Problem #1
A 1.50 kg snowball is shot upward at an angle of 340 to the horizontal with an initial speed of 20.0 m/s. (a) What is its initial kinetic energy? (b) By how much does the gravitational potential energy of the snowball–Earth system change as the snowball moves from the launch point to the point of maximum height? (c) What is that maximum height?
Answer:
(a) Since the initial speed of the snowball is 20.0 m/s , we have its initial kinetic energy:
Ki = ½ mv02 = ½ (1.50 kg)(20.0 m/s)2 = 300 J
(b) We need to remember that since this projectile was not fired straight up, it will still have some kinetic energy when it gets to maximum height! That means we have to think a little harder before applying energy principles to answer this question.
At maximum height, we know that the y component of the snowball’s velocity is zero. The x component is not zero.
But we do know that since a projectile has no horizontal acceleration, the x component will remain constant; it will keep its initial value of
v0x = v0 cos θ0 = (20.0 m/s) cos 340 = 16.6 m/s
so the speed of the snowball at maximum height is 16.6 m/s . At maximum height, (the final position) the kinetic energy is
Kf = ½ mvf2 = ½ (1.50 kg)(16.6 m/s)2 = 206 J
In this problem there are only conservative forces (namely, gravity). The mechanical energy is conserved:
Ki + Ui = Kf + Uf
We already found the initial kinetic energy of the snowball: Ki = 300 J. Using Ugrav = mgy (with y = 0 at ground level), the initial potential energy is Ui = 0. Then we can find the final potential energy of the snowball:
Uf = Ki + Ui −Kf
= 300 J + 0 – 206 J
= 94 J
The final gravitational potential energy of the snowball–earth system (a long–winded way of saying what U is!) is then 94 J. (Since its original value was zero, this is the answer to part (b).)
(c) If we call the maximum height of the snowball h, then we have
Uf = mgh
Solve for h:
h = Uf/mg = (94 J)/(1.5 kg)(9.80 m/s2) = 6.38 m
The maximum height of the snowball is 6.38 m.
Problem #2
Pendulum consists of a 2.0kg stone on a 4.0m string of negligible mass. The stone has a speed of 8.0 m/s when it passes its lowest point. (a) What is the speed when the string is at 600 to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone’s motion? (c) If the potential energy of the pendulum–Earth system is taken to be zero at the stone’s lowest point, what is the total mechanical energy of the system?
Answer:
(a) The condition of the pendulum when the stone passes the lowest point is shown in Fig. 1(a). Throughout the problem we will measure the height y of the stone from the bottom of its swing. Then at the bottom of the swing the stone has zero potential energy, while its kinetic energy is
Ki = ½ mv02 = ½ (2.0 kg)(8.0 m/s)2 = 64 J
When the stone has swung up by 600 (as in Fig. 1(b)) it has some potential energy. To figure out how much, we need to calculate the height of the stone above the lowest point of the swing. By simple geometry, the stone’s position is
(4.0 m) cos 600 = 2.0 m
down from the top of the string, so it must be
4.0 m − 2.0 m = 2.0 m
up from the lowest point. So its potential energy at this point is
Uf = mgy = (2.0 kg)(9.80 m/s2)(2.0 m) = 39.2 J
It will also have a kinetic energy Kf = ½ mvf2, where vf is the final speed.
Now in this system there are only a conservative force acting on the particle of interest, i.e. the stone. (We should note that the string tension also acts on the stone, but since it always pulls perpendicularly to the motion of the stone, it does no work.) So the total mechanical energy of the stone is conserved:
Ki + Ui = Kf + Uf
We can substitute the values found above to get:
64.0 J + 0 = ½ (2.0kg)vf2 + 39.2J
which we can solve for vf :
(1.0kg)vf2 = 64.0 J − 39.2 J = 24.8 J
vf2 = 24.8 m2/s2
and then:
vf = 5.0 m/s
The speed of the stone at the 600 position will be 5.0 m/s.
(b) Clearly, since the stone is still in motion at an angle of 600 it will keep moving to greater angles and larger heights above the bottom position. For all we know, it may keep rising until it gets to some angle θ above the position where the string is horizontal, as shown in Fig. 2. We do assume that the string will stay straight until this point, but that is a reasonable assumption.
Now at this point of maximum height, the speed of the mass is instantaneously zero. So in this final position, the kinetic energy is Kf = 0. Its height above the starting position is
y = 4.0 m + (4.0 m) sin θ = (4.0 m)(1 + sin θ) (1)
so that its potential energy there is
Uf = mgyf = (2.0 kg)(9.80 m/s2)(4.0 m)(1 + sin θ) = (78.4 J)(1 + sin θ)
We use the conservation of mechanical energy (from the position at the bottom of the swing)to find θ: Ki + Ui = Kf + Uf , so:
Uf = Ki + Ui − Kf
(78.4 J)(1 + sin θ) = 64 J+ 0 − 0
This gives us:
1 + sin θ = 78.4J/64 J = 1.225
sin θ = 0.225
and finally
θ = 130
We do get a sensible answer of θ so we were right in writing down Eq. 1. Actually this equation would also have been correct if θ were negative and the pendulum reached its highest point with the string below the horizontal.
Problem #3
A bead slides without friction on a loop–the–loop track (see Fig. 3). Ifthe bead is released from a height h = 3.50R, what is its speed at point A? How large is the normal force on it if its mass is 5.00 g?
Answer:
In this problem, there are no friction forces acting on the particle (the bead). Gravity acts on it and gravity is a conservative force. The track will exert a normal forces on the bead, but this force does no work. So the total energy of the bead —kinetic plus (gravitational) potential energy— will be conserved.
At the initial position, when the bead is released, the bead has no speed; Ki = 0. But it is at a height h above the bottom of the track. If we agree to measure height from the bottom of the track, then the initial potential energy of the bead is
Ui = mgh
where m = 5.00 g is the mass of the bead.
At the final position (A), the bead has both kinetic and potential energy. If the bead’s speed at A is v, then its final kinetic energy is Kf = ½ mv2. At position A its height is 2R (it is a full diameter above the “ground level” of the track) so its potential energy is
Uf = mg(2R) = 2mgR .
The total energy of the bead is conserved: Ki + Ui = Kf + Uf . This gives us:
0 + mgh = ½ mv2 + 2mgR ,
where we want to solve for v (the speed at A). The mass m cancels out, giving:
gh = ½v2 = 2gR
½v2 = gh − 2gR = g(h − 2R)
and then
v2 = 2g(h − 2R) = 2g(3.50R − 2R) = 2g(1.5R) = 3.0gR (2)
and finally
v = [3.0gR]1/2
Since we don’t have a numerical value for R, that’s as far as we can go. In the next part of the problem, we think about the forces acting on the bead at point A.
These are diagrammed in Fig. 4. Gravity pulls down on the bead with a force mg. There is also a normal force from the track which I have drawn as having a downward component N. But it is possible for the track to be pushing upward on the bead; if we get a negative value for N we’ll know that the track was pushing up.
At the top of the track the bead is moving on a circular path of radius R, with speed v. So it is accelerating toward the center of the circle, namely downward. We know that the downward forces must add up to give the centripetal force mv2/R:
mg + N = mv2/R
n = mv2/R − mg = m(v2/R – g)
.
But we can use our result from Eq. 2 to substitute for v2. This gives:
N = m[3.0gR/R – g] = m(2g) = 2mg
Plug in the numbers:
N = 2(5.00 × 10−3 kg)(9.80 m/s2) = 9.80 × 10−2 N
At point A the track is pushing downward with a force of 9.80 × 10−2 N.
Problem #4
Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring –loaded gun that is mounted on the table. The target box is 2.20m horizontally from the edge of the table; see Fig. 5. Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?
Answer:
Let’s put the origin of our coordinate system (for the motion of the marble) at the edge of the table. With this choice of coordinates, the object of the game is to insure that the x coordinate of the marble is 2.20m when it reaches the level of the floor.
There are many things we are not told in this problem! We don’t know the spring constant for the gun, or the mass of the marble. We don’t know the height of the table above the floor, either!
When the gun propels the marble, the spring is initially compressed and the marble is motionless (see Fig. 6(a).) The energy of the system here is the energy stored in the spring, Ei = ½ kx2, where k is the force constant of the spring and x is the amount of compression of the spring.) When the spring has returned to its natural length and has given the marble a speed v, then the energy of the system is Ef = ½ mv2. If we can neglect friction then mechanical energy is conserved during the firing, so that Ef = Ei, which gives us:
½ mv2 = ½ kx2
v = [kx2/m]1/2 = x[k/m]1/2
We will let x and v be the compression and initial marble speed for Bobby’s attempt. Then
we have:
v = (1.10 × 10−2 m)[k/m] (3)
The marble’s trip from the edge of the table to the floor is (by now!) a fairly simple kinematics problem. If the time the marble spends in the air is t and the height of the table is h then the equation for vertical motion tells us:
h = ½ gt2
(This is true because the marble’s initial velocity is all horizontal. We do know that on Bobby’s try, the marble’s x coordinate at impact was
x = 2.20 m − 0.27 m = 1.93 m
and since the horizontal velocity of the marble is v, we have:
vt = 1.93m
There are too many unknowns to solve for k, v, h and t. . . but let’s go on.
Let’s suppose that Rhoda compresses the spring by an amount x’ so that the marble is given a speed v’. As before, we have
½ mv’2 = ½ kx’2
(it’s the same spring and marble so that k and m are the same) and this gives:
v' = x’[k/m]
Now when Rhoda’s shot goes off the table and through the air, then if its time of flight is t’ then the equation for vertical motion gives us:
h = ½ gt’2
This is the same equation as for t, so that the times of flight for both shots is the same: t’ = t. Since the x coordinate of the marble for Rhoda’s shot will be x = 2.20 m, the equation for horizontal motion gives us
v’t = 2.20m (6)
What can we do with these equations? If we divide Eq. 6 by Eq. 4 we get:
v‘t/vt = v’/v = 2.20/1.93 = 1.14
If we divide Eq. 5 by Eq. 3 we get:
v'/v = x’[k/m]1/2 /{(1.10 × 10−2 m)[k/m]1/2 = x’/(1.10 × 10−2 m)
With these last two results, we can solve for x’. Combining these equations gives:
1.14 = x’/(1.10 × 10−2 m)
x' = 1.14(1.10 × 10−2 m) = 1.25 cm
Rhoda should compress the spring by 1.25 cm in order to score a direct hit.
Problem #5
A small boy sits at the top of a frictionless hemisphere of radius R, as shown in Fig. 7.(a). He starts to slip down one side of it and at some angle θ measured from the vertical he loses contact with the surface, as shown in Fig. 7.(b). Find the angle θ.
Answer:
This is a classic and somewhat challenging problem but it does not require any more math than simple trig. We focus on the point at which the loss of contact occurs. While the boy starts from rest at the top of the sphere, his speed is v at this point. As mentioned, his position is given by the angle θ as measured form the vertical so that while his initial height was R it is now R cos θ; see Fig. 8. Since there are no friction forces acting, the boy’s total energy is conserved between the initial position at the top and the final position at θ, that is:
Ki + Ui = Kf + Uf
Using U = mgy this is
0 + mgR = ½mv2 + mgR cosθ
where m is the mass of the boy. Do a little algebra on this and get:
½ mv2 = mgR – mgR cosθ = mgR(1 – cos θ)
v2 = 2gR(1− cos θ)
The condition that the boy “loses contact” with the surface means that there is no normal force from the surface on the boy at some point. We know that the surface exerts no sideways (tangential) force on the boy because there is no friction. But in general the surface can push outward on the boy with some force FN. (It cannot pull inward, and that is why the boy will fly off at some point.) The forces on the boy when he is at position θ are shown in Fig. 8. Gravity mg pulls downward and the normal force FN pushes outward.
The force of gravity has been split into components: Doing a little geometry, we can see that a component mgcosθ points inward toward the center of the sphere and a component mgsinθ points tangentially. Now at the moment the boy loses contact he is following the path of a circle (with speed v) so that the net force in the inward direction is the centripetal force, mv2/R. (It is true that there is also a tangential force but that doesn’t matter.) According to our force diagram the net force in the inward direction is
Fc = mg cos θ − FN
and so
mv2/R = mg cos θ − FN
But at the moment he loses contact, FN = 0 so we get
mv2/R = mg cosθ
v2 = Rg cosθ
Comparing our results for energy conservation and the centripetal force, we got two expressions for v2. If we equate them, we get
2gR(1 − cosθ) = Rg cosθ
Cancel the Rg and do some algebra:
2(1 – cos θ) = cos θ
2 = 3 cosθ
Finally,
Cos θ = 2/3
θ = 48.2◦
A 1.50 kg snowball is shot upward at an angle of 340 to the horizontal with an initial speed of 20.0 m/s. (a) What is its initial kinetic energy? (b) By how much does the gravitational potential energy of the snowball–Earth system change as the snowball moves from the launch point to the point of maximum height? (c) What is that maximum height?
Answer:
(a) Since the initial speed of the snowball is 20.0 m/s , we have its initial kinetic energy:
Ki = ½ mv02 = ½ (1.50 kg)(20.0 m/s)2 = 300 J
(b) We need to remember that since this projectile was not fired straight up, it will still have some kinetic energy when it gets to maximum height! That means we have to think a little harder before applying energy principles to answer this question.
At maximum height, we know that the y component of the snowball’s velocity is zero. The x component is not zero.
But we do know that since a projectile has no horizontal acceleration, the x component will remain constant; it will keep its initial value of
v0x = v0 cos θ0 = (20.0 m/s) cos 340 = 16.6 m/s
so the speed of the snowball at maximum height is 16.6 m/s . At maximum height, (the final position) the kinetic energy is
Kf = ½ mvf2 = ½ (1.50 kg)(16.6 m/s)2 = 206 J
In this problem there are only conservative forces (namely, gravity). The mechanical energy is conserved:
Ki + Ui = Kf + Uf
We already found the initial kinetic energy of the snowball: Ki = 300 J. Using Ugrav = mgy (with y = 0 at ground level), the initial potential energy is Ui = 0. Then we can find the final potential energy of the snowball:
Uf = Ki + Ui −Kf
= 300 J + 0 – 206 J
= 94 J
The final gravitational potential energy of the snowball–earth system (a long–winded way of saying what U is!) is then 94 J. (Since its original value was zero, this is the answer to part (b).)
(c) If we call the maximum height of the snowball h, then we have
Uf = mgh
Solve for h:
h = Uf/mg = (94 J)/(1.5 kg)(9.80 m/s2) = 6.38 m
The maximum height of the snowball is 6.38 m.
Problem #2
 |
| Fig.1 |
Answer:
(a) The condition of the pendulum when the stone passes the lowest point is shown in Fig. 1(a). Throughout the problem we will measure the height y of the stone from the bottom of its swing. Then at the bottom of the swing the stone has zero potential energy, while its kinetic energy is
Ki = ½ mv02 = ½ (2.0 kg)(8.0 m/s)2 = 64 J
When the stone has swung up by 600 (as in Fig. 1(b)) it has some potential energy. To figure out how much, we need to calculate the height of the stone above the lowest point of the swing. By simple geometry, the stone’s position is
(4.0 m) cos 600 = 2.0 m
down from the top of the string, so it must be
4.0 m − 2.0 m = 2.0 m
up from the lowest point. So its potential energy at this point is
Uf = mgy = (2.0 kg)(9.80 m/s2)(2.0 m) = 39.2 J
It will also have a kinetic energy Kf = ½ mvf2, where vf is the final speed.
Now in this system there are only a conservative force acting on the particle of interest, i.e. the stone. (We should note that the string tension also acts on the stone, but since it always pulls perpendicularly to the motion of the stone, it does no work.) So the total mechanical energy of the stone is conserved:
Ki + Ui = Kf + Uf
We can substitute the values found above to get:
64.0 J + 0 = ½ (2.0kg)vf2 + 39.2J
which we can solve for vf :
(1.0kg)vf2 = 64.0 J − 39.2 J = 24.8 J
vf2 = 24.8 m2/s2
and then:
vf = 5.0 m/s
The speed of the stone at the 600 position will be 5.0 m/s.
 |
| Fig.2 |
Now at this point of maximum height, the speed of the mass is instantaneously zero. So in this final position, the kinetic energy is Kf = 0. Its height above the starting position is
y = 4.0 m + (4.0 m) sin θ = (4.0 m)(1 + sin θ) (1)
so that its potential energy there is
Uf = mgyf = (2.0 kg)(9.80 m/s2)(4.0 m)(1 + sin θ) = (78.4 J)(1 + sin θ)
We use the conservation of mechanical energy (from the position at the bottom of the swing)to find θ: Ki + Ui = Kf + Uf , so:
Uf = Ki + Ui − Kf
(78.4 J)(1 + sin θ) = 64 J+ 0 − 0
This gives us:
1 + sin θ = 78.4J/64 J = 1.225
sin θ = 0.225
and finally
θ = 130
We do get a sensible answer of θ so we were right in writing down Eq. 1. Actually this equation would also have been correct if θ were negative and the pendulum reached its highest point with the string below the horizontal.
Problem #3
 |
| Fig.3 |
Answer:
In this problem, there are no friction forces acting on the particle (the bead). Gravity acts on it and gravity is a conservative force. The track will exert a normal forces on the bead, but this force does no work. So the total energy of the bead —kinetic plus (gravitational) potential energy— will be conserved.
At the initial position, when the bead is released, the bead has no speed; Ki = 0. But it is at a height h above the bottom of the track. If we agree to measure height from the bottom of the track, then the initial potential energy of the bead is
Ui = mgh
where m = 5.00 g is the mass of the bead.
At the final position (A), the bead has both kinetic and potential energy. If the bead’s speed at A is v, then its final kinetic energy is Kf = ½ mv2. At position A its height is 2R (it is a full diameter above the “ground level” of the track) so its potential energy is
Uf = mg(2R) = 2mgR .
The total energy of the bead is conserved: Ki + Ui = Kf + Uf . This gives us:
0 + mgh = ½ mv2 + 2mgR ,
where we want to solve for v (the speed at A). The mass m cancels out, giving:
gh = ½v2 = 2gR
½v2 = gh − 2gR = g(h − 2R)
and then
v2 = 2g(h − 2R) = 2g(3.50R − 2R) = 2g(1.5R) = 3.0gR (2)
and finally
v = [3.0gR]1/2
Since we don’t have a numerical value for R, that’s as far as we can go. In the next part of the problem, we think about the forces acting on the bead at point A.
 |
| Fig.4 |
At the top of the track the bead is moving on a circular path of radius R, with speed v. So it is accelerating toward the center of the circle, namely downward. We know that the downward forces must add up to give the centripetal force mv2/R:
mg + N = mv2/R
n = mv2/R − mg = m(v2/R – g)
.
But we can use our result from Eq. 2 to substitute for v2. This gives:
N = m[3.0gR/R – g] = m(2g) = 2mg
Plug in the numbers:
N = 2(5.00 × 10−3 kg)(9.80 m/s2) = 9.80 × 10−2 N
At point A the track is pushing downward with a force of 9.80 × 10−2 N.
Problem #4
Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring –loaded gun that is mounted on the table. The target box is 2.20m horizontally from the edge of the table; see Fig. 5. Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?
 |
| Fig.5 |
Let’s put the origin of our coordinate system (for the motion of the marble) at the edge of the table. With this choice of coordinates, the object of the game is to insure that the x coordinate of the marble is 2.20m when it reaches the level of the floor.
 |
| Fig.6 |
When the gun propels the marble, the spring is initially compressed and the marble is motionless (see Fig. 6(a).) The energy of the system here is the energy stored in the spring, Ei = ½ kx2, where k is the force constant of the spring and x is the amount of compression of the spring.) When the spring has returned to its natural length and has given the marble a speed v, then the energy of the system is Ef = ½ mv2. If we can neglect friction then mechanical energy is conserved during the firing, so that Ef = Ei, which gives us:
½ mv2 = ½ kx2
v = [kx2/m]1/2 = x[k/m]1/2
We will let x and v be the compression and initial marble speed for Bobby’s attempt. Then
we have:
v = (1.10 × 10−2 m)[k/m] (3)
The marble’s trip from the edge of the table to the floor is (by now!) a fairly simple kinematics problem. If the time the marble spends in the air is t and the height of the table is h then the equation for vertical motion tells us:
h = ½ gt2
(This is true because the marble’s initial velocity is all horizontal. We do know that on Bobby’s try, the marble’s x coordinate at impact was
x = 2.20 m − 0.27 m = 1.93 m
and since the horizontal velocity of the marble is v, we have:
vt = 1.93m
There are too many unknowns to solve for k, v, h and t. . . but let’s go on.
Let’s suppose that Rhoda compresses the spring by an amount x’ so that the marble is given a speed v’. As before, we have
½ mv’2 = ½ kx’2
(it’s the same spring and marble so that k and m are the same) and this gives:
v' = x’[k/m]
Now when Rhoda’s shot goes off the table and through the air, then if its time of flight is t’ then the equation for vertical motion gives us:
h = ½ gt’2
This is the same equation as for t, so that the times of flight for both shots is the same: t’ = t. Since the x coordinate of the marble for Rhoda’s shot will be x = 2.20 m, the equation for horizontal motion gives us
v’t = 2.20m (6)
What can we do with these equations? If we divide Eq. 6 by Eq. 4 we get:
v‘t/vt = v’/v = 2.20/1.93 = 1.14
If we divide Eq. 5 by Eq. 3 we get:
v'/v = x’[k/m]1/2 /{(1.10 × 10−2 m)[k/m]1/2 = x’/(1.10 × 10−2 m)
With these last two results, we can solve for x’. Combining these equations gives:
1.14 = x’/(1.10 × 10−2 m)
x' = 1.14(1.10 × 10−2 m) = 1.25 cm
Rhoda should compress the spring by 1.25 cm in order to score a direct hit.
Problem #5
 |
| Fig.7 |
Answer:
 |
| Fig.8 |
Ki + Ui = Kf + Uf
Using U = mgy this is
0 + mgR = ½mv2 + mgR cosθ
where m is the mass of the boy. Do a little algebra on this and get:
½ mv2 = mgR – mgR cosθ = mgR(1 – cos θ)
v2 = 2gR(1− cos θ)
The condition that the boy “loses contact” with the surface means that there is no normal force from the surface on the boy at some point. We know that the surface exerts no sideways (tangential) force on the boy because there is no friction. But in general the surface can push outward on the boy with some force FN. (It cannot pull inward, and that is why the boy will fly off at some point.) The forces on the boy when he is at position θ are shown in Fig. 8. Gravity mg pulls downward and the normal force FN pushes outward.
 |
| Fig.9 |
Fc = mg cos θ − FN
and so
mv2/R = mg cos θ − FN
But at the moment he loses contact, FN = 0 so we get
mv2/R = mg cosθ
v2 = Rg cosθ
Comparing our results for energy conservation and the centripetal force, we got two expressions for v2. If we equate them, we get
2gR(1 − cosθ) = Rg cosθ
Cancel the Rg and do some algebra:
2(1 – cos θ) = cos θ
2 = 3 cosθ
Finally,
Cos θ = 2/3
θ = 48.2◦
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