is the momentum of the cannon after firing? pbefore = pafter pafter = pb + pc 0 = + 425 kg m/s + pc - 425 kg m/s - 425 kg m/s Momentum of the cannon is the same as the momentum of the ball, but in the opposite direction pc = - 425 kg m/s OR 425 kg m/s west

Before vb = 85 m/s east vc = ? mb = 5.0 kg After pb = + 425 kg m/s pc = - 425 kg m/s mc = 425 kg (d) What is the velocity of the cannon after firing? pc = mc vc mc mc pc - 425 kg m/s vc = = = vc = - 1.0 m/s mc 425 kg OR 1.0 m/s west

2. A gun of mass 3. 0 kg fired a bullet of mass 10 2. A gun of mass 3.0 kg fired a bullet of mass 10.0 g with a velocity of 625 m/s. Determine the recoil velocity of the gun. mb = 10.0 g = 0.010 kg mg = 3.0 kg vb = + 625 m/s vg = ? pb = mb vb = ( 0.010 kg )( +625 m/s ) = +6.25 kg m/s pg = mg vg = ( 3.0 kg ) vg pbefore = pafter 0 = pb + pg 0 = +6.25 kg m/s + ( 3.0 kg ) vg -6.25 kg m/s -6.25 kg m/s -6.25 kg m/s = ( 3.0 kg ) vg 3.0 kg 3.0 kg vg = - 2.1 m/s

2. A gun of mass 3. 0 kg fired a bullet of mass 10 2. A gun of mass 3.0 kg fired a bullet of mass 10.0 g with a velocity of 625 m/s. Determine the recoil velocity of the gun. mb = 10.0 g = 0.010 kg mg = 3.0 kg vb = + 625 m/s vg = ? pbefore = pafter 0 = pb + pg pb = mb vb pg = mg vg 0 = mb vb + mg vg vg = ? - mb vb - mb vb - mb vb - mb vb = mg vg vg = mg mg mg - ( 0.010 kg )( + 625 m/s ) = 3.0 kg vg = - 2.1 m/s

3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a stationary pin of mass 4.4 kg. After the collision, the pin flies forward at 5.9 m/s. Find the velocity of the bowling ball after the collision. m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s pbefore = pafter pbefore = p1b + p2b p2b = 0 = p1b = m1 v1b = ( 7.0 kg )( + 4.8 m/s ) pbefore = + 33.6 kg m/s

3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a stationary pin of mass 4.4 kg. After the collision, the pin flies forward at 5.9 m/s. Find the velocity of the bowling ball after the collision. m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s pbefore = pafter = 33.6 kg m/s p2a = m2 v2a = ( 4.4 kg )( + 5.9 m/s ) pafter = p1a + p2a = + 25.96 kg m/s + 33.6 kg m/s = p1a + 25.96 kg m/s - 25.96 kg m/s - 25.96 kg m/s + 7.64 kg m/s = p1a

3. A bowling ball of mass 7.0 kg moving at 4.8 m/s collides with a stationary pin of mass 4.4 kg. After the collision, the pin flies forward at 5.9 m/s. Find the velocity of the bowling ball after the collision. m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s p1a = m1 v1a p1a = + 7.64 kg m/s m1 m1 p1a + 7.64 kg m/s v1a = = = v1a = + 1.1 m/s m1 7.0 kg

m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s pbefore = pafter p1b + p2b = p1a + p2a p2b = 0 m1 v1b = m1 v1a + m2 v2a

m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s pbefore = pafter p1b + p2b = p1a + p2a p2b = 0 m1 v1b = m1 v1a + m2 v2a - m2 v2a - m2 v2a m1 v1b - m2 v2a = m1 v1a m1 m1 m1 v1b - m2 v2a v1a = m1

m2 = 4.4 kg m1 = 7.0 kg Before v1b = + 4.8 m/s After v1a = ? v2a = + 5.9 m/s m1 v1b - m2 v2a v1a = m1 ( 7.0 kg )( 4.8 m/s ) – ( 4.4 kg )( 5.9 m/s ) = 7.0 kg v1a = + 1.1 m/s

4. A golf ball of mass 0.045 kg moving at 28.0 m/s collides with a stationary baseball of mass 0.150 kg. The baseball moves forward at 12.9 m/s. Find the velocity of the golf ball after the collision. m2 = 0.150 kg m1 = 0.045 kg Before v1b = + 28.0 m/s v2a = + 12.9 m/s After v1a = ? pbefore = p1b + 0 p2a = m2 v2a = + 1.935 kg m/s = + 1.26 kg m/s pbefore = pafter = p1a + p2a p1a = - 0.675 kg m/s p1a = m1 v1a v1a = - 15 m/s

pbefore = pafter p1b + p2b = p1a + p2a p2b = 0 m2 = 0.150 kg m1 = 0.045 kg v1b = + 28 m/s v2a = + 12.9 m/s v1a = ? pbefore = pafter p1b + p2b = p1a + p2a p2b = 0 m1 v1b = m1 v1a + m2 v2a - m2 v2a - m2 v2a m1 v1b - m2 v2a = m1 v1a m1 v1b - m2 v2a v1a = m1 m1 m1 (0.045 kg)(28 m/s) – (0.150 kg)(12.9 m/s) = v1a = - 15 m/s v1a = 0.045 kg

5. A train car of mass 2000 kg and moving at 2.5 m/s collides with a stationary car of mass 1200 kg. The cars couple together and move down the track with velocity v. Find v. Before 2000 kg 1200 kg v1b = + 2.5 m/s After va = ? pbefore = pafter M = 1200 kg + 2000 kg = 3200 kg p1b + p2b = pa m1 v1b = M va M M m1 v1b ( 2000 kg )( 2.5 m/s ) va = = = va = 1.6 m/s M 3200 kg

6. A running back has a mass of 105 kg and is moving in a straight line at 9.4 m/s. He is hit head-on by a linebacker that has a mass of 115 kg and is moving in the opposite direction at 6.4 m/s. After the inelastic collision, what is the velocity of the running back-linebacker system? v1b = + 9.4 m/s m1 = 105 kg m2 = 115 kg v2b = - 6.4 m/s M = 105 kg + 115 kg = 220 kg pbefore = pafter p1b + p2b = pa m1 v1b + m2 v2b = M va M M

v1b = + 9.4 m/s m1 = 105 kg m2 = 115 kg v2b = - 6.4 m/s M = 105 kg + 115 kg = 220 kg m1 v1b + m2 v2b va = M ( 105 kg )( + 9.4 m/s ) + ( 115 kg )( - 6.4 m/s ) = 220 kg va = + 1.1 m/s