Study Questions/Problems Week 6

Transcription

1 Study Questions/Problems Week 6 Chapters 9 explores further elaborations on Newton s laws that lead to another conservation principle conservation of linear momentum. These problems will help you appreciate the power of both energy and momentum principles. Chapter 9 Conceptual Questions, 3, 4, 7, 8, 3, 5, 7, 8 Conceptual Exercises 9, 0,, 3, 4 Problems, 3, 7, 9,, 7, 9, 4, 6, 9, 3, 36, 4, 47, 50 Answers/solutions for even numbered CQs and CEs, and for all the problems listed above are on the following pages don t peek until you have done your best to solve a problem.

2 Chapter 9 Linear Momentum and Collisions Answers to Even-Numbered Conceptual Questions. Doubling an object s speed increases its kinetic energy by a factor of four, and its momentum by a factor of two. 4. No. Consider, for example, a system of two particles. The total momentum of this system will be zero if the particles move in opposite directions with equal momentum. The kinetic energy of each particle is positive, however, and hence the total kinetic energy is also positive. 6. Yes. Just point the fan to the rear of the boat. The resulting thrust will move the boat forward. 8. (a) The force due to braking which ultimately comes from friction with the road reduces the momentum of the car. The momentum lost by the car does not simply disappear, however. Instead, it shows up as an increase in the momentum of the Earth. (b) As with braking, the ultimate source of the force accelerating the car is the force of static friction between the tires and the road. 0. Yes. For example, we know that in a one-dimensional elastic collision of two objects of equal mass the objects swap speeds. Therefore, if one object is at rest before the collision, it is possible for one object to be at rest after the collision as well. See Figure 9-7(a).. No. Any collision between cars will be at least partially inelastic, due to denting, sound production, heating, and other effects. 4. The center of mass of the hourglass starts at rest in the upper half of the glass and ends up at rest in the lower half. Therefore, the center of mass accelerates downward when the sand begins to fall to get it moving downward and then accelerates upward when most of the sand has fallen to bring it to rest again. It follows from equation 9-8 that the weight read by the scale is less than Mg when the sand begins falling, but is greater than Mg when most of the sand has fallen. 6. (a) Assuming a very thin base, we conclude that the center of mass of the glass is at its geometric center of the glass. (b) In the early stages of filling, the center of mass is below the center of the glass. When the glass is practically full, the center of mass is again at the geometric center of the glass. Thus, as water is added, the center of mass first moves downward, then turns around and moves back upward to its initial position. 8. As this jumper clears the bar, a significant portion of his body extends below the bar due to the extreme arching of his back. Just as the center of mass of a donut can lie outside the donut, the center of mass of the jumper can be outside his body. In extreme cases, the center of mass can even be below the bar at all times during the jump. Answers to Even-Numbered Conceptual Exercises. Recall that kinetic energy is related to momentum as follows K p /m. It follows, then, that p mk. Given that the kinetic energy of each of these objects is the same, their momentum varies as the square root of the mass. Applying these considerations, we have the following ranking C < A < B. 4. It is better if the collision is inelastic, because then the light pole gives your car only enough impulse to bring it to rest. If the collision is elastic, the impulse given to your car is about twice as much. This additional impulse which acts over a very short period of time could cause injury.

3 6. The rate of change in momentum is the same for both objects. As a result, the rate of change in velocity for the less massive object (the pebble) must be greater than it is for the more massive object (the boulder). Alternatively, we know that the acceleration (rate of change of velocity) of an object is proportional to the force acting on it and inversely proportional to its mass. These objects experience the same force, and therefore the less massive object has the greater acceleration. 8. The speed of the ball after bouncing off the elephant will be greater than the speed it had before the collision. The situation is similar to that shown in Figure 9-7 (c), except that the small mass has a nonzero speed before the collision. 0. The center of mass is higher than the midway point between the tip of the stalactite and the cave floor. The reason is that as the drops fall, their separations increase (see Conceptual Checkpoint -5). With the drops more closely spaced on the upper half of their fall, the center of mass is shifted above the halfway mark.. The scale supports the juggler and the three balls for an extended period of time. Therefore, we conclude that the average reading of the scale is equal to the weight of the juggler plus the weight of the three balls. 4. The two halves of the triangle have the same mass, but the left side is, on average, farther from the fulcrum than the right side. As a result, the center of mass is to the left of the fulcrum, and hence the triangle will tip to the left when released. See following pages for Problem Solutions

4 Solutions to Ch. 9 Problems. Picture the Problem The car and the baseball each travel in a straight line at constant speed. The data given in the Exercise 9- include p car 5,800 kg m/s and m ball 0.4 kg. Strategy Use equation 9- to set the magnitude of the momentum of the baseball equal to the magnitude of the momentum of the car, and solve for m ball. Solution Set p ball p car and solve for v ball p ball m ball v ball p car v ball p car 5,800 kg m/s. 06 m 3600 s m ball 0.4 kg s h mi 609 m mi/h Insight The huge required speed for the baseball (3,40 times the speed of sound!) is due to the huge mass difference between the two objects. 3. Picture the Problem The owner walks slowly toward the northeast while the cat runs eastward and the dog runs northward. Strategy Sum the momenta of the dog and cat using the component method. Use the known components of the total momentum to find its magnitude and direction. Let north be in the y direction, east in the x direction. Solution. Use the component method of vector addition to find the owner s momentum. Divide the owner s momentum by his mass to get the components of the owner s velocity 3. Use the known components to find the direction and magnitude of the owner s velocity p total p d + p c m vd d vc c ( 0.0 kg) (.50 m/s ŷ) kg p total ( 5.0 kg m/s)ˆx + ( 50.0 kg m/s)ŷ p 0 m v0 0 p total p v 0 total ( 5.0 kg m/s)ˆx + ( 50.0 kg m/s)ŷ m kg ( 0.4 m/s)ˆx + ( 0.74 m/s)ŷ tan v 0 ( 0.43 m/s) + ( m/s) m/s ( 3.00 m/s ˆx ) Insight We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The owner is moving much slower than either the cat or the dog because of his larger mass. 7. Picture the Problem The individual momenta and final momentum vectors are depicted at right. Strategy The momenta of the two objects are perpendicular. Because of this we can say that the momentum of object is equal to the x-component of the total momentum and the momentum of object is equal to the y-component of the total momentum. Find the momenta of objects and in this manner and divide by their speeds to determine the masses.

5 Solution. Find p total, x and divide by p p total, x p total cos ( 7.6 kg m/s) ( cos 66.5 ) 7.0 kg m/s m p 7.0 kg m/s.5 kg.80 m/s. Find p total, y and divide by v p p total sin 7.6 kg m/s v 3.0 m/s v ( sin 66.5 ) 5. kg Insight Note that object has the larger momentum because the total momentum points mostly in the ŷ direction. The two objects have similar speeds, so object must have the larger mass in order to have the larger momentum. 9. Picture the Problem The golf club exerts an impulse on the ball, imparting momentum. Strategy it according to equation 9-3. Solution Apply equation 9-3 directly Find the change of momentum of the golf ball and use it to find the force exerted on F av p t mv f v i t ( kg) 67 m/s s 3.0 kn Insight This is a force equivalent to 670 lb! A ball launched with a speed of 67 m/s launched at 45 will land 457 m (500 yd) down range in the absence of air friction. A par 5 hole in one?. Picture the Problem The marble drops straight down from rest and rebounds from the floor. Strategy Use equation - to find the speed of the marble just before it hits the floor. Use the same equation and the known rebound height to find the rebound speed. Then use equation 9-6 to find the impulse delivered to the marble by the floor. Let upward be the positive direction, so that the marble hits the floor with speed v i and rebounds upward with speed v f. Solution. (a) Find v i using equation - v i v 0 gy v i v 0 gy m/s.44 m 5.3 m/s. Find v f using equation - again v v f gy v f v + gy m/s 3.54 ( 5.3) m/s 3. Use equation 9-6 to find I I mv mv f v i kg m 3.54 m/s 0.33 kg m/s 4. (b) If the marble had bounced to a greater height, its rebound speed would have been larger and the impulse would have been greater than the impulse found in part (a). Insight By Newton s Third Law the marble also delivers a downward impulse on the floor. The Earth theoretically moves in response to this impulse, but its change of velocity is tiny (.0-6 m/s) due to its enormous mass.

6 7. Picture the Problem The two skaters push apart and move in opposite directions without friction. Strategy By applying the conservation of momentum we conclude that the total momentum of the two skaters after the push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for. Let the velocity v point in the negative direction, v in the positive direction. Solution Set p total 0 and solve for p x + p x 0 m x + v x m v x ( 45 kg) ( 0.6 m/s) 3 kg v x 0.89 m/s Insight An alternative way to find the mass is to use the equations of kinematics in a manner similar to that described in Example Picture the Problem The two pieces fly in opposite directions at different speeds.. Strategy As long as there is no friction the total momentum of the two pieces must remain zero, as it was before the explosion. Combine the conservation of momentum with the given kinetic energy ratio to determine the ratio of the masses. Let m represent the piece with the smaller kinetic energy. Solution. Set p 0 and solve for p + p 0 m + v m. Set K K 3. Combine the expressions from steps and m v K K v v v m m 4. The piece with the smaller kinetic energy has the larger mass. m v v m m Insight The smaller mass carries the larger kinetic energy because kinetic energy increases with the square of the velocity but is linear with mass. Its higher speed more than compensates for its smaller mass. 4. Picture the Problem The car and the minivan collide and stick together, as indicated in the figure at right. The data given in Example 9-6 include m 950 kg and 300 kg. Strategy If we ignore frictional forces the total momentum of the two vehicles before the crash equals the momentum after the collision. The vehicles stick together so they behave as a single object with the sum of their masses. Use conservation of momentum to find the initial speed v of the minivan and the final speed v f of the two vehicles after the crash. Solution. Conserve momentum in the x direction p x m + 0 ( m + )v f cos

7 . Conserve momentum in the y direction 3. Divide the equation from step by the equation for step and solve for v 4. Solve the equation in step for v f p y v + 0 ( m + )v f sin v ( m + )v f sin m ( m + )v f cos tan v m tan 0.0 m/s ( 950 kg) ( 300 kg) ( 0.0 m/s) tan 40.0 m/s m v f ( m + )cos 950 kg m/s ( 950 kg kg)cos 40.0 Insight In real life the assumption that there is no friction is quite incorrect unless the collision occurs on very slick, icy pavement. Substantial deviations from the conservation of momentum will be observed if the pavement is dry. 6. Picture the Problem A 00-kg car moving at.5 m/s is struck in the rear by a 600-kg truck moving at 6. m/s. The vehicles stick together after the collision and move together with a speed of 5.0 m/s. Strategy In an inelastic collision we expect a loss of kinetic energy. Use equation 7-6 to find the kinetic energies before and after the collision and verify the loss. Solution. (a) The final kinetic energy of the car and truck together is less than the sum of their initial kinetic energies. Some of the initial kinetic energy is converted to heat, sound, and the permanent deformations in the materials of the car and truck.. (b) Use equation 7-6 to find the initial kinetic energy 3. Use equation 7-6 to find the final kinetic energy K i c v c + t v t ( 00 kg).5 m/s K f ( c t )v f 00 kg kg + ( 600 kg) ( 6. m/s) 54 kj ( 5.0 m/s) 48 kj Insight The kinetic energy loss is only % because the two vehicles are traveling in the same direction initially. If this were a head-on collision, the final speed would be 3.5 m/s in the direction the truck was traveling, and K f 3 kj, a loss of 58%. In fact, two identical objects traveling at the same speed and colliding together will lose 00% of their kinetic energy if they stick together after the collision. 9. Picture the Problem The putty is thrown horizontally, strikes the side of the block, and sticks to it. The putty and the block move together in the horizontal direction immediately after the collision, compressing the spring. Strategy Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the collision, then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximum compression of the spring after the collision. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of the putty will be converted to heat, sound, and permanent deformation of material during the inelastic collision.

8 Solution. Set p i p f and solve for v f. Set E after E rest after the collision m p v p ( m b p )v f v f m b p m p m b p K after U rest v p kx max m p v m b p p 3. Solve the resulting expression for x max x max p v p k( m p b ) ( kg) (.30 m/s) ( 0.0 N/m) ( kg) m 3.7 cm Insight The putty-block conglomerate will compress the spring even further if v p is larger or if m p is larger. 3. Picture the Problem The hammer strikes the nail with an elastic collision. The collision drives the nail in the forward direction and slows down the speed of the hammer. Strategy This is a one-dimensional, elastic collision where one of the objects (the nail) is initially at rest. Therefore, equations 9- apply and can be used to find the final speed of the nail. Once the nail speed is found, its kinetic energy is determined by equation 7-6. Let m be the mass of the hammer, be the mass of the nail, and v 0 be the initial speed of the hammer. Solution. Use equations 9- to find v,f. Use equation 7-6 to find K nail K nail mv nail ( 550 g ) m v,f m v 0 ( 4.5 m/s) 8.8 m/s v g nail ( 0.0 kg) 8.8 m/s 0.46 J Insight The elastic collision produces a bigger jolt for the car. If instead the two vehicles stuck together, the final speed of the car (and the truck) would be 0.0 m/s. 36. Picture the Problem The cart 4m collides with the cart m, which is given kinetic energy as a result and later collides with the cart m. Strategy In each case a moving cart collides with a cart that is at rest, so application of equations 9- will yield the final velocities of all the carts. First apply equations 9- to the collision between carts 4m and m, then to the collision between m and m. Let the 4m cart be called cart 4, the m cart be called cart, and the m cart be called cart

9 Solution. (a) Apply equations 9- to the first collision v 4,f m 4 m m 4 v 4,i 4m m 4m v 0 v 3 0 m v,f 4 m 4 v 4,i 4m 4m v 0 4 v 3 0. Apply equations 9- to the second collision. In this case cart has an initial speed of 4 v (b) Verify that K i K f by writing using equation 7-6 and dividing both sides by mv 0 v,f m v,i m m m ( 4 v 3 0 ) 4 v 9 0 m,f v,i m 4 ( m 3 v 0 ) 6 v 9 0 ( 4m )v? 0 ( 4m ) ( v 3 0 ) + ( m ) 4 ( v 9 0 ) + ( m ) 6 ( v 9 0 ) Insight Note that due to the transfer of kinetic energy via collisions, the cart with the smallest mass ends up with the largest speed. 4. Picture the Problem The box with no top rests on a flat surface. Its center of mass is measured relative to the geometric center of the box, a distance L above the bottom surface. Strategy Place the origin at the center of the box with the plane of the missing top perpendicular to the positive z-axis. Due to symmetry, X cm Y cm 0. Use a version of equation 9-4 written in the z direction to determine the location of the center of mass. Assume each side of the box has mass m. Let z 0 correspond to the center of the box. Solution Apply equation 9-4 directly Z cm mz M mz z z 3 z 4 z bottom 5m The center of mass is L/0 units below the center of the box. m ( L ) L 5m 0 Insight Placing some other masses at the bottom of the box will lower the center of mass even further. 47. Picture the Problem The physical arrangement for this problem is depicted in the figure at right. Strategy Before the string breaks, the reading on the scale is the total weight of the ball and the liquid. After the string breaks, the ball falls with constant speed, so that the center of mass of the ball liquid system does not accelerate. Solution. (a) Find the total weight of the ball and the liquid Mg ( kg) ( 9.8 m/s ) 3. N.. (b) After the string breaks, the reading is 3. N. Because the center of mass of the ball-liquid system undergoes no net acceleration, the scale must not exert any net force on it (Newton s Second Law), and by Newton s Third Law it exerts no net force on the scale. Therefore, the reading will not change. Insight If the ball were to accelerate, the center of mass of the ball-liquid system would accelerate

10 downward, and the scale force would decrease. Another way of looking at the problem is to realize that the water must exert exactly the same upward force on the ball (.47 N) when it is at rest as when it is falling at constant speed. 50. Picture the Problem The child throws rocks in the backward direction, propelling the wagon in the forward direction. Strategy Use Newton s Second Law to set the forward thrust equal to the backward friction force because the acceleration of the wagon is zero. Solve for the rate of mass ejection by the rock-throwing girl. Solution Use Newton s Second Law and equation 9-9 to find m t F x thrust f k 0 thrust m t m t v f k f k v 3.4 N kg rock 60 s rocks/min m/s s 0.50 kg min Insight If the girl had a 40-lb (8-kg) pile of rocks she could keep the wagon moving for about minute.