Final Exam

Physics 6A

3/18/2003

You must show your work in order to obtain credit. Write your name and include this sheet with your exam papers.

1. A stone is thrown vertically upward. On its way up it passes point $A$ with speed $v$, and point $B$, $6 m$ higher than $A$ with speed $v/6$. Calculate

(a)(10 points) the speed $v$ and

(b)(10 points) the maximum height reached by the stone above point $B$ (that is, the distance between the maximum height at point B).

2. The two blocks shown below are free to move. The coefficient of static friction between the two blocks is $\mu_s = 0.47$ but the surface beneath $M$ is frictionless. Take $m = 8 kg$ and $M = 81 kg$.

(a) (5 points) Draw free body diagrams for the two blocks.

(b) (15 points) What is the minimum horizontal force $F$ required to hold $m$ against $M$?

3.(20 points) A uniform cylinder of radius $R = 0.75 m$ and mass $M = 5 kg$ rotates about a vertical axis on frictionless bearings as shown below. A light cord passes around the cylinder and then over a uniform cylindrical pulley having a moment of inertia $I = 0.02 kgm^2$ and radius $r = 0.13 m$. The cord is attached to a small object of mass $m = 3 kg$ that falls under the influence of gravity. Calculate the speed of the object after it has fallen a height $h = 10 m$ from rest. Hint: use conservation of energy.

4. A block of mass $M = 0.9 kg$, at rest on a horizontal frictionless surface, is attached to a rigid support by a spring of constant $k = 176 N/m$. A bullet of mass $m = 0.12 kg$ and velocity $v = 395 m/s$ strikes the block as shown. The bullet remains embedded in the block. Determine

(a)(10 points) the velocity of the block immediately after the collision and

(b)(10 points) the amplitude of the resulting simple harmonic motion. (Ignore any possibility of the block hitting a wall.)

5.(20 points) (You must explain your reasoning to obtain credit). A system of masses and pulleys is shown below. All pulleys and ropes are massless and frictionless. The tension in the rope connected to $m_1$ is

(a)

$g (m_1+m_2)$

(b)

$m_1 g$

(c)

$g(m_1+m_2+m_3+m_4)/4$

(d)

${2g \over 1/m_1 +1/m_2}$

(e)

${4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$

Hint: Don't try to work this out from scratch! Eliminate answers by looking at various limits, e.g. what would happen if $m_1$ were to equal ....

USEFUL INFORMATION Moment of inertia of a solid cylinder of radius $R$ and mass $M$ about center of mass $= \ MR^2 /2$. The acceleration of gravity $g = 9.8 m/s^2$.

Solutions

1. Here $n = 6$ and $h = 6$

(a) Find speed.

Use conservation of energy. $E_i = {1\over 2}m v^2 = E_f = mgh + {1\over 2}m (v/n)^2$. Solving for $v$:

$$v^2 = {2gh\over 1-({1\over n})^2}$$ (1)

or $v = 11m/s$.

(b) Again, use conservation of energy. At the maximum height $v=0$ so $E_i = {1\over 2}m v^2 = mgh_{max}$. So

$$h_{max} = {v^2\over 2g} = {2gh\over 1-({1\over 6})^2}{1 \over 2g} = {h\over 1-({1\over 6})^2} = 6.17m.$$ (2)

Relative to point B, we must subtract $h$ giving the answer

$${h\over n^2 -1} = 0.171m.$$ (3)

2.

Here $m = 8$, $\mu_s = 0.47$, and $M = 81$. (a)

(b) From part (a) and $F_{net} = ma$ in component form, for mass $m$, in the $y$ direction: $f_s - mg = ma_y = 0$, and in the $x$ direction: $F-N_1 = ma$.

For mass $M$ in the $x$ direction: $F_{net,x,M} = N_1 = Ma$. Eliminating $N_1$ $F-Ma = ma$ or

$$a = {F\over M+m}$$ (4)

Solving for $N_1$:

$$N_1 = Ma = {M\over M+m}F$$ (5)

At the point of slipping, $f_s = \mu_s N_1$, so from above, $\mu_s N_1 = mg$ or

$$\mu_s {M\over M+m}F = mg$$ (6)

so

$$F = {M+m\over M} {mg\over \mu_s} = (1+{m\over M}) {mg\over \mu_s} = 183N.$$ (7)

3. Here $m = 3$, $M = 5$, $r = 0.13$, $R = 0.75$, $I = 0.02$, $h = 10$.

Initially, the potential energy $U_i = 0$ and the kinetic energy $K_i = 0$. So the total energy is zero.

Finally, $E_f = 0 = U_f + K_f$

$U_f = -mgh$. The kinetic energy is the sum of three terms. The kinetic energy of mass $m$ and the rotational kinetic energy of the two cylinders. Rotational kinetic energy is of the form ${1\over 2}I \omega^2$. Here the velocities of the rims of the two cylinders are the same, though their angular velocities may be different. So $v = \omega r$, and solving for $\omega$ we can write the rotational kinetic energy as ${1\over 2}(I/r^2)v^2$. The moment of inertia of cylinder is ${1\over 2}M R^2$. So the total kinetic energy is ${1\over 2}mv^2 + {1\over 2}I {v^2\over r^2} + {1\over 2}({1\over 2}M R^2) {v^2\over R^2}$. Adding $-mgh$ to this at setting the whole expression to zero, we can solve for $v$:

$$v = \sqrt{2mgh\over m + (I/r^2) + {1\over 2}M} = 9.38 m/s$$ (8)

4. Here $M = 0.9$, $k = 176$, $m = 0.12$, and $v = 395$.

(a) Call the velocities before and after the collision $v_b$ and $v_a$ respectively. Conservation of momentum gives:

395 $m v_b = (m+M)v_a$ so

$$v_a = {m\over m+M}v_b = {m\over m+M}v =46.5m/s$$ (9)

(b) After this collision, energy is conserved. At maximum compression the distance gone is at its maximum, hence this is the amplitude $A$ of oscillation. At this point, the kinetic energy is zero. So ${1\over 2}k A^2 = {1\over 2}(M+m)v_a^2$. Solving for $A$:

$$A = \sqrt{M+m\over k} v_a = \sqrt{M+m\over k} {m\over M+m} v = 3.54m$$ (10)

5. If $m_1$ or $m_2$ or $m_3$ or $m_4$ equals 0, the system will have an acceleration of $g$ and there will be no tension in the ropes. The only answer consistent with this is

$$T= {4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$$ (11)

Final Exam

Physics 6A

3/18/2003

You must show your work in order to obtain credit. Write your name and include this sheet with your exam papers.

1. A stone is thrown vertically upward. On its way up it passes point $A$ with speed $v$, and point $B$, $1 m$ higher than $A$ with speed $v/6$. Calculate

(a)(10 points) the speed $v$ and

(b)(10 points) the maximum height reached by the stone above point $B$ (that is, the distance between the maximum height at point B).

2. The two blocks shown below are free to move. The coefficient of static friction between the two blocks is $\mu_s = 0.34$ but the surface beneath $M$ is frictionless. Take $m = 17 kg$ and $M = 80 kg$.

(a) (5 points) Draw free body diagrams for the two blocks.

(b) (15 points) What is the minimum horizontal force $F$ required to hold $m$ against $M$?

3.(20 points) A uniform cylinder of radius $R = 0.96 m$ and mass $M = 11 kg$ rotates about a vertical axis on frictionless bearings as shown below. A light cord passes around the cylinder and then over a uniform cylindrical pulley having a moment of inertia $I = 0.21 kgm^2$ and radius $r = 0.63 m$. The cord is attached to a small object of mass $m = 2 kg$ that falls under the influence of gravity. Calculate the speed of the object after it has fallen a height $h = 6 m$ from rest. Hint: use conservation of energy.

4. A block of mass $M = 0.5 kg$, at rest on a horizontal frictionless surface, is attached to a rigid support by a spring of constant $k = 116 N/m$. A bullet of mass $m = 0.1 kg$ and velocity $v = 351 m/s$ strikes the block as shown. The bullet remains embedded in the block. Determine

(a)(10 points) the velocity of the block immediately after the collision and

(b)(10 points) the amplitude of the resulting simple harmonic motion. (Ignore any possibility of the block hitting a wall.)

5.(20 points) (You must explain your reasoning to obtain credit). A system of masses and pulleys is shown below. All pulleys and ropes are massless and frictionless. The tension in the rope connected to $m_1$ is

(a)

${4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$

(b)

$g(m_1+m_2+m_3+m_4)/4$

(c)

${2g \over 1/m_1 +1/m_2}$

(d)

$g (m_1+m_2)$

(e)

$m_1 g$

Hint: Don't try to work this out from scratch! Eliminate answers by looking at various limits, e.g. what would happen if $m_1$ were to equal ....

USEFUL INFORMATION Moment of inertia of a solid cylinder of radius $R$ and mass $M$ about center of mass $= \ MR^2 /2$. The acceleration of gravity $g = 9.8 m/s^2$.

Solutions

1. Here $n = 6$ and $h = 1$

(a) Find speed.

Use conservation of energy. $E_i = {1\over 2}m v^2 = E_f = mgh + {1\over 2}m (v/n)^2$. Solving for $v$:

$$v^2 = {2gh\over 1-({1\over n})^2}$$ (12)

or $v = 4.49m/s$.

(b) Again, use conservation of energy. At the maximum height $v=0$ so $E_i = {1\over 2}m v^2 = mgh_{max}$. So

$$h_{max} = {v^2\over 2g} = {2gh\over 1-({1\over 6})^2}{1 \over 2g} = {h\over 1-({1\over 6})^2} = 1.03m.$$ (13)

Relative to point B, we must subtract $h$ giving the answer

$${h\over n^2 -1} = 0.0286m.$$ (14)

2.

Here $m = 17$, $\mu_s = 0.34$, and $M = 80$. (a)

(b) From part (a) and $F_{net} = ma$ in component form, for mass $m$, in the $y$ direction: $f_s - mg = ma_y = 0$, and in the $x$ direction: $F-N_1 = ma$.

For mass $M$ in the $x$ direction: $F_{net,x,M} = N_1 = Ma$. Eliminating $N_1$ $F-Ma = ma$ or

$$a = {F\over M+m}$$ (15)

Solving for $N_1$:

$$N_1 = Ma = {M\over M+m}F$$ (16)

At the point of slipping, $f_s = \mu_s N_1$, so from above, $\mu_s N_1 = mg$ or

$$\mu_s {M\over M+m}F = mg$$ (17)

so

$$F = {M+m\over M} {mg\over \mu_s} = (1+{m\over M}) {mg\over \mu_s} = 594N.$$ (18)

3. Here $m = 2$, $M = 11$, $r = 0.63$, $R = 0.96$, $I = 0.21$, $h = 6$.

Initially, the potential energy $U_i = 0$ and the kinetic energy $K_i = 0$. So the total energy is zero.

Finally, $E_f = 0 = U_f + K_f$

$U_f = -mgh$. The kinetic energy is the sum of three terms. The kinetic energy of mass $m$ and the rotational kinetic energy of the two cylinders. Rotational kinetic energy is of the form ${1\over 2}I \omega^2$. Here the velocities of the rims of the two cylinders are the same, though their angular velocities may be different. So $v = \omega r$, and solving for $\omega$ we can write the rotational kinetic energy as ${1\over 2}(I/r^2)v^2$. The moment of inertia of cylinder is ${1\over 2}M R^2$. So the total kinetic energy is ${1\over 2}mv^2 + {1\over 2}I {v^2\over r^2} + {1\over 2}({1\over 2}M R^2) {v^2\over R^2}$. Adding $-mgh$ to this at setting the whole expression to zero, we can solve for $v$:

$$v = \sqrt{2mgh\over m + (I/r^2) + {1\over 2}M} = 5.41 m/s$$ (19)

4. Here $M = 0.5$, $k = 116$, $m = 0.1$, and $v = 351$.

(a) Call the velocities before and after the collision $v_b$ and $v_a$ respectively. Conservation of momentum gives:

351 $m v_b = (m+M)v_a$ so

$$v_a = {m\over m+M}v_b = {m\over m+M}v =58.5m/s$$ (20)

(b) After this collision, energy is conserved. At maximum compression the distance gone is at its maximum, hence this is the amplitude $A$ of oscillation. At this point, the kinetic energy is zero. So ${1\over 2}k A^2 = {1\over 2}(M+m)v_a^2$. Solving for $A$:

$$A = \sqrt{M+m\over k} v_a = \sqrt{M+m\over k} {m\over M+m} v = 4.21m$$ (21)

5. If $m_1$ or $m_2$ or $m_3$ or $m_4$ equals 0, the system will have an acceleration of $g$ and there will be no tension in the ropes. The only answer consistent with this is

$$T= {4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$$ (22)

Final Exam

Physics 6A

3/18/2003

You must show your work in order to obtain credit. Write your name and include this sheet with your exam papers.

1. A stone is thrown vertically upward. On its way up it passes point $A$ with speed $v$, and point $B$, $12 m$ higher than $A$ with speed $v/5$. Calculate

(a)(10 points) the speed $v$ and

(b)(10 points) the maximum height reached by the stone above point $B$ (that is, the distance between the maximum height at point B).

2. The two blocks shown below are free to move. The coefficient of static friction between the two blocks is $\mu_s = 0.2$ but the surface beneath $M$ is frictionless. Take $m = 7 kg$ and $M = 79 kg$.

(a) (5 points) Draw free body diagrams for the two blocks.

(b) (15 points) What is the minimum horizontal force $F$ required to hold $m$ against $M$?

3.(20 points) A uniform cylinder of radius $R = 0.18 m$ and mass $M = 7 kg$ rotates about a vertical axis on frictionless bearings as shown below. A light cord passes around the cylinder and then over a uniform cylindrical pulley having a moment of inertia $I = 0.01 kgm^2$ and radius $r = 0.12 m$. The cord is attached to a small object of mass $m = 2 kg$ that falls under the influence of gravity. Calculate the speed of the object after it has fallen a height $h = 2 m$ from rest. Hint: use conservation of energy.

4. A block of mass $M = 0.7 kg$, at rest on a horizontal frictionless surface, is attached to a rigid support by a spring of constant $k = 156 N/m$. A bullet of mass $m = 0.18 kg$ and velocity $v = 306 m/s$ strikes the block as shown. The bullet remains embedded in the block. Determine

(a)(10 points) the velocity of the block immediately after the collision and

(b)(10 points) the amplitude of the resulting simple harmonic motion. (Ignore any possibility of the block hitting a wall.)

5.(20 points) (You must explain your reasoning to obtain credit). A system of masses and pulleys is shown below. All pulleys and ropes are massless and frictionless. The tension in the rope connected to $m_1$ is

(a)

$g(m_1+m_2+m_3+m_4)/4$

(b)

$g (m_1+m_2)$

(c)

$m_1 g$

(d)

${4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$

(e)

${2g \over 1/m_1 +1/m_2}$

Hint: Don't try to work this out from scratch! Eliminate answers by looking at various limits, e.g. what would happen if $m_1$ were to equal ....

USEFUL INFORMATION Moment of inertia of a solid cylinder of radius $R$ and mass $M$ about center of mass $= \ MR^2 /2$. The acceleration of gravity $g = 9.8 m/s^2$.

Solutions

1. Here $n = 5$ and $h = 12$

(a) Find speed.

Use conservation of energy. $E_i = {1\over 2}m v^2 = E_f = mgh + {1\over 2}m (v/n)^2$. Solving for $v$:

$$v^2 = {2gh\over 1-({1\over n})^2}$$ (23)

or $v = 15.7m/s$.

(b) Again, use conservation of energy. At the maximum height $v=0$ so $E_i = {1\over 2}m v^2 = mgh_{max}$. So

$$h_{max} = {v^2\over 2g} = {2gh\over 1-({1\over 5})^2}{1 \over 2g} = {h\over 1-({1\over 5})^2} = 12.5m.$$ (24)

Relative to point B, we must subtract $h$ giving the answer

$${h\over n^2 -1} = 0.5m.$$ (25)

2.

Here $m = 7$, $\mu_s = 0.2$, and $M = 79$. (a)

(b) From part (a) and $F_{net} = ma$ in component form, for mass $m$, in the $y$ direction: $f_s - mg = ma_y = 0$, and in the $x$ direction: $F-N_1 = ma$.

For mass $M$ in the $x$ direction: $F_{net,x,M} = N_1 = Ma$. Eliminating $N_1$ $F-Ma = ma$ or

$$a = {F\over M+m}$$ (26)

Solving for $N_1$:

$$N_1 = Ma = {M\over M+m}F$$ (27)

At the point of slipping, $f_s = \mu_s N_1$, so from above, $\mu_s N_1 = mg$ or

$$\mu_s {M\over M+m}F = mg$$ (28)

so

$$F = {M+m\over M} {mg\over \mu_s} = (1+{m\over M}) {mg\over \mu_s} = 373N.$$ (29)

3. Here $m = 2$, $M = 7$, $r = 0.12$, $R = 0.18$, $I = 0.01$, $h = 2$.

Initially, the potential energy $U_i = 0$ and the kinetic energy $K_i = 0$. So the total energy is zero.

Finally, $E_f = 0 = U_f + K_f$

$U_f = -mgh$. The kinetic energy is the sum of three terms. The kinetic energy of mass $m$ and the rotational kinetic energy of the two cylinders. Rotational kinetic energy is of the form ${1\over 2}I \omega^2$. Here the velocities of the rims of the two cylinders are the same, though their angular velocities may be different. So $v = \omega r$, and solving for $\omega$ we can write the rotational kinetic energy as ${1\over 2}(I/r^2)v^2$. The moment of inertia of cylinder is ${1\over 2}M R^2$. So the total kinetic energy is ${1\over 2}mv^2 + {1\over 2}I {v^2\over r^2} + {1\over 2}({1\over 2}M R^2) {v^2\over R^2}$. Adding $-mgh$ to this at setting the whole expression to zero, we can solve for $v$:

$$v = \sqrt{2mgh\over m + (I/r^2) + {1\over 2}M} = 3.56 m/s$$ (30)

4. Here $M = 0.7$, $k = 156$, $m = 0.18$, and $v = 306$.

(a) Call the velocities before and after the collision $v_b$ and $v_a$ respectively. Conservation of momentum gives:

306 $m v_b = (m+M)v_a$ so

$$v_a = {m\over m+M}v_b = {m\over m+M}v =62.6m/s$$ (31)

(b) After this collision, energy is conserved. At maximum compression the distance gone is at its maximum, hence this is the amplitude $A$ of oscillation. At this point, the kinetic energy is zero. So ${1\over 2}k A^2 = {1\over 2}(M+m)v_a^2$. Solving for $A$:

$$A = \sqrt{M+m\over k} v_a = \sqrt{M+m\over k} {m\over M+m} v = 4.7m$$ (32)

5. If $m_1$ or $m_2$ or $m_3$ or $m_4$ equals 0, the system will have an acceleration of $g$ and there will be no tension in the ropes. The only answer consistent with this is

$$T= {4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$$ (33)

Final Exam

Physics 6A

3/18/2003

You must show your work in order to obtain credit. Write your name and include this sheet with your exam papers.

1. A stone is thrown vertically upward. On its way up it passes point $A$ with speed $v$, and point $B$, $7 m$ higher than $A$ with speed $v/4$. Calculate

(a)(10 points) the speed $v$ and

(b)(10 points) the maximum height reached by the stone above point $B$ (that is, the distance between the maximum height at point B).

2. The two blocks shown below are free to move. The coefficient of static friction between the two blocks is $\mu_s = 0.57$ but the surface beneath $M$ is frictionless. Take $m = 15 kg$ and $M = 78 kg$.

(a) (5 points) Draw free body diagrams for the two blocks.

(b) (15 points) What is the minimum horizontal force $F$ required to hold $m$ against $M$?

3.(20 points) A uniform cylinder of radius $R = 0.4 m$ and mass $M = 14 kg$ rotates about a vertical axis on frictionless bearings as shown below. A light cord passes around the cylinder and then over a uniform cylindrical pulley having a moment of inertia $I = 0.13 kgm^2$ and radius $r = 0.62 m$. The cord is attached to a small object of mass $m = 1 kg$ that falls under the influence of gravity. Calculate the speed of the object after it has fallen a height $h = 8 m$ from rest. Hint: use conservation of energy.

4. A block of mass $M = 0.8 kg$, at rest on a horizontal frictionless surface, is attached to a rigid support by a spring of constant $k = 196 N/m$. A bullet of mass $m = 0.16 kg$ and velocity $v = 362 m/s$ strikes the block as shown. The bullet remains embedded in the block. Determine

(a)(10 points) the velocity of the block immediately after the collision and

(b)(10 points) the amplitude of the resulting simple harmonic motion. (Ignore any possibility of the block hitting a wall.)

5.(20 points) (You must explain your reasoning to obtain credit). A system of masses and pulleys is shown below. All pulleys and ropes are massless and frictionless. The tension in the rope connected to $m_1$ is

(a)

$g(m_1+m_2+m_3+m_4)/4$

(b)

${4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$

(c)

$m_1 g$

(d)

${2g \over 1/m_1 +1/m_2}$

(e)

$g (m_1+m_2)$

Hint: Don't try to work this out from scratch! Eliminate answers by looking at various limits, e.g. what would happen if $m_1$ were to equal ....

USEFUL INFORMATION Moment of inertia of a solid cylinder of radius $R$ and mass $M$ about center of mass $= \ MR^2 /2$. The acceleration of gravity $g = 9.8 m/s^2$.

Solutions

1. Here $n = 4$ and $h = 7$

(a) Find speed.

Use conservation of energy. $E_i = {1\over 2}m v^2 = E_f = mgh + {1\over 2}m (v/n)^2$. Solving for $v$:

$$v^2 = {2gh\over 1-({1\over n})^2}$$ (34)

or $v = 12.1m/s$.

(b) Again, use conservation of energy. At the maximum height $v=0$ so $E_i = {1\over 2}m v^2 = mgh_{max}$. So

$$h_{max} = {v^2\over 2g} = {2gh\over 1-({1\over 4})^2}{1 \over 2g} = {h\over 1-({1\over 4})^2} = 7.47m.$$ (35)

Relative to point B, we must subtract $h$ giving the answer

$${h\over n^2 -1} = 0.467m.$$ (36)

2.

Here $m = 15$, $\mu_s = 0.57$, and $M = 78$. (a)

(b) From part (a) and $F_{net} = ma$ in component form, for mass $m$, in the $y$ direction: $f_s - mg = ma_y = 0$, and in the $x$ direction: $F-N_1 = ma$.

For mass $M$ in the $x$ direction: $F_{net,x,M} = N_1 = Ma$. Eliminating $N_1$ $F-Ma = ma$ or

$$a = {F\over M+m}$$ (37)

Solving for $N_1$:

$$N_1 = Ma = {M\over M+m}F$$ (38)

At the point of slipping, $f_s = \mu_s N_1$, so from above, $\mu_s N_1 = mg$ or

$$\mu_s {M\over M+m}F = mg$$ (39)

so

$$F = {M+m\over M} {mg\over \mu_s} = (1+{m\over M}) {mg\over \mu_s} = 307N.$$ (40)

3. Here $m = 1$, $M = 14$, $r = 0.62$, $R = 0.4$, $I = 0.13$, $h = 8$.

Initially, the potential energy $U_i = 0$ and the kinetic energy $K_i = 0$. So the total energy is zero.

Finally, $E_f = 0 = U_f + K_f$

$U_f = -mgh$. The kinetic energy is the sum of three terms. The kinetic energy of mass $m$ and the rotational kinetic energy of the two cylinders. Rotational kinetic energy is of the form ${1\over 2}I \omega^2$. Here the velocities of the rims of the two cylinders are the same, though their angular velocities may be different. So $v = \omega r$, and solving for $\omega$ we can write the rotational kinetic energy as ${1\over 2}(I/r^2)v^2$. The moment of inertia of cylinder is ${1\over 2}M R^2$. So the total kinetic energy is ${1\over 2}mv^2 + {1\over 2}I {v^2\over r^2} + {1\over 2}({1\over 2}M R^2) {v^2\over R^2}$. Adding $-mgh$ to this at setting the whole expression to zero, we can solve for $v$:

$$v = \sqrt{2mgh\over m + (I/r^2) + {1\over 2}M} = 4.34 m/s$$ (41)

4. Here $M = 0.8$, $k = 196$, $m = 0.16$, and $v = 362$.

(a) Call the velocities before and after the collision $v_b$ and $v_a$ respectively. Conservation of momentum gives:

362 $m v_b = (m+M)v_a$ so

$$v_a = {m\over m+M}v_b = {m\over m+M}v =60.3m/s$$ (42)

(b) After this collision, energy is conserved. At maximum compression the distance gone is at its maximum, hence this is the amplitude $A$ of oscillation. At this point, the kinetic energy is zero. So ${1\over 2}k A^2 = {1\over 2}(M+m)v_a^2$. Solving for $A$:

$$A = \sqrt{M+m\over k} v_a = \sqrt{M+m\over k} {m\over M+m} v = 4.22m$$ (43)

5. If $m_1$ or $m_2$ or $m_3$ or $m_4$ equals 0, the system will have an acceleration of $g$ and there will be no tension in the ropes. The only answer consistent with this is

$$T= {4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$$ (44)

Final Exam

Physics 6A

3/18/2003

You must show your work in order to obtain credit. Write your name and include this sheet with your exam papers.

1. A stone is thrown vertically upward. On its way up it passes point $A$ with speed $v$, and point $B$, $3 m$ higher than $A$ with speed $v/3$. Calculate

(a)(10 points) the speed $v$ and

(b)(10 points) the maximum height reached by the stone above point $B$ (that is, the distance between the maximum height at point B).

2. The two blocks shown below are free to move. The coefficient of static friction between the two blocks is $\mu_s = 0.44$ but the surface beneath $M$ is frictionless. Take $m = 5 kg$ and $M = 77 kg$.

(a) (5 points) Draw free body diagrams for the two blocks.

(b) (15 points) What is the minimum horizontal force $F$ required to hold $m$ against $M$?

3.(20 points) A uniform cylinder of radius $R = 0.61 m$ and mass $M = 10 kg$ rotates about a vertical axis on frictionless bearings as shown below. A light cord passes around the cylinder and then over a uniform cylindrical pulley having a moment of inertia $I = 0.01 kgm^2$ and radius $r = 0.11 m$. The cord is attached to a small object of mass $m = 5 kg$ that falls under the influence of gravity. Calculate the speed of the object after it has fallen a height $h = 5 m$ from rest. Hint: use conservation of energy.

4. A block of mass $M = 0.9 kg$, at rest on a horizontal frictionless surface, is attached to a rigid support by a spring of constant $k = 136 N/m$. A bullet of mass $m = 0.14 kg$ and velocity $v = 318 m/s$ strikes the block as shown. The bullet remains embedded in the block. Determine

(a)(10 points) the velocity of the block immediately after the collision and

(b)(10 points) the amplitude of the resulting simple harmonic motion. (Ignore any possibility of the block hitting a wall.)

5.(20 points) (You must explain your reasoning to obtain credit). A system of masses and pulleys is shown below. All pulleys and ropes are massless and frictionless. The tension in the rope connected to $m_1$ is

(a)

$g(m_1+m_2+m_3+m_4)/4$

(b)

$g (m_1+m_2)$

(c)

$m_1 g$

(d)

${2g \over 1/m_1 +1/m_2}$

(e)

${4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$

Hint: Don't try to work this out from scratch! Eliminate answers by looking at various limits, e.g. what would happen if $m_1$ were to equal ....

USEFUL INFORMATION Moment of inertia of a solid cylinder of radius $R$ and mass $M$ about center of mass $= \ MR^2 /2$. The acceleration of gravity $g = 9.8 m/s^2$.

Solutions

1. Here $n = 3$ and $h = 3$

(a) Find speed.

Use conservation of energy. $E_i = {1\over 2}m v^2 = E_f = mgh + {1\over 2}m (v/n)^2$. Solving for $v$:

$$v^2 = {2gh\over 1-({1\over n})^2}$$ (45)

or $v = 8.13m/s$.

(b) Again, use conservation of energy. At the maximum height $v=0$ so $E_i = {1\over 2}m v^2 = mgh_{max}$. So

$$h_{max} = {v^2\over 2g} = {2gh\over 1-({1\over 3})^2}{1 \over 2g} = {h\over 1-({1\over 3})^2} = 3.38m.$$ (46)

Relative to point B, we must subtract $h$ giving the answer

$${h\over n^2 -1} = 0.375m.$$ (47)

2.

Here $m = 5$, $\mu_s = 0.44$, and $M = 77$. (a)

(b) From part (a) and $F_{net} = ma$ in component form, for mass $m$, in the $y$ direction: $f_s - mg = ma_y = 0$, and in the $x$ direction: $F-N_1 = ma$.

For mass $M$ in the $x$ direction: $F_{net,x,M} = N_1 = Ma$. Eliminating $N_1$ $F-Ma = ma$ or

$$a = {F\over M+m}$$ (48)

Solving for $N_1$:

$$N_1 = Ma = {M\over M+m}F$$ (49)

At the point of slipping, $f_s = \mu_s N_1$, so from above, $\mu_s N_1 = mg$ or

$$\mu_s {M\over M+m}F = mg$$ (50)

so

$$F = {M+m\over M} {mg\over \mu_s} = (1+{m\over M}) {mg\over \mu_s} = 119N.$$ (51)

3. Here $m = 5$, $M = 10$, $r = 0.11$, $R = 0.61$, $I = 0.01$, $h = 5$.

Initially, the potential energy $U_i = 0$ and the kinetic energy $K_i = 0$. So the total energy is zero.

Finally, $E_f = 0 = U_f + K_f$

$U_f = -mgh$. The kinetic energy is the sum of three terms. The kinetic energy of mass $m$ and the rotational kinetic energy of the two cylinders. Rotational kinetic energy is of the form ${1\over 2}I \omega^2$. Here the velocities of the rims of the two cylinders are the same, though their angular velocities may be different. So $v = \omega r$, and solving for $\omega$ we can write the rotational kinetic energy as ${1\over 2}(I/r^2)v^2$. The moment of inertia of cylinder is ${1\over 2}M R^2$. So the total kinetic energy is ${1\over 2}mv^2 + {1\over 2}I {v^2\over r^2} + {1\over 2}({1\over 2}M R^2) {v^2\over R^2}$. Adding $-mgh$ to this at setting the whole expression to zero, we can solve for $v$:

$$v = \sqrt{2mgh\over m + (I/r^2) + {1\over 2}M} = 6.73 m/s$$ (52)

4. Here $M = 0.9$, $k = 136$, $m = 0.14$, and $v = 318$.

(a) Call the velocities before and after the collision $v_b$ and $v_a$ respectively. Conservation of momentum gives:

318 $m v_b = (m+M)v_a$ so

$$v_a = {m\over m+M}v_b = {m\over m+M}v =42.8m/s$$ (53)

(b) After this collision, energy is conserved. At maximum compression the distance gone is at its maximum, hence this is the amplitude $A$ of oscillation. At this point, the kinetic energy is zero. So ${1\over 2}k A^2 = {1\over 2}(M+m)v_a^2$. Solving for $A$:

$$A = \sqrt{M+m\over k} v_a = \sqrt{M+m\over k} {m\over M+m} v = 3.74m$$ (54)

5. If $m_1$ or $m_2$ or $m_3$ or $m_4$ equals 0, the system will have an acceleration of $g$ and there will be no tension in the ropes. The only answer consistent with this is

$$T= {4g \over 1/m_1 +1/m_2 + 1/m_3 + 1/m_4}$$ (55)

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