Chapter 2: Kinematics in One Dimension

Example: A juggler throws a ball straight up with an initial speed of 10 m/s. With what speed would she need to throw a second ball, half a second later, starting from the same position as the first ball, so that the second ball hits the first ball at the top of the first ball's trajectory?

Solution:

Exam ReviewTuesday, September 17, 2013 10:00 PM

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Example: A car travelling at 30 m/s runs out of gas while travelling up a 5.0 degree slope. How far will it coast before starting to roll back down?

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Chapter 3: Kinematics in Two Dimensions

Example: A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 s, then coasts for 2.0 s, then slows down at a rate of 1.5 m/s2

until the next stop sign is reached. Determine the distance between the stop signs.

Solution:

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c. Repeat the calculation for Earth data, and you'll find that both the time of flight and the horizontal distance are 6 times greater on the Moon.

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Example: On the Apollo 14 mission to the Moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the surface of the Moon is 1/6th that of the Earth. Suppose he hit the ball with an initial speed of 25 m/s at an angle of 30 degrees above the horizontal. Determine how long the ball was in flight.a.Determine how far the ball travelled (horizontally).b.Ignoring air resistance, how much farther would the ball travel on the Moon than on the Earth?

c.

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Example: A spring-loaded gun, fired vertically, shoots a marble 6.0 m straight up. Determine the marble's range if it is fired horizontally from 1.5 m above the ground.

Solution: First determine the initial velocity of the marble.

Example: A car moves around a circular road that has radius 110 m.Determine the car's acceleration if the car moves through the curve at a constant speed of 40 m/s.

a.

At what speed would the car's acceleration be double the value calculated in Part a?

b.

Solution:

Next, determine the time of flight for the second motion.

Therefore, the range is

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Example: A child slides along frictionless ice at a speed of 4 m/s relative to the ice at an angle of 31 degrees north of east. The child throws a puck along the ice at a speed of 2 m/s relative to the child at an angle of 26 degrees south of east. Determine the velocity of the puck relative to the ice.

Solution: The easiest way to add the velocity vectors of the child and the puck is to separate each of them into components. Here are the details:

Example: Car A moves at a constant speed of 80 km/h East relative to the ground, and Car B moves at a constant speed of 70 km/h north relative to the ground. Determine the velocity of Car B relative to Car A.

If you prefer magnitude and direction:

Thus, the velocity of the puck relative to the ice is (5.23, 1.18) m/s, which is

equivalent to 5.36 m/s in the direction E13◦N._____________________________________________________________

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Solution:

If you prefer magnitude and direction:

Thus, the velocity of Car B relative to Car A is (−80, 70) km/h, which is

equivalent to 106 km/h in the direction E41◦N.___________________________________________________________

Example: A person stands 10 m away from the base of a wall that is 8 m high. She can throw comfortably at an angle of 60 degrees above the horizontal. Determine the minimum initial speed she must give a tennis ball so that it will clear the wall.

Solution: Draw a diagram!

Consider the points on the ball's path labelled A and B; they are the key points of the path. Because we know the coordinates of both points, it makes sense to use the displacement equations, as follows:

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Example: A 500 kg piano is being lowered into position by a crane while two people steady it with ropes pulling to the sides. Bob's rope pulls to the left, 15 degrees below the horizontal, with 500 N of tension. Ellen's rope pulls to the right, 25 degrees below the horizontal.Determine the tension in Ellen's rope if the piano descends vertically at a constant speed.

a.

Determine the tension in the main cable supporting the piano.b.

Solution:

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Chapter 4: Forces and Newton's Laws of Motion

makes sense to use the displacement equations, as follows:

We have two equations for the two unknown quantities, so we have enough to solve the problem. One way to solve the equations is to solve equation (2) for t and substitute the resulting expression into equation (1), which can then be solved for the required initial speed:

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Example: Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m.Determine the constant force Bob exerts on the rock.a.If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?b.

Solution: Bob clearly has a "rifle-arm."

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Example: A person with compromised pinch strength in his fingers can only exert a normal force of 6.0 N to either side of a pinch-held object, such as a book. Determine the heaviest book he can hold if the coefficient of static friction between his fingers and the surface of the book is 0.80.

Solution:

Example: A wood block is sliding up a wood ramp. If the ramp is very steep, the block will reverse direction at its highest point and slide back down. If the ramp is shallow, the block will stop when it reaches its highest point. Determine the smallest ramp angle, measured from the horizontal, for which the block will slide back down. (Note that the coefficient of static friction is 0.5 and the coefficient of kinetic friction is 0.2.)

Solution: Once the block reaches its highest point, the forces acting on it along the slope are a component of its weight and the frictional force. If the slope is not large enough, then static friction will be able to balance the component of

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For larger angles, the block will slide back down, as the static friction force will not be large enough to balance the component of the weight directed down the slope.

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Here are the free-body diagrams for a problem on a previous test:

The full solution to the problem is posted online; examine it carefully if you wish.

not large enough, then static friction will be able to balance the component of weight along the slope. When the static friction force is maximum,

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wish. ___________________________________________________________

Example: In the drawing, the rope and the pulleys are massless, and there is no friction. Find (a) the tension in the rope and (b) the acceleration of the 10.0-kg block. (Hint: The larger mass moves twice as far as the smaller mass.)

Solution: Draw free-body diagrams for each block!

Let x1 represent the displacement of the more massive block, and let x2

represent the displacement of the less massive block. The hint means that (using the chosen positive directions):

Because each block begins from rest, it follows that

Applying Newton's second law of motion to each block (with the help of the free-body diagrams), we obtain

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Chapter 5: Dynamics of Uniform Circular Motion

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Example: A car drives over the top of a hill that has a radius of 50 m. Determine the car's maximum speed so that it does not fly off the road at the top of the hill.

Solution:

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Example: A 100 g ball on a 60-cm-long string is swung in a vertical circle whose centre is 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. Determine the tension in the string an instant before it broke.

Solution:

Example: A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0075 m/s2 less than at sea level. Determine the observatory's altitude.

Solution:

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Example: Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall acceleration at the surface be three times its current value?

Solution:

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Example: Determine the speed and altitude of a geostationary satellite orbiting Mars. Mars rotates on its axis once every 24.8 h, has a mass of 6.42 × 1023 kg, and has a radius of 3370 km.

Solution:

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Chapter 6: Work and Energy

Example: A swing is made from a rope that can support a maximum tension of 800 N without breaking. Initially, the swing hangs vertically. The swing is then pulled back to an angle of 60 degrees with respect to the vertical and released from rest. Determine the mass of the heaviest person that can ride the swing without breaking the rope.

Solution: Draw a free-body diagram! But at which point of the swing? I'm not sure, so I'll draw a free-body diagram for a random point of the swing. Then I'll write down Newton's second law of motion for the radial component and the tangential component of the motion:

We are asked to determine something related to the maximum tension in the rope. You can see the relation between the maximum tension and the maximum mass allowable most easily by solving equation (1) for the tension:

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Can you see that the tension is at its maximum when the person is at the lowest point of the motion? That is the point when the speed is greatest, and it's also the point where the cosine of the angle is the greatest, because the cosine of 0 degrees is 1. Thus,

Now if we only had a way of knowing the maximum speed (i.e., the speed of the mass when it reaches the lowest point of its motion), then we could make further progress. The other problem is that we don't know r, the length of the rope. I'm not sure what to do about this latter problem, but for the former problem, I would definitely try energy methods. For instance, upon reading the problem again, I notice that we haven't used the fact that the initial angle is 60 degrees. This calls for another diagram, and an application of the principle of conservation of mechanical energy. (Recall that the tension force in this case does no work on the person, because the tension force is always radial, which is perpendicular to the motion.)

Now substitute this relation into the key equation from above,

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Chapter 7: Impulse and Momentum

and we obtain:

The swing is clearly not save, even for children, much less for more massive adults.______________________________________________________________

Example: A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 20 m/s. The third piece has twice the mass as the other two. Determine the speed and direction of the third piece.

Solution:

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Example: A 10 g bullet is fired into a 10 kg wood block that is at rest on a wooden table. The block, with the bullet embedded, slides 5.0 cm across thetable. Determine the speed of the bullet. (The coefficient of friction is 0.20.)

Solution:

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Example: A 1500 kg weather rocket accelerates upward at 10.0 m/s2. It explodes 2.00 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

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Example: The figure shows a collision between three balls of clay. The three hit simultaneously and stick together. Determine the speed and direction of the resulting blob of clay.

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Example: A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

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Example: Two blocks, A and B, slide on a frictionless surface. Block A has an initial velocity of 8 m/s at an angle of 20 degrees south of east, and Block B has an initial velocity v at an angle of 30 degrees north of east. The blocks collide; after the collision, Block A has a velocity of 5 m/s at an angle of 50 degrees north of east and Block B has a velocity of 7 m/s at an angle of 40 degrees south of east. The mass of Block B is 1.3 kg. Determine the mass m of Block A and the speed v of Block B before the collision.

Solution: Use the principle of conservation of momentum.

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We have two equations in two unknowns, which is enough to solve the problem. One way to do this is to solve each equation for the unknown speed, and then equate the two expressions:

Setting the expressions in equations (3) and (4) equal to each other, and solving for m, we obtain

Substituting the value for the mass of Block A into equation (3), we obtain the speed of Block B before the collision:

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Example: The 2.5 kg object shown in the figure has a moment of inertia about

the rotation axis of 0.085 kg m2. The rotation axis is horizontal. When released, what will be the magnitude of the object's initial angular acceleration?

Chapter 8: Rotational Kinematics, and Chapter 9: Rotational Dynamics

Thus, the mass of Block A is 1.1 kg, and the speed of Block B before the collision is 2.0 m/s.__________________________________________________________________

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Example: A computer disk is 8.0 cm in diameter. A reference dot on the edge of the disk is initially located at an angle of 45 degrees. The disk accelerates steadily for 0.50 s, reaching 2000 rpm, then coasts at a steady angular velocity for another 0.50 s. Determine the tangential acceleration of the reference dot after 0.25 s. a.Determine the centripetal acceleration of the reference dot after 0.25 s.b.Determine the angular position of the reference dot after 1.0 s.c.Determine the speed of the reference dot after 1.0 s.d.

Solution:

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Example: The ropes in the figure are each wrapped around a cylinder, and the cylinders are fastened together. The smaller cylinder has a diameter of 10 cm and a mass of 5.0 kg; the larger cylinder has a diameter of 20 cm and a mass of 20 kg. Determine the angular acceleration of the cylinders assuming they turn

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Example: The 20-cm-diameter disk in the figure can rotate on an axle through its centre. Determine the net torque about the axle.

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20 kg. Determine the angular acceleration of the cylinders assuming they turn on a frictionless axle.

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Example: A bicycle is rolling down a circular portion of a path, as shown in the figure. This portion of the path has radius 9.00 m. The angular displacement of the bicycle is 0.96 rad. Each bicycle wheel has radius 0.400 m. Determine the angle through which each bicycle wheel turns.

Solution: First determine the distance travelled by the bicycle:

Now note that the distance travelledby a spot on one of the tires is the same as the distance just calculated.This allows us to calculate the anglethat each tire turns through, as follows:

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Example: Consider two identical coins lying flat on a table. One coin is fixed in place and the second coin is touching the first coin. The movable coin is rotated in such a way that it always touches the fixed coin, and rolls along it without slipping. When the movable coin is moved all the way around the fixed coin so that it returns to its starting position, through what angle has the moving coin turned?

Solution: The moving coin makes two complete rotations, so it moves through an angle of 720 degrees. Study the following diagrams:

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Example: A tennis ball, starting from rest, rolls (without slipping) down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of 35 degrees with respect to the horizontal. Treat the ball as a thin-walled spherical shell, and determine the range x.

Discussion: Our goal is to determine the translational velocity of the ball at point 2. If we can do this, then the rest of the problem is just a projectile motion problem, of the type that we have solved back in Chapter 3.

Solution: The moment of inertia of the ball is

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Solution: The moment of inertia of the ball is

Assuming that the ball's mechanical energy is conserved between points 1 and 2 in the diagram,

Because of the no-slipping condition,

Thus,

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Chapter 10: Simple Harmonic Motion

Example: How far must you stretch a spring with stiffness constant 1000 N/m to store 200 J of energy?

Solution:

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Now that we know the translational speed at position 2, and we know the projection angle, we can solve the projectile motion problem to determine the range.

First determine the time of flight, and then use it to determine the range.

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Some conceptual questions

When a ball is thrown vertically upward, the force acting on the ball in a verticallyupward direction gradually decreases as the ball's speed decreases.

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Solution:

Example: A 10 kg runaway grocery cart runs into a spring with stiffness constant 250 N/m and compresses it by 60 cm. What was the speed of the cart just before it hit the spring?

Solution:

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When a ball is thrown vertically upward it gradually slows down, momentarilystops, and then falls down again. When it momentarily stops, the net force actingon the ball is zero.

A cargo plane flies West at 900 km/h. When the plane is directly over the Brocktower, it drops a package of physics textbooks. The package lands to the West ofthe tower.

When a heavy truck collides with a light car, the force that the truck exerts onthe car is greater than the force that the car exerts on the truck.

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An elevator is lifted up at a constant speed by a steel cable. The force that thecable exerts on the elevator cabin is greater than the force that gravity exerts onthe elevator cabin.

A passenger in a car moving very fast around a circular curve feels that he orshe is "thrown" towards the outside of the curve because of a force that pushesobjects away from the centre of the circular curve.

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If we assume no air resistance, then for a projectile motion the net force in thehorizontal direction is constant but the net force in the vertical direction is notconstant.

A horse is hitched to a cart. The driver says, "Giddyup!", but the horse doesn't move. He argues (yes, he's a talking horse) that there is no point in exerting any effort, because no motion is possible, because of Newton's third law. "After all," says the horse, "when I exert a force on the cart, by Newton's third law the cart exerts an equal and opposite force on me. The net force is therefore zero, and nothing moves." Explain.

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A 250-pound linebacker collides with a 150-pound wide receiver. The receiver is thrown to the ground violently. Obviously the force that the linebacker exerts on the wide receiver is greater than the force that the wide receiver exerts on the linebacker.

When you fire a rifle, you should hold the butt of the rifle tightly against your shoulder. Why?

(a) Explain how a propeller (boat or airplane) works. (b) How do rockets work?

If the acceleration vector of a moving object is non-zero and NOT in the samedirection as the velocity vector, then the object's speed is NOT increasing.

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direction as the velocity vector, then the object's speed is NOT increasing.

A ball is thrown straight up. At the peak of its motion, the ball stops momentarily.Because the ball is stopped momentarily, it experiences no net force for thatinstant.

In a collision between a truck and a car, if the truck is twice as massive as thecar then the truck exerts a force on the car that is twice as large as the force thatthe car exerts on the truck.

Rockets work in space in the same way that garden hoses recoil when water surgesthrough them; that is, the hot gases ejected from the rocket engine nozzles at veryhigh speed cause the rocket to move in the opposite direction.

For an object moving along a curved path, the net force acts along a tangent tothe curve, because this force is needed to push the object along its path.

If it's di�cult to turn a bolt, then it's helpful to switch to a longer wrench, becausethe longer wrench is more massive and therefore has greater force.

The angular speed of a spinning skater increases when she brings her arms in; thisillustrates the principle of conservation of angular momentum.

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In space stations that are above the Earth's atmosphere, astronauts are weightlessbecause they are beyond the reach of Earth's gravity.

According to the special theory of relativity, everything is relative, not absoluteas was previously thought in Newtonian mechanics.

Energy is always conserved, but momentum is not always conserved.

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