Quiz 6

Physics 6A

2/19/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A small piece of putty of mass $m_p = 0.13 kg$ hits a ball of mass $m_b = 0.77 kg$ that is attached to a rigid rod of length $L = 3 m$ as shown below. The putty collides with the ball at point A. The other end of the rod is secured to a rigid support and allowed to freely rotate. The ball and rod are initially at rest and hanging vertically. The putty hits the ball moving in the horiontal direction with velocity $v_p = 98.9 m/s$ and sticks to it. After that the ball and putty move together. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect the frictional forces that may be present as the rod swings, and assume that the size of the putty, ball and the mass of the rod are all negligible.

(a) (5 points) Does the rod manage to move 180 degrees from its starting point, that is to point B?

(b) (5 points) What is the speed of the putty at its highest point of motion?

(c) (5 points) What is the tension in the rod at this point?

Solution

(a): Does it make it to the top?

This is an inelastic collision. Use conservation of momentum.

Initial momentum: $p_i = m_p v_p = 0.13 kg \times 98.9 m/s = 12.86kg m/s$.

This must equal the momentum after the collision: $p_a = (m_b+m_b)v_a$. Solving we obtain $v_a = m_p v_p/(m_b+m_p) = 14.29 m/s$.

Call the total mass $m_t = m_b+m_p$. If we reckon the potential energy from point A, then the height at point B is $2L$. So it can make it to the top if the kinetic energy after the collision is at least $m_t g (2L)$. So the criterion for making it to B is $K = (1/2)m_t v_a^2 = m_t g 2L$ or $v_a = 14.29m/s > 2 \sqrt{gL} = 10.84m/s$.

So it does make it to point B.

(b): What is the speed of the putty at the highest point?

Use conservation energy:

${1\over 2}m_t v_a^2 = {1\over 2}m_t v_f^2 + m_t g (2L)$ or solving for $v_f$:

$v_f = \sqrt{v_a^2 - 4 g L} = 9.30m/s$

(c): What is the tension at this point?

$F_{net} = m_t a$. $F_{net} = -m_t g -T$ and $a = -v_f^2/L$. Therefore $-m_t g -T = -m_t v_f^2/L$. So $T = m_t(v_f^2/L -g) = m_t (v_a^2/L - 5g) = 17.12N$

Quiz 6

Physics 6A

2/19/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A small piece of putty of mass $m_p = 0.1 kg$ hits a ball of mass $m_b = 0.64 kg$ that is attached to a rigid rod of length $L = 3 m$ as shown below. The putty collides with the ball at point A. The other end of the rod is secured to a rigid support and allowed to freely rotate. The ball and rod are initially at rest and hanging vertically. The putty hits the ball moving in the horiontal direction with velocity $v_p = 114.06 m/s$ and sticks to it. After that the ball and putty move together. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect the frictional forces that may be present as the rod swings, and assume that the size of the putty, ball and the mass of the rod are all negligible.

(a) (5 points) Does the rod manage to move 180 degrees from its starting point, that is to point B?

(b) (5 points) What is the speed of the putty at its highest point of motion?

(c) (5 points) What is the tension in the rod at this point?

Solution

(a): Does it make it to the top?

This is an inelastic collision. Use conservation of momentum.

Initial momentum: $p_i = m_p v_p = 0.1 kg \times 114.06 m/s = 11.41kg m/s$.

This must equal the momentum after the collision: $p_a = (m_b+m_b)v_a$. Solving we obtain $v_a = m_p v_p/(m_b+m_p) = 15.41 m/s$.

Call the total mass $m_t = m_b+m_p$. If we reckon the potential energy from point A, then the height at point B is $2L$. So it can make it to the top if the kinetic energy after the collision is at least $m_t g (2L)$. So the criterion for making it to B is $K = (1/2)m_t v_a^2 = m_t g 2L$ or $v_a = 15.41m/s > 2 \sqrt{gL} = 10.84m/s$.

So it does make it to point B.

(b): What is the speed of the putty at the highest point?

Use conservation energy:

${1\over 2}m_t v_a^2 = {1\over 2}m_t v_f^2 + m_t g (2L)$ or solving for $v_f$:

$v_f = \sqrt{v_a^2 - 4 g L} = 10.95m/s$

(c): What is the tension at this point?

$F_{net} = m_t a$. $F_{net} = -m_t g -T$ and $a = -v_f^2/L$. Therefore $-m_t g -T = -m_t v_f^2/L$. So $T = m_t(v_f^2/L -g) = m_t (v_a^2/L - 5g) = 22.34N$

Quiz 6

Physics 6A

2/19/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A small piece of putty of mass $m_p = 0.17 kg$ hits a ball of mass $m_b = 0.5 kg$ that is attached to a rigid rod of length $L = 3 m$ as shown below. The putty collides with the ball at point A. The other end of the rod is secured to a rigid support and allowed to freely rotate. The ball and rod are initially at rest and hanging vertically. The putty hits the ball moving in the horiontal direction with velocity $v_p = 55.64 m/s$ and sticks to it. After that the ball and putty move together. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect the frictional forces that may be present as the rod swings, and assume that the size of the putty, ball and the mass of the rod are all negligible.

(a) (5 points) Does the rod manage to move 180 degrees from its starting point, that is to point B?

(b) (5 points) What is the speed of the putty at its highest point of motion?

(c) (5 points) What is the tension in the rod at this point?

Solution

(a): Does it make it to the top?

This is an inelastic collision. Use conservation of momentum.

Initial momentum: $p_i = m_p v_p = 0.17 kg \times 55.64 m/s = 9.46kg m/s$.

This must equal the momentum after the collision: $p_a = (m_b+m_b)v_a$. Solving we obtain $v_a = m_p v_p/(m_b+m_p) = 14.12 m/s$.

Call the total mass $m_t = m_b+m_p$. If we reckon the potential energy from point A, then the height at point B is $2L$. So it can make it to the top if the kinetic energy after the collision is at least $m_t g (2L)$. So the criterion for making it to B is $K = (1/2)m_t v_a^2 = m_t g 2L$ or $v_a = 14.12m/s > 2 \sqrt{gL} = 10.84m/s$.

So it does make it to point B.

(b): What is the speed of the putty at the highest point?

Use conservation energy:

${1\over 2}m_t v_a^2 = {1\over 2}m_t v_f^2 + m_t g (2L)$ or solving for $v_f$:

$v_f = \sqrt{v_a^2 - 4 g L} = 9.04m/s$

(c): What is the tension at this point?

$F_{net} = m_t a$. $F_{net} = -m_t g -T$ and $a = -v_f^2/L$. Therefore $-m_t g -T = -m_t v_f^2/L$. So $T = m_t(v_f^2/L -g) = m_t (v_a^2/L - 5g) = 11.68N$

Quiz 6

Physics 6A

2/19/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A small piece of putty of mass $m_p = 0.14 kg$ hits a ball of mass $m_b = 0.87 kg$ that is attached to a rigid rod of length $L = 2 m$ as shown below. The putty collides with the ball at point A. The other end of the rod is secured to a rigid support and allowed to freely rotate. The ball and rod are initially at rest and hanging vertically. The putty hits the ball moving in the horiontal direction with velocity $v_p = 89.81 m/s$ and sticks to it. After that the ball and putty move together. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect the frictional forces that may be present as the rod swings, and assume that the size of the putty, ball and the mass of the rod are all negligible.

(a) (5 points) Does the rod manage to move 180 degrees from its starting point, that is to point B?

(b) (5 points) What is the speed of the putty at its highest point of motion?

(c) (5 points) What is the tension in the rod at this point?

Solution

(a): Does it make it to the top?

This is an inelastic collision. Use conservation of momentum.

Initial momentum: $p_i = m_p v_p = 0.14 kg \times 89.81 m/s = 12.57kg m/s$.

This must equal the momentum after the collision: $p_a = (m_b+m_b)v_a$. Solving we obtain $v_a = m_p v_p/(m_b+m_p) = 12.45 m/s$.

Call the total mass $m_t = m_b+m_p$. If we reckon the potential energy from point A, then the height at point B is $2L$. So it can make it to the top if the kinetic energy after the collision is at least $m_t g (2L)$. So the criterion for making it to B is $K = (1/2)m_t v_a^2 = m_t g 2L$ or $v_a = 12.45m/s > 2 \sqrt{gL} = 8.85m/s$.

So it does make it to point B.

(b): What is the speed of the putty at the highest point?

Use conservation energy:

${1\over 2}m_t v_a^2 = {1\over 2}m_t v_f^2 + m_t g (2L)$ or solving for $v_f$:

$v_f = \sqrt{v_a^2 - 4 g L} = 8.75m/s$

(c): What is the tension at this point?

$F_{net} = m_t a$. $F_{net} = -m_t g -T$ and $a = -v_f^2/L$. Therefore $-m_t g -T = -m_t v_f^2/L$. So $T = m_t(v_f^2/L -g) = m_t (v_a^2/L - 5g) = 28.77N$

Quiz 6

Physics 6A

2/19/03

Please read the question carefully before attempting it. You will not be given any credit if you only write down the final answers. You must show your work.

A small piece of putty of mass $m_p = 0.11 kg$ hits a ball of mass $m_b = 0.74 kg$ that is attached to a rigid rod of length $L = 2 m$ as shown below. The putty collides with the ball at point A. The other end of the rod is secured to a rigid support and allowed to freely rotate. The ball and rod are initially at rest and hanging vertically. The putty hits the ball moving in the horiontal direction with velocity $v_p = 88.02 m/s$ and sticks to it. After that the ball and putty move together. Take the acceleration of gravity to be $9.8 m/s^2$. Neglect the frictional forces that may be present as the rod swings, and assume that the size of the putty, ball and the mass of the rod are all negligible.

(a) (5 points) Does the rod manage to move 180 degrees from its starting point, that is to point B?

(b) (5 points) What is the speed of the putty at its highest point of motion?

(c) (5 points) What is the tension in the rod at this point?

Solution

(a): Does it make it to the top?

This is an inelastic collision. Use conservation of momentum.

Initial momentum: $p_i = m_p v_p = 0.11 kg \times 88.02 m/s = 9.68kg m/s$.

This must equal the momentum after the collision: $p_a = (m_b+m_b)v_a$. Solving we obtain $v_a = m_p v_p/(m_b+m_p) = 11.39 m/s$.

Call the total mass $m_t = m_b+m_p$. If we reckon the potential energy from point A, then the height at point B is $2L$. So it can make it to the top if the kinetic energy after the collision is at least $m_t g (2L)$. So the criterion for making it to B is $K = (1/2)m_t v_a^2 = m_t g 2L$ or $v_a = 11.39m/s > 2 \sqrt{gL} = 8.85m/s$.

So it does make it to point B.

(b): What is the speed of the putty at the highest point?

Use conservation energy:

${1\over 2}m_t v_a^2 = {1\over 2}m_t v_f^2 + m_t g (2L)$ or solving for $v_f$:

$v_f = \sqrt{v_a^2 - 4 g L} = 7.17m/s$

(c): What is the tension at this point?

$F_{net} = m_t a$. $F_{net} = -m_t g -T$ and $a = -v_f^2/L$. Therefore $-m_t g -T = -m_t v_f^2/L$. So $T = m_t(v_f^2/L -g) = m_t (v_a^2/L - 5g) = 13.49N$

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