Momentum and Energy Key

Chapter 3 Review Questions 1. Two monkeys are connected by a rope of negligible mass, which passes over a pulley. Friction in the pulley is negligible. One monkey has a mass of 20.0 kg while the other has a mass of 22.0 kg. At what rate will the monkeys accelerate? What will be the tension in the rope connecting the monkeys?   Fnet = m*a 22g - 20g = (22+20)a a = 0.467 m/s² T - 20g = 20a T = 20(0.467 + 9.8) = 196 N 2. The following system is set up in your science lab. Assuming no friction: (a) At what rate will the system accelerate? Fnet = 1.2g - 0.8g = 0.4g a = Fnet/m = 0.4g/2 = 0.2g = 1.96 m/s² (b) What is the tension in the string while the system accelerates? T - 0.8g = 0.8a T = 0.8(1.96 + 9.8) = 9.4 N 3. How much momentum does a 0.500 kg rock have when thrown at 25.0 m/s? P = mv = (0.5)(25) = 12.5 kg m/s 4. What impulse must be imparted to a 100.0 g ball to get it moving at 40.0 m/s? Impulse = Δp = mv = (0.1)(40) = 4.00 N·s 5. A 2.4 kg cart moving 0.64 m/s collides with a stationary 1.8 kg cart. The carts lock together. What is their combined velocity after the collision?   m₁v₁ + m₂v₂ = (m₁ + m₂)v_f (2.4)(0.64) = (2.4 + 1.8)v_f v_f = 0.37 m/s 6. A 1.5 x 10³ kg car collides head-on with a 1.2 x 10³ kg car. Both cars were travelling at the same speed, 20.0 m/s, but in opposite directions. What will the velocity of the combined wreckage be immediately after the collision?   m₁v₁ + m₂v₂ = (m₁ + m₂)v_f (1.5 x 10³)(20) + (1.2 x 10³)(-20) = (1.5 + 1.2) x 10³v_f v_f = 2.2 m/s 7. A 52 kg girl is coasting along the floor on a large 2.0 kg skateboard. If she is moving 1.8 m/s when a 46 kg boy jumps on to the same skateboard, what is the speed of the skateboard immediately after the boy jumps on it?   m₁v₁ = (m₁ + m₂)v_f (52 kg)(1.8 m/s) = (52 + 46)v_f v_f = 0.97 m/s 8. A 30.0 kg hockey player travelling 1.0 m/s toward the south collides with a 25.0 kg hockey player moving 0.50 m/s toward the west. They become tangled and they move off together. With what speed and in what direction will the two players move immediately after the collision?   p_x = 30(0) + 25(0.5) = 12.5 kg m/s p_y = 30(-1) + 25(0) = -30 kg m/s p_r = √(30² + 12.5²) = 32.5 kg m/s tanθ = 12.5/30 θ = 23° W of S v_f = p_r / m = 32.5 / (30 + 25) = 0.59 m/s 9. A 5.00 x 10⁻³ kg steel ball moving 1.20 m/s collides elastically with an identical, stationary steel ball. The incident ball is deflected 30° from its original path. (a) Draw a vector diagram showing the paths of both balls after the perfectly elastic collision. (b) What is the velocity of the incident ball after the collision?   cos 30° = P'_i / P_i P'_i = 5 x 10⁻³ x 1.2 x cos 30° v'_1 = 1.0392 ≈ 1.04 m/s (c) What is the velocity of the struck ball after the collision? sin 30° = P'_2 / P_i 0.5 = m₂v'_2 / m₁v₁ v'_2 = 0.6 m/s (d) How much of the incident ball's kinetic energy was transferred to the struck ball? E_i = 1/2 mv² = 1/2 (5 x 10⁻³) (1.2)² = 0.0036 J E'_2 = 1/2 mv'_2² = 1/2 (5 x 10⁻³) (0.6)² = 0.0009 J 10. If an amateur astronomer carries a 50 kg telescope to the top of a 250 m hill, how much work will she do on the telescope? W = Fd = mgh = (50)(9.8)(250) = 1.2 x 10⁵ J = 120 kJ 11. A child pulls a toy wagon along the road for a distance of 250 m. The force he exerts along the handle is 32 N. The handle makes an angle of 30° with the horizontal road. How much work does the child do on the wagon? Express your answer in kl. 3200535 W=Fd= (32cos30°(23)m) 6928 J = 6.9 KJ = 12. How much kinetic energy does a 1.0 x 10 kg car, travelling at 90 km/h, have? F(1x10rg) (90km 2 K 3.1x105 J 100m 2 36005 13. How much gravitational potential energy does a 75 kg skier have when at the top of a hill 2.0 x 10 m high? Ep=mgh =(75g) (9.8)(2×10m) = 1.5×10J/ = 1.5 MJ 14. A large ball of modelling clay is dropped from a height of 2 m to the floor. The ball does not bounce back up in the air. Explain this in terms of the conservation laws discussed in this chapter. The potential converted to energy is Kinetic. The Kinetic E is used to deform the clay and the friction of clay's internal motion heat's the clay. Edvantage Interactive 2013 15. A girl lifts a 30. kg box from the ground up to a table 1.0 m high in a time of 1.5 s. A weightlifter = friend carries a 100. kg load across the gym in 5.0 s, without dropping it or lifting it higher. Who has the greater power output? Explain. The girl has greater power output since the weight lifter did zero work on the luuleg lood (the force and distance have to be parallel, not perpendicular) 16. How much work can a 6.0 hp motor do in an 8.0 h working day? Express your answer in megajoules. (1 MJ=10J) 1 hp = 750J (60px)(2)xx3600) - 130. MJ P 17. An 80 kg rider on a 120 kg motorbike, travelling 25 m/s in a direction 12° west of north, enters a highway without properly checking that the road is clear. He collides with a 1200 kg car travelling north at 20 m/s. The two vehicles and rider become entangled. In what direction and at what speed will the wreckage move immediately after the collision? (Assume the car's mass includes its occupants.) Pe P=(1200 (20) = 24000 Pm=(2001) (2546) = 5000 P P=Pe²+Pm²-2(2XPm)cos 168" =24000+2000-2(24000) (300) 00) 168 PA=1835755424.2 = sin ke Pe P 9:9.9° 120-9.9 2.1° 28909 kg V = 2890920.65-1 12004700 21m 2.1°WdN/ Momentum a Chapter 3 Momentum and Energy 115 1.2