CHAPTER 26 LINEAR MOMENTUM AND IMPULSE

Transcription

1 CHAPTER 26 LINEAR MOMENTUM AND IMPULSE EXERCISE 118, Page Determine the momentum in a mass of 50 kg having a velocity of 5 m/s. Momentum = mass velocity = 50 kg 5 m/s = 250 kg m/s downwards 2. A milling machine and its component have a combined mass of 400 kg. Determine the momentum of the table and component when the feed rate is 360 mm/min. Momentum = mass velocity = 400 kg m/s = 2.4 kg m/s downwards 3. The momentum of a body is 160 kg m/s when the velocity is 2.5 m/s. Determine the mass of the body. Momentum = mass velocity Hence, 160 = mass 2.5 from which, mass = 160 = 64 kg Calculate the momentum of a car of mass 750 kg moving at a constant velocity of 108 km/h. Momentum = mass velocity Mass = 750 kg and velocity = 108 km/h = m/s = 30 m/s. Hence, momentum = 750 kg 30 m/s = 22,500 kg m/s 5. A football of mass 200 g has a momentum of 5 kg m/s. What is the velocity of the ball in km/h. 312

2 Momentum = mass velocity Hence, 5 = 0.2 v from which, velocity, v = 5 = 25 m/s 0.2 = km/h = 90 km/h 6. A wagon of mass 8 t is moving at a speed of 5 m/s and collides with another wagon of mass 12 t, which is stationary. After impact, the wagons are coupled together. Determine the common velocity of the wagons after impact. Mass m 1 = 8 t = 8000 kg, m 2 = kg and velocity = 5 m/s, = 0. Total momentum before impact = m 1 = (8000 5) + ( ) = kg m/s Let the common velocity of the wagons after impact be v m/s Since total momentum before impact = total momentum after impact: = m 1 v v = v(m 1 ) = v(20000) Hence v = = 2 m/s i.e. the common velocity after impact is 2 m/s in the direction in which the 8 t wagon is initially travelling. 7. A car of mass 800 kg was stationary when hit head-on by a lorry of mass 2000 kg travelling at 15 m/s. Assuming no brakes are applied and the car and lorry move as one, determine the speed of the wreckage immediately after collision. Mass m 1 = 800 kg, m 2 = 2000 kg and velocity = 0, = 15 m/s 313

3 Total momentum before impact = m 1 = (800 0) + ( ) = kg m/s Let the common velocity of the wagons after impact be v m/s Since total momentum before impact = total momentum after impact: = m 1 v v = v(m 1 ) = v(2800) Hence v = = m/s 2800 i.e. the speed of the wreckage immediately after collision is m/s in the direction in which the lorry is initially travelling. 8. A body has a mass of 25 g and is moving with a velocity of 30 m/s. It collides with a second body which has a mass of 15 g and which is moving with a velocity of 20 m/s. Assuming that the bodies both have the same speed after impact, determine their common velocity (a) when the speeds have the same line of action and the same sense, and (b) when the speeds have the same line of action but are opposite in sense. Mass m 1 = 25 g = kg, m 2 = 15 g = kg, velocity = 30 m/s and = 20 m/s. (a) When the velocities have the same line of action and the same sense, both and are considered as positive values Total momentum before impact = m 1 = ( ) + ( ) = = 1.05 kg m/s Let the common velocity after impact be v m/s Total momentum before impact = total momentum after impact i.e = m 1 v v = v(m 1 ) 1.05 = v( ) 314

4 from which, common velocity, v = are initially travelling = m/s in the direction in which the bodies (b) When the velocities have the same line of action but are opposite in sense, one is considered as positive and the other negative. Taking the direction of mass m 1 as positive gives: velocity = +30 m/s and = - 20 m/s Total momentum before impact = m 1 = ( ) + ( ) = = kg m/s and since it is positive this indicates a momentum in the same direction as that of mass m 1. If the common velocity after impact is v m/s then from which, common velocity, v = initially travelling = v(m 1 ) = v(0.040) = m/s in the direction that the 25 g mass is 315

5 EXERCISE 119, Page The sliding member of a machine tool has a mass of 200 kg. Determine the change in momentum when the sliding speed is increased from 10 mm/s to 50 mm/s. Change of linear momentum = mass change of velocity Hence, change in momentum = 200 kg (50 10) 10 3 m/s = 8 kg m/s 2. A force of 48 N acts on a body of mass 8 kg for 0.25 s. Determine the change in velocity. Impulse = applied force time = change in linear momentum i.e. 48 N 0.25 s = mass change in velocity = 8 kg change in velocity from which, change in velocity = 48 N 0.25s 8kg = 1.5 m/s (since 1 N = 1 kg m/s 2 ) 3. The speed of a car of mass 800 kg is increased from 54 km/h to 63 km/h in 2 s. Determine the average force in the direction of motion necessary to produce the change in speed. Change of momentum = applied force time i.e. mass change of velocity = applied force time i.e. 800 kg m/s = applied force 2 s from which, applied force = 36. = 1000 N or 1kN 2 316

6 4. A 10 kg mass is dropped vertically on to a fixed horizontal plane and has an impact velocity of 15 m/s. The mass rebounds with a velocity of 5 m/s. If the contact time of mass and plane is s, calculate (a) the impulse, and (b) the average value of the impulsive force on the plane. (a) Impulse = change in momentum = m( - v 1 ) where = impact velocity = 15 m/s and v 1 = rebound velocity = - 5 m/s (v 1 is negative since it acts in the opposite direction to ) Thus, impulse = m( - v 1 ) = 10 kg ( ) m/s = = 200 kg m/s (b) Impulsive force = impulse 200kg m / s = 8000 N or 8 kn time 0.025s 5. The hammer of a pile driver of mass 1.2 t falls 1.4 m on to a pile. The blow takes place in 20 ms and the hammer does not rebound. Determine the average applied force exerted on the pile by the hammer. Initial velocity, u = 0, acceleration due to gravity, g = 9.81 m/s 2 and distance, s = 1.4 m. Using the equation of motion: v 2 = + 2gs gives: v 2 = (9.81)(1.4) from which, impact velocity, v = ( 2)( 9. 81)( 1. 4) = m/s Neglecting the small distance moved by the pile and hammer after impact, momentum lost by hammer = the change of momentum = mv = 1200 kg m/s Rate of change of momentum = changeof momentum changeof time = 3 = N Since the impulsive force is the rate of change of momentum, the average force exerted on the pile is kn 317

7 6. A tennis ball of mass 60 g is struck from rest with a racket. The contact time of ball on racket is 10 ms and the ball leaves the racket with a velocity of 25 m/s. Calculate (a) the impulse, and (b) the average force exerted by a racket on the ball. (a) Impulse = change of momentum = mv = (0.060 kg)(25 m/s) = 1.5 kg m/s impulse 1.5kg m / s (b) Impulsive force = 3 time = 150 N 7. In a press-tool operation, the tool is in contact with the work piece for 40 ms. If the average force exerted on the work piece is 90 kn, determine the change in momentum. Change in momentum = applied force time = N = 3600 kg m/s EXERCISE 120, Page 267 Answers found from within the text of the chapter, pages 262 to 267. EXERCISE 121, Page (d) 2. (b) 3. (f) 4. (c) 5. (a) 6. (c) 7. (a) 8. (g) 9. (f) 10. (f) 11. (b) 12. (e) 318