Slide 1 / 42 Momentum Free Response Problems
Slide 2 / 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. 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.
Slide 3 / 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. a. Determine the momentum of block 1 before the collision. p 1 = m 1 v 1 = (0.5 kg)(5 m/s) = 2.5 kg m/s
Slide 4 / 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. b. Determine the kinetic energy of block 1 before the collision. 2 = ½(0.5 kg)(5 m/s) 2 KE o = ½m 1 v 1 = 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. p 1 + p 2 = p' p' = m 1 v 1 + m 2 v 2 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. p 1 = (m 1 + m 2 )v' v' = p'/(m 1 + m 2 ) v' = (2.5 kg m/s)/(0.5 kg + 1.5kg) = 1.25 m/s
Slide 7 / 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. e. Determine the kinetic energy of the system two blocks after the collision. KE F = ½Mv' 2 KE F = ½(m1 + m2)(v') 2 KE F = 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. E o + W = E f KE o = EPE F 2 KE o = ½kx x = (2KE o /k) 1/2 1/2 = 0.12 m x = [(2)(1.56 J)/(200 N/m)]
Slide 9 / 42 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 of 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.
Slide 10 / 42 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 of a string. a. Calculate the momentum of the piece of clay before the collision. p 1 = m 1 v 1 = (0.02 kg)(15 m/s) = 0.3 kg m/s
Slide 11 / 42 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 of a string. b. Calculate the kinetic energy of the piece of clay before the collision. KE o = ½(0.02 kg)(15 m/s) 2 = 2.25 J
Slide 12 / 42 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 of a string. c. What is the momentum of two objects after the collision? p 1 + p 2 = p' p' = p 1 + p 2 p' = m 1 v 1 + m 2 v 2 p' = (0.02 kg)(15 m/s) 2 = 0.3 kg m/s
Slide 13 / 42 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 of a string. d. Calculate the velocity of the combination of two objects after the collision. p' = Mv' p' = (m 1 + m 2 )v' v' = p'/(m 1 + m 2 ) v' = (0.3 kg m/s)/(0.02 kg + 0.9 kg) = 0.33 m/s
Slide 14 / 42 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 of a string. e. Calculate the kinetic energy of the combination of two objects after the collision. KE F = ½(m 1 + m 2 )v' KE F = ½(0.02 kg + 0.9 kg)(0.33 m/s) 2 = 0.05 J
Slide 15 / 42 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 of a string. f. Calculate the change in kinetic energy during the collision. # KE = KE F - KE o # KE = (0.05 J) - (2.25 J) = -2.2 J
Slide 16 / 42 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 of a string. g. Calculate the maximum vertical height of the combination of two objects after the collision. E o + W = E f E o = E F KE = Mgh h = KE/Mg h = KE/(m 1 + m 2 )g h = (0.05 J)/(0.02 kg + 0.90 kg)(9.8 m/s 2 ) = 0.006 m
Slide 17 / 42 3. A 10 g bullet moves at a constant speed of 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.
Slide 18 / 42 3. A 10 g bullet moves at a constant speed of 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. p 1 = m 1 v 1 = (0.01 kg)(500 m/s) = 5 kg m/s
Slide 19 / 42 3. A 10 g bullet moves at a constant speed of 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. KE o = KE 1 + KE 2 2 KE o = ½m 1 v 1 KE o = ½(0.01 kg)(500 m/s) 2 KE o = 1,250 J
Slide 20 / 42 3. A 10 g bullet moves at a constant speed of 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' = p 1 + p 2 (m 1 + m 2 )v' = m 1 v 1 v' = (m 1 v 1 )/(m 1 + m 2 ) v' = (0.01 kg)(500 m/s)/(0.01 kg + 1.6 kg) v' = 3.3 m/s
Slide 21 / 42 3. A 10 g bullet moves at a constant speed of 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. KE F = ½(m 1 + m 2 )v' 2 KE F = ½(0.01 kg + 1.5 kg)(3.3 m/s) 2 KE F = 8.2 J
Slide 22 / 42 3. A 10 g bullet moves at a constant speed of 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 = KE F - KE o ΔKE = 8.2 J - 1250 J ΔKE = -1241.8 J
Recommend
More recommend
