Momentum Free Response Problems Slide 2 / 42 1. Block 1 with a - - PowerPoint PPT Presentation

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Momentum Free Response Problems

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• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• a. Determine the momentum of block 1 before the

collision.

• b. Determine the kinetic energy of block 1 before the

collision.

• c. Determine the momentum of the system of two

blocks after the collision.

• d. Determine the velocity of the system of two blocks

after the collision.

• e. Determine the kinetic energy of the system two

blocks after the collision. f. Determine the maximum compression in the spring after the collision.

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• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• a. Determine the momentum of block 1 before the

collision. p1 = m1v1 = (0.5 kg)(5 m/s) = 2.5 kg m/s

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• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• b. Determine the kinetic energy of block 1 before the

collision. KEo = ½m1v1

2 = ½(0.5 kg)(5 m/s)2 = 6.25 J

Slide 5 / 42

• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• c. Determine the momentum of the system of two

blocks after the collision.

p1 + p2 = p' p' = m1v1 + m2v2 p' = (0.5 kg)(5 m/s) = 2.5 kg m/s

Slide 6 / 42

• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• d. Determine the velocity of the system of two blocks

after the collision.

p1 = (m1 + m2)v' v' = p'/(m1 + m2) v' = (2.5 kg m/s)/(0.5 kg + 1.5kg) = 1.25 m/s

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• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m.

• e. Determine the kinetic energy of the system two

blocks after the collision.

KEF = ½Mv'2 KEF = ½(m1 + m2)(v')2 KEF = ½(0.5 kg + 1.5 kg)(1.25 m/s)2 = 1.56 J

Slide 8 / 42

• 1. Block 1 with a mass of 500 g moves at a constant

speed of 5 m/s on a horizontal frictionless track and collides and sticks to a stationary block 2 mass of 1.5

• kg. Block 2 is attached to an unstretched spring with a

spring constant 200 N/m. f. Determine the maximum compression in the spring after the collision.

Eo + W = Ef KEo = EPEF KEo = ½kx

2

x = (2KEo/k)1/2 x = [(2)(1.56 J)/(200 N/m)]

1/2 = 0.12 m

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

a. Calculate the momentum of the piece of clay before the collision. b. Calculate the kinetic energy of the piece of clay before the collision. c. What is the momentum of two objects after the collision? d. Calculate the velocity of the combination of two objects after the collision. e. Calculate the kinetic energy of the combination of two objects after the collision. f. Calculate the change in kinetic energy during the collision. g. Calculate the maximum vertical height of the combination of two objects after the collision.

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

a. Calculate the momentum of the piece of clay before the collision. p1 = m1v1 = (0.02 kg)(15 m/s) = 0.3 kg m/s

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

b. Calculate the kinetic energy of the piece of clay before the collision. KEo = ½(0.02 kg)(15 m/s)2 = 2.25 J

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

c. What is the momentum of two objects after the collision? p1 + p2 = p' p' = p1 + p2 p' = m1v1 + m2v2 p' = (0.02 kg)(15 m/s)2 = 0.3 kg m/s

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

d. Calculate the velocity of the combination of two

• bjects after the collision.

p' = Mv' p' = (m1 + m2)v' v' = p'/(m1 + m2) v' = (0.3 kg m/s)/(0.02 kg + 0.9 kg) = 0.33 m/s

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

e. Calculate the kinetic energy of the combination

• f two objects after the collision.

KEF = ½(m1 + m2)v' KEF = ½(0.02 kg + 0.9 kg)(0.33 m/s)2 = 0.05 J

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

f. Calculate the change in kinetic energy during the collision. #KE = KEF - KEo #KE = (0.05 J) - (2.25 J) = -2.2 J

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• 2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of

clay collides and sticks to a massive ball of mass 900 g suspended at the end

• f a string.

g. Calculate the maximum vertical height of the combination of two objects after the collision. Eo + W = Ef Eo = EF KE = Mgh h = KE/Mg h = KE/(m1 + m2)g h = (0.05 J)/(0.02 kg + 0.90 kg)(9.8 m/s2) = 0.006 m

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. a. Find the momentum of the bullet before the collision. b. Find the kinetic energy of the bullet before the collision. c. Find the velocity of the bullet-block system after the collision. d. Find the kinetic energy of the bullet-block after the collision. e. Find the change in kinetic energy during the collision. f. How much time it takes the bullet-block system to reach the floor? g. Find the maximum horizontal distance between the table and the striking point on the floor.

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. a. Find the momentum of the bullet before the collision. p1 = m1v1 = (0.01 kg)(500 m/s) = 5 kg m/s

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. b. Find the kinetic energy of the bullet before the collision. KEo = KE1 + KE2 KEo = ½m1v1

2

KEo = ½(0.01 kg)(500 m/s)2 KEo = 1,250 J

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. c. Find the velocity of the bullet-block system after the collision. p' = p1 + p2 (m1 + m2)v' = m1v1 v' = (m1v1)/(m1 + m2) v' = (0.01 kg)(500 m/s)/(0.01 kg + 1.6 kg) v' = 3.3 m/s

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. d. Find the kinetic energy of the bullet-block after the collision. KEF = ½(m1 + m2)v'2 KEF = ½(0.01 kg + 1.5 kg)(3.3 m/s)2 KEF = 8.2 J

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. e. Find the change in kinetic energy during the collision. ΔKE = KEF - KEo ΔKE = 8.2 J - 1250 J ΔKE = -1241.8 J

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. f. How much time it takes the bullet-block system to reach the floor? y = yo + vot + ½at2 y = ½at2 t = (2y/g)1/2 t = [(2)(0.7 m)/(9.8 m/s2)]1/2 t = 0.38 s

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• 3. A 10 g bullet moves at a constant speed
• f 500 m/s and collides with a 1.5 kg

wooden block initially at rest. The surface of the table is frictionless and 70 cm above the floor level. After the collision the bullet becomes embedded into the block. The bullet-block system slides off the top of the table and strikes the floor. g. Find the maximum horizontal distance between the table and the striking point on the floor. Δx = vxt Δx = (3.3 m/s)(0.38s) Δx = 1.25 m

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. a. What is the speed of block A just before it hits block B? b. What is the speed of the system of two blocks after the collision? c. What is the kinetic energy of the system of two blocks the collision? d. How much energy is lost due to the collision? e. What is the stopping distance X of the system of two blocks?

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. a. What is the speed of block A just before it hits block B? Eo + W = EF mgh = ½mv2 v = (2gh)1/2 v = (2gr)1/2

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. b. What is the speed of the system of two blocks after the collision? p = p' mv + 0 = v'(m + m) mv = 2mv' v' = v/2 since v = (2gr)1/2, v' = [(2gr)1/2]/2

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. c. What is the kinetic energy of the system of two blocks after the collision? KE' = ½(m + m)v'2 KE' = ½(2m)[(2gr)1/2/2]2 KE' = 2grm/4 KE' = grm/2

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. d. How much energy is lost due to the collision? ΔE = EF - Eo ΔE = mgr/2 - mgr ΔE = -mgr/2 (energy lost)

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• 4. Block A with a mass of m is released

from the top of the curved track of radius

• r. Block A slides down the track without

friction and collides inelastically with an identical block B initially at rest. After the collision the two blocks move distance X to the right on the rough horizontal part

• f the track with a coefficient of kinetic

friction µ. e. What is the stopping distance X of the system of two blocks? W = F#X ΔX = W/F ΔX = (mgr/2)/µmg ΔX = r/2µ

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram.

a. What is the x-component of the initial momentum of disc m1? b. What is the y-component of the initial momentum of disc m1? c. What is the x-component of the initial momentum of disc m2? d. What is the y-component of the initial momentum of disc m2? e. What is the x-component of the final momentum of disc m1? j. What is the final vector velocity of disc m1? h. What is the final vector velocity of m2? g. What is the y-component of the final momentum of disc m2? i. What is the y-component of the final momentum of disc m1? f. What is the x-component of the final momentum of disc m2?

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. a. What is the x-component of the initial momentum of disc m1? p = (2 kg)(8 m/s) p = 16 kg m/s

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. b. What is the y-component of the initial momentum of disc m1?

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. c. What is the x-component of the initial momentum of disc m2?

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. d. What is the y-component of the initial momentum of disc m2?

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. e. What is the x-component of the final momentum of disc m1?

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. f.

What is the x-component of the final momentum of disc m2?

30o p2' p2y' p2x' px + I = px' p1x + 0 = 0 + p2x' p2x' = p1x = 16 kg m/s

Slide 38 / 42

• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram. g.

What is the y-component of the final momentum of disc m2?

30o p2' p2y' 16 kg m/s tan 30o = p2y'/p2x' p2y' = p2x'(tan30o) p2y' = (16 kg m/s)(tan 30o) p2y' = 9.2 kg m/s

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram.

h. What is the final vector velocity of m2?

30o p2' 9.2 kg m/s 16 kg m/s p2' = (p2x'2 + p2y'2)1/2 p2' = [(16 kg m/s)2 + (9.2 kg m/s)2]1/2 p2' = 18.5 kg m/s p2' = m2v2' v2' = p2'/m2 v2' = (18.5 kg m/s)/(8 kg) v2' = 2.3 m/s

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram.

i. What is the y-component of the final momentum of disc m1?

py + I = py' 0 = p1y' + p2y' p1y' = -p2y' p1y' = -9.2 kg m/s

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• 5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal

frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x direction and disc m2 is initially at rest. The collision of two discs is perfectly elastic and the directions of two velocities presented by the diagram.

j. What is the final vector velocity of disc m1?

p1' = m1v1' v1' = p1'/m1 v1' = (-9.2 kg m/s)/(2 kg) v1' = -4.6 m/s

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