Week 5 homework

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WebAssign Problem 1: Consult Interactive Solution 7.9 at www.wiley.com/college /cutnell for a review of problem-solving skills that are involved in this problem.
Week 6 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem 1: Consult Interactive Solution 7.9 at www.wiley.com/college/cutnell for a review of problem-solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of + 16.0 m/s, while the exiting water stream has a velocity of – 16.0 m/s. The mass of water per second that strikes the blade is 30.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

REASONING During the time interval ∆t, a mass m of water strikes the turbine blade. The incoming water has a momentum mv0 and that of the outgoing water is mvf. In order to change the momentum of the water, an impulse ( Σ F ) ∆ t is applied

to it by the stationary turbine blade. Now ( Σ F ) ∆ t = F ∆ t , since only the force of the blade is assumed to act on the water in the horizontal direction. These variables are related by the impulse-momentum theorem, F ∆ t = mvf − mv0, which can be solved to find the average force F exerted on the water by the blade. SOLUTION Solving the impulse-momentum theorem for the average force gives F=

mv f − mv 0 m = ( v − v0 ) ∆t ∆t f

The ratio m / ( ∆ t ) is the mass of water per second that strikes the blade, or 30.0 kg/s, so the average force is F=

m ( v − v0 ) = ( 30.0 kg/s )  ( − 16.0 m/s ) − ( + 16.0 m/s )  = − 960 N ∆t f

The magnitude of the average force is 960 N .

WebAssign Problem 2: A 55-kg swimmer is standing on a stationary 210-kg .oating raft. The swimmer then runs off the raft horizontally with a velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water. REASONING The sum of the external forces acting on the swimmer/raft system is zero, because the weight of the swimmer and raft is balanced by a corresponding normal force and friction is negligible. The swimmer and raft constitute an isolated system, so the principle of conservation of linear momentum applies. We will use this principle to find the recoil velocity of the raft. SOLUTION As the swimmer runs off the raft, the total linear momentum of the swimmer/raft system is conserved: ms vs + mr vr = 14243

Total momentum after swimmer runs off raft

0 {

Total momentum before swimmer starts running

where ms and vs are the mass and final velocity of the swimmer, and mr and vr are the mass and final velocity of the raft. Solving for v gives r vr = −

ms vs mr

= −

( 55 kg ) ( + 4.6 m/s ) 210 kg

= − 1.2 m/s

WebAssign Problem 3: The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is 0.010 kg and its velocity is + 720 m/s. Her mass (including the gun) is 51 kg. (a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place? (b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of at a velocity of + 720 m/s? REASONING For the system consisting of the female character, the gun and the bullet, the sum of the external forces is zero, because the weight of each object is balanced by a corresponding upward (normal) force, and we are ignoring friction. The female character, the gun and the bullet, then, constitute an isolated system, and the principle of conservation of linear momentum applies.

SOLUTION a. The total momentum of the system before the gun is fired is zero, since all parts of the system are at rest. Momentum conservation requires that the total momentum remains zero after the gun has been fired. m1 v f1 + m2 v f2 =  0      Total momentum before Total momentum after gun is fired

gun is fired

where the subscripts 1 and 2 refer to the woman (plus gun) and the bullet, respectively. Solving for vf1, the recoil velocity of the woman (plus gun), gives v f1 = –

m2 v f2 m1

=

– (0.010 kg)(720 m / s) = – 0.14 m / s 51 kg

b. Repeating the calculation for the situation in which the woman shoots a blank cartridge, we have m v – (5.0 × 10 –4 kg)(720 m / s) v f1 = – 2 f2 = = – 7.1 × 10 –3 m / s m1 51 kg In both cases, the minus sign means that the bullet and the woman move in opposite directions when the gun is fired. The total momentum of the system remains zero, because momentum is a vector quantity, and the momenta of the bullet and the woman have equal magnitudes, but opposite directions.

WebAssign Problem 4: Multiple-Concept Example 7 presents a model for solving problems such as this one. A 1055-kg van, stopped at a traffic light, is hit directly in the rear by a 715-kg car traveling with a velocity of +2.25 m/s. Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of (a) the car and (b) the van? REASONING Since the collision is an elastic collision, both the linear momentum and kinetic energy of the two-vehicle system are conserved. The final velocities of the car and van are given in terms of the initial velocity of the car by Equations 7.8a and 7.8b. SOLUTION a. The final velocity vf1 of the car is given by Equation 7.8a as  m − m2  vf 1 =  1  v01 m + m 2   1 where m1 and m2 are, respectively, the masses of the car and van, and v01 is the initial velocity of the car. Thus,

 715 kg − 1055 kg  vf 1 =   ( + 2.25 m/s ) = − 0.432 m/s  715 kg + 1055 kg  b. The final velocity of the van is given by Equation 7.8b:  2m1  vf 2 =   v01 =  m1 + m2 

  2 ( 715 kg )   ( + 2.25 m/s ) = + 1.82 m/s  715 kg + 1055 kg 

WebAssign Problem 5: Consult Multiple-Concept Example 8 for background pertinent to this problem. A 2.50-g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 215 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position? REASONING a. During the collision between the bullet and the wooden block, linear momentum is conserved, since no net external force acts on the bullet and the block. The weight of each is balanced by the tension in the suspension wire, and the forces that the bullet and block exert on each other are internal forces. This conservation law will allow us to find the speed of the bullet/block system immediately after the collision. b. Just after the collision, the bullet/block rise up, ultimately reaching a final height hf before coming to a momentary rest. During this phase, the tension in the wire (a nonconservative force) does no work, since it acts perpendicular to the motion. Thus, the work done by nonconservative forces is zero, and the total mechanical energy of the system is conserved. An application of this conservation law will enable us to determine the height hf. SOLUTION a. The principle of conservation of linear momentum states that the total momentum after the collision is equal to that before the collision. m + m )v (144 42444 3 bullet

block

f

Momentum after collision

= mbullet v0,bullet + mblock v0,block 14444244443 Momentum before collision

Solving this equation for the speed vf of the bullet/block system just after the collision gives

vf = =

mbullet v0,bullet + mblock v0,block mbullet + mblock

( 0.00250 kg ) ( 425 m/s ) + ( 0.215 kg ) ( 0 m/s ) 0.00250 kg + 0.215 kg

= 4.89 m/s

b. Just after the collision, the total mechanical energy of the system is all kinetic energy, since we take the zero-level for the gravitational potential energy to be at the initial height of the block. As the bullet/block system rises, kinetic energy is converted into potential energy. At the highest point, the total mechanical energy is all gravitational potential energy. Since the total mechanical energy is conserved, we have m + m ) gh = ( m + m )v (144 42444 3 14442444 3 bullet

block

f

Total mechanical energy at the top of the swing, all potential

1 2

bullet

block

2 f

Total mechanical energy at the bottom of the swing, all kinetic

Solving this expression for the height hf gives hf =

1 2

vf2 = g

1 2

( 4.89 m/s ) 9.80 m/s 2

2

= 1.22 m

WebAssign Problem 6: Multiple-Concept Example 9 reviews the concepts that play roles in this problem. The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025 kg and is moving along the x axis with a velocity of + 5.5 m/s. It makes a collision with puck B, which has a mass of 0.050 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a) puck A and (b) puck B.

REASONING The net external force acting on the two-puck system is zero (the weight of each ball is balanced by an upward normal force, and we are ignoring friction due

to the layer of air on the hockey table). Therefore, the two pucks constitute an isolated system, and the principle of conservation of linear momentum applies. SOLUTION Conservation of linear momentum requires that the total momentum is the same before and after the collision. Since linear momentum is a vector, the x and y components must be conserved separately. Using the drawing in the text, momentum conservation in the x direction yields mA v0A = mA vfA ( cos 65° ) + mBvfB ( cos 37° )

(1)

while momentum conservation in the y direction yields 0 = mA vfA ( sin 65° ) − mBvfB ( sin 37° )

(2)

Solving equation (2) for vfB, we find that vfB =

mA vfA ( sin 65° ) mB ( sin 37° )

(3)

Substituting equation (3) into Equation (1) leads to  m v ( sin 65° )  mA v0A = mA vfA ( cos 65° ) +  A fA  ( cos 37° ) sin 37°   a. Solving for vfA gives vfA =

v0A + 5.5 m/s = = 3.4 m/s  sin 65°   sin 65°  cos 65° +   cos 65° +    tan 37°   tan 37° 

b. From equation (3), we find that vfB =

( 0.025 kg ) ( 3.4 m/s ) ( sin 65° ) ( 0.050 kg ) ( sin 37° )

= 2.6 m/s

WebAssign Problem 7: Consider the two moving boxcars in Example 5. Determine the velocity of their center of mass (a) before and (b) after the collision. (c) Should your answer in part (b) be less than, greater than, or equal to the common velocity vf of the two coupled cars after the collision? Justify your answer.

REASONING AND SOLUTION The velocity of the center of mass of a system is given by Equation 7.11. Using the data and the results obtained in Example 5, we obtain the following: a. The velocity of the center of mass of the two-car system before the collision is

( vcm ) before =

m1v01 + m2v02 m1 + m2

(65 × 103 kg)( + 0.80 m/s) + (92 × 103 kg)( +1.2 m/s) = = +1.0 m/s 65 × 103 kg + 92 × 103 kg b. The velocity of the center of mass of the two-car system after the collision is

( vcm ) after =

m1vf + m2vf = vf = +1.0 m/s m1 + m2

c. The answer in part (b) should be the same as the common velocity vf. Since the cars are coupled together, every point of the two-car system, including the center of mass, must move with the same velocity. WebAssign Problem 8: When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75-kg man just before contact with the ground has a speed of 6.4 m/s. (a) In a stiff-legged landing he comes to a halt in 2.0 ms. Find the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.10 s. Find the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the force of the ground on the man in parts (a) and (b). REASONING We will apply the impulse momentum theorem as given in Equation 7.4 to solve this problem. From this theorem we know that, for a given change in momentum, greater forces are associated with shorter time intervals. Therefore, we expect that the force in the stiff-legged case will be greater than in the knees-bent case. SOLUTION a. Assuming that upward is the positive direction, we find from the impulse-momentum theorem that ΣF=

mv f − mv 0 ∆t

=

( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( − 6.4 m/s ) 2.0 × 10

−3

s

= + 2.4 × 105 N

b. Again using the impulse-momentum theorem, we find that ΣF=

mvf − mv 0 ∆t

=

( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( − 6.4 m/s ) 0.10 s

= + 4.8 × 103 N

c. The net average force acting on the man is Σ F = FGround + W , where FGround is the average upward force exerted on the man by the ground and W is the downwardacting weight of the man. It follows, then, that FGround = Σ F − W . Since the weight is W = −mg, we have Stiff − legged

FGround = Σ F − W

(

)

(

)

= + 2.4 × 105 N −  − ( 75 kg ) 9.80 m/s 2  = + 2.4 × 105 N   Knees − bent

FGround = Σ F − W = + 4.8 × 103 N −  − ( 75 kg ) 9.80 m/s 2  = + 5.5 × 103 N  

WebAssign Problem 9: To view an interactive solution to a problem that is similar to this one, go to www.wiley.com/college/cutnell and select Interactive Solution 7.55. A 0.015-kg bullet is fired straight up at a falling wooden block that has a mass of 1.8 kg. The bullet has a speed of 810 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurred. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t. REASONING We will divide the problem into two parts: (a) the motion of the freely falling block after it is dropped from the building and before it collides with the bullet, and (b) the collision of the block with the bullet. During the falling phase we will use an equation of kinematics that describes the velocity of the block as a function of time (which is unknown). During the collision with the bullet, the external force of gravity acts on the system. This force changes the momentum of the system by a negligibly small amount since the collision occurs over an extremely short time interval. Thus, to a good approximation, the sum of the external forces acting on the system during the collision is negligible, so the linear momentum of the system is conserved. The principle of conservation of linear momentum can be used to provide a relation between the momenta of the system before and after the collision. This relation will enable us to find a value for the time it takes for the bullet/block to reach the top of the building.

SOLUTION Falling from rest (v0, block = 0 m/s), the block attains a final velocity vblock just before colliding with the bullet. This velocity is given by Equation 2.4 as =

vblock {

Final velocity of block just before bullet hits it

v0, block + a t 123

Initial velocity of block at top of building

2

where a is the acceleration due to gravity (a = −9.8 m/s ) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is vblock = a t (1) During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that m + m )v (144 42444 3 bullet

block

f

Total linear momentum after collision

= mbullet vbullet + mblock vblock 1444 424444 3

(2)

Total linear momentum before collision

Here vf is the final velocity of the bullet/block system after the collision, and vbullet and vblock are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that vf , the final velocity of the bullet/block system after the collision must have the same magnitude as vblock, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that vf = −vblock. Substituting this relation and Equation (1) into Equation (2) gives

(m

bullet

+ mblock ) ( − a t ) = mbullet vbullet + mblock ( a t )

Solving for the time, we find that t=

− mbullet vbullet

a ( mbullet + 2mblock )

=

− ( 0.015 kg ) ( + 810 m/s )

( − 9.80 m/s )  ( 0.015 kg ) + 2 ( 1.8 kg )  2

= 0.34 s

Practice conceptual problems: 3. Two objects have the same momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same magnitudes? Give your reasoning in each

case. REASONING AND SOLUTION a. Yes. Momentum is a vector, and the two objects have the same momentum. This means that the direction of each object’s momentum is the same. Momentum is mass times velocity, and the direction of the momentum is the same as the direction of the velocity. Thus, the velocity directions must be the same. b. No. Momentum is mass times velocity. The fact that the objects have the same momentum means that the product of the mass and the magnitude of the velocity is the same for each. Thus, the magnitude of the velocity of one object can be smaller, for example, as long as the mass of that object is proportionally greater to keep the product of mass and velocity unchanged. 4. (a) Can a single object have kinetic energy but no momentum? (b) Can a system of two or more objects have a total kinetic energy that is not zero but a total momentum that is zero? Account for your answers. REASONING AND SOLUTION a. If a single object has kinetic energy, it must have a velocity; therefore, it must have linear momentum as well. b. In a system of two or more objects, the individual objects could have linear momenta that cancel each other. In this case, the linear momentum of the system would be zero. The kinetic energies of the objects, however, are scalar quantities that are always positive; thus, the total kinetic energy of the system of objects would necessarily be nonzero. Therefore, it is possible for a system of two or more objects to have a total kinetic energy that is not zero but a total momentum that is zero. 9. The drawing shows a garden sprinkler that whirls around a vertical axis. From each of the three arms of the sprinkler, water exits through a tapered nozzle. Because of this nozzle, the water leaves each nozzle with a speed that is greater than the speed inside the arm. (a) Apply the impulse–momentum theorem to deduce the direction of the force applied to the water. (b) Then, with the aid of Newton’s third law, explain how the water causes the whirling motion. REASONING AND SOLUTION a. Since the water leaves each nozzle with a speed that is greater than the speed inside the arm, the quantity mv f − mv 0 is positive. From the impulse-momentum theorem, ( Σ F ) ∆ t = mv f − mv 0 , and we can deduce that there is a net positive or outward impulse. Therefore, a net outward force is exerted on the water.

b. From Newton's third law, the water must exert a net force that is equal in magnitude, but negative and directed toward the nozzle. The nozzle and the arm, in

turn, move. whirl.

Since each arm is free to rotate about the vertical axis, the arm will

18. Where would you expect the center of mass of a doughnut to be located? Why? . REASONING AND SOLUTION Many objects have a point, a line, or a plane of symmetry. If the mass of the system is uniformly distributed, the center of mass of such an object lies at that point, on that line, or in that plane. The point of symmetry of a doughnut is at the geometric center of the hole. Thus, the center of mass of a doughnut is at the center of the hole.