Conservation of Momentum Problems Answers 1. A cannonball of mass 4.0 kg is placed in a cannon of mass 650 kg. After firing, the cannonball is moving to the east at 76 m/s. Before 76 m/s east 4.0 kg After 650 kg mcannon = mc = 650 kg mball = mb = 4.0 kg vball = vb = + 76 m/s

Before vb = 76 m/s east mb = 4.0 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 = 76 m/s east mb = 4.0 kg After mc = 650 kg (b) What is the momentum of the cannonball after firing? pb = mb vb = ( 4.0 kg )( + 76 m/s ) pb = + 304 kg m/s OR 304 kg m/s east

Before vb = 76 m/s east mb = 4.0 kg After pb = + 304 kg m/s mc = 650 kg (c) What is the momentum of the cannon after firing? pbefore = pafter pbefore = 0 part (a) pafter = pball + pcannon = pb + pc pb = + 304 kg m/s part (b)

Before vb = 76 m/s east mb = 4.0 kg After pb = + 304 kg m/s mc = 425 kg (c) What is the momentum of the cannon after firing? pbefore = pafter pafter = pb + pc 0 = + 304 kg m/s + pc - 304 kg m/s - 304 kg m/s Momentum of the cannon is the same as the momentum of the ball, but in the opposite direction pc = - 304 kg m/s OR 304 kg m/s west

Before vb = 76 m/s east vc = ? mb = 4.0 kg After pb = + 304 kg m/s pc = - 304 kg m/s mc = 650 kg (d) What is the velocity of the cannon after firing? pc = mc vc mc mc pc - 304 kg m/s vc = = = vc = - 0.47 m/s mc 650 kg OR 0.47 m/s west

2. A cannonball of mass 4.0 kg is fired from a cannon of mass 2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s to the north. Calculate the velocity of the cannon after firing. mb = 4.0 kg mc = 2.0 x 103 kg vb = + 125 m/s (north) pb = mb vb = ( 4.0 kg )( + 125 m/s ) = + 500 kg m/s (north) Momentum of cannon is same as that of the ball, but in the opposite direction So momentum of cannon = - 500 kg m/s (south) pc = mc vc mc mc pc - 500 kg m/s vc = = = vc = - 0.25 m/s mc 2000 kg OR 0.25 m/s south

2. A cannonball of mass 4.0 kg is fired from a cannon of mass 2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s to the north. Calculate the velocity of the cannon after firing. mb = 4.0 kg mc = 2.0 x 103 kg vb = + 125 m/s (north) pbefore = pafter pbefore = 0 pafter = pb + pc 0 = pb + pc pb = mb vb 0 = + 500 kg m/s + pc = ( 4.0 kg )( + 125 m/s ) - 500 kg m/s - 500 kg m/s = + 500 kg m/s pc = - 500 kg m/s pc = mc vc mc mc pc - 500 kg m/s vc = = = vc = - 0.25 m/s mc 2000 kg OR 0.25 m/s south

2. A cannonball of mass 4.0 kg is fired from a cannon of mass 2.0 x 103 kg. After firing, the velocity of the cannonball is 125 m/s to the north. Calculate the velocity of the cannon after firing. mb = 4.0 kg mc = 2.0 x 103 kg vb = + 125 m/s (north) pbefore = pafter 0 = pb + pc pb = mb vb pc = mc vc 0 = mb vb + mc vc vc = ? - mb vb - mb vb - mb vb - mb vb = mc vc vc = mc mc mc - ( 4.0 kg )( +125 m/s ) = 2.0 x 103 kg vc = - 0.25 m/s

3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a stationary brass ball of mass 2.0 kg. After the collision, the brass ball moves forward at 10.0 m/s. Find the velocity of the steel ball after the collision. v1b = + 7.5 m/s m2 = 2.0 kg Before m1 = 4.0 kg After v1a = ? v2a = + 10.0 m/s pbefore = pafter pbefore = p1b + p2b p2b = 0 pbefore = p1b = m1 v1b = ( 4.0 kg )( + 7.5 m/s ) pbefore = + 30 kg m/s

3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a stationary brass ball of mass 2.0 kg. After the collision, the brass ball moves forward at 10.0 m/s. Find the velocity of the steel ball after the collision. v1b = + 7.5 m/s m2 = 2.0 kg Before m1 = 4.0 kg After v1a = ? v2a = + 10.0 m/s pbefore = pafter = + 30 kg m/s pafter = p1a + p2a = m1 v1a + m2 v2a = ( 4.0 kg ) v1a = ( 2.0 kg )( + 10.0 m/s ) = + 20 kg m/s

3. A steel ball of mass 4.0 kg and moving at 7.5 m/s collides with a stationary brass ball of mass 2.0 kg. After the collision, the brass ball moves forward at 10.0 m/s. Find the velocity of the steel ball after the collision. v1b = + 7.5 m/s m2 = 2.0 kg Before m1 = 4.0 kg After v1a = ? v2a = + 10.0 m/s pbefore = pafter = + 30 kg m/s pafter = p1a + p2a = m1 v1a + m2 v2a + 30 kg m/s = ( 4.0 kg ) v1a + 20 kg m/s - 20 kg m/s - 20 kg m/s + 10 kg m/s = ( 4.0 kg ) v1a 4.0 kg 4.0 kg v1a = + 2.5 m/s

steel ball of mass 0.150 kg. After the collision, the billiard ball 4. A billiard ball of mass 0.225 kg collides elastically with a stationary steel ball of mass 0.150 kg. After the collision, the billiard ball continues forward with a velocity of 1.5 m/s, while the steel ball is propelled forward at 9.0 m/s. Find the speed of the billiard ball before the collision. m1 = 0.225 kg m2 = 0.150 kg Before v1b = ? v2a = + 9.0 m/s After v1a = + 1.5 m/s pbefore = pafter p1b + p2b = p1a + p2a p2b = 0 m1 v1b = m1 v1a + m2 v2a m1 m1

m1 v1a + m2 v2a v1b = m1 (0.225 kg)(1.5 m/s) + (0.150 kg)(9.0 m/s) = v2a = + 9.0 m/s v1a = + 1.5 m/s m1 v1a + m2 v2a v1b = m1 (0.225 kg)(1.5 m/s) + (0.150 kg)(9.0 m/s) = 0.225 kg v1b = + 7.5 m/s

8. A train car of mass 1800 kg is carrying a metric ton (1000 kg) of coal. It moves down the track at 3.6 m/s and collides inelastically with a stationary car of mass 3500 kg. Find the velocity of the cars after the collision. Before 2800 kg 3500 kg v1b = + 3.6 m/s va = ? After pbefore = pafter M = 2800 kg + 3500 kg = 6300 kg p1b + p2b = pa m1 v1b = M va M M m1 v1b ( 2800 kg )( 3.6 m/s ) va = = = va = 1.6 m/s M 6300 kg

6. A block of wood of mass 1.5 kg is hanging from a tree by a string. A bullet of mass 15.0 g is fired at 180 m/s into the block. Find the speed at which the bullet/block system moves forward after the bullet is embedded into the block. m1 = 15.0 g va = ? Before After 1.5 kg v1b = + 180 m/s m1 = 15.0 g = 0.0150 kg M = 1.5 kg + 0.015 kg = 1.515 kg pbefore = pafter p1b + p2b = pa m1 v1b = M va M M m1 v1b ( 0.015 kg )( 180 m/s ) va = = = va = 1.8 m/s M 1.515 kg