Physics 2211 Quiz #4 Solutions Spring 2011 G Universal Gravitational ...
Physics 2211 Spring 2011
Quiz #4
Solutions
G Universal Gravitational Constant g Earth’s Gravitational Acceleration Unless otherwise directed, problems take place on Earth, pulleys and ropes are ideal, and drag should be neglected. Any integrals in free-response problems must be evaluated. Questions about magnitudes will state so explicitly. I. (16 points) Block A, with mass mA , is at rest on a frictionless table and firmly attached to a spring with Hooke’s Law constant k. The spring is at its natural or relaxed length, with the other end anchored to the wall. Block B, with mass mB , is sliding toward block A with speed v0 . What is the maximum compression of the spring after block B hits and sticks to block A? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . There are no net external forces on the blocks during their collision, so momentum is conserved. P~i = P~f
⇒
mA vAi + mB vBi = mA vAf + mB vBf
Block A is initially at rest, vBi = v0 , and the two blocks have the same velocity after they stick together, so 0 + mB v0 = (mA + mB ) vf
⇒
vf =
mB v0 mA + mB
Use the Work-Energy Theorem as the spring compresses. h Wext = ∆K + ∆U =
2 1 2 mvf
i h i 2 2 − 12 mvi2 + 12 k (∆sf ) − 21 k (∆si )
No work is done by external forces on the system consisting of the blocks and the spring. The mass m is the sum of the block masses mA + mB . This initial compression of the spring is zero, and the speed of the blocks is zero at maximum compression. h 0= 0−
1 2
i h i 2 (mA + mB ) vi2 + 12 k (∆sf ) − 0)2
µ ⇒
∆s2f =
mA + mB k
¶ vi2
But the initial speed of the blocks as the spring compresses is just the final speed of the blocks after the collision. µ ¶µ ¶2 mA + mB mB v0 m B v0 ∆s2f = ∆sf = p ⇒ k mA + mB k (mA + mB )
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II. (16 points) Two blocks, with masses m1 and m2 , are connected by an ideal cord passing over an ideal pulley, as shown. They are held at rest, with block m2 a distance d above the floor, and block m1 on a frictionless surface and connected to a spring with Hooke’s Law constant k. If the spring is at its natural or relaxed length in this configuration, what is the speed of block m2 just before it hits the floor after being released? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . . Use the Work-Energy Theorem. Choose a system consisting of both blocks, the spring, and the Earth. h Wext = ∆K +∆U =
2 2 1 1 2 m1 v1f − 2 m1 v1i
i h i h i h i 2 2 2 2 + 12 m2 v2f − 12 m2 v2i + 12 k (∆sf ) − 21 k (∆si ) + m2 ghf −m2 ghi
No work is done by external forces on the system. The initial speeds of both blocks is zero. The final speeds of both blocks is the same, so v1f = v2f = vf . If we let the final height of block be zero, then its initial height is d. The spring is initially unstretched, and is finally stretched a distance d. h 0=
2 1 2 m1 vf
i h i h i h i − 0 + 12 m2 vf2 − 0 + 12 kd2 − 0 + 0 − m2 gd
Solve for vf .
s m2 gd − 21 kd2 =
1 2
(m1 + m2 ) vf2
⇒
vf =
2m2 gd − kd2 m1 + m2
1. (6 points) In the problem above, how does the magnitude of the tension in the cord compare to the gravitational force on the block m2 as it descends? . . . . . . . . . . . . . . . . . . . . . . . As the block initially accelerates downward, the net force must be downward, and tension must initially be less than the gravitational force on block 2. However, the force applied by the spring increases as the spring stretches, so the tension is not constant. The tension magnitude is initially less than m2 g, and changes as the block descends.
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III. (16 points) Natalie (mass mn ) is standing at the left of a cart with length L and mass mc . The cart has frictionless wheels and is at rest on a frictionless track. Suddenly, Natalie starts running along the cart with a speed vnc relative to the cart. How far will Natalie have traveled relative to the ground when she reaches the right end of the cart? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . Use constant-acceleration kinematics in the reference frame of the cart to find how much time is required for Natalie to reach the end of the cart. Let Natalie travel in the positive direction, starting at x = 0. Note that Natalie’s acceleration is zero. 2
x = x0 + v0 ∆t + 21 a (∆t)
⇒
L = 0 + vnc ∆t + 0
⇒
∆t =
L vnc
There are no net external forces on the system consisting of Natalie and the cart, so momentum is conserved. Natalie and the cart are both initially stationary in the reference frame of the ground. P~i = P~f
⇒
0 = mn vng + mc vcg
where vng is the velocity of Natalie with respect to the ground, and vcg is the velocity of the cart with respect to the ground. Relate the velocity of Natalie with respect to the ground to her velocity with respect to the cart. vng = vnc + vcg ⇒ vcg = vng − vnc Substitute to eliminate the velocity of the cart with respect to the ground, and find the velocity of Natalie with respect to the ground. 0 = mn vng + mc (vng − vnc ) = mn vng + mc vng − mc vnc
⇒
vng =
mc vnc mn + mc
Use constant-acceleration kinematics to find how far Natalie travels at this velocity with respect to the ground, knowing that she travels for the time found above. µ 2
x = x0 + v0 ∆t + 12 a (∆t) = 0 +
mc vnc mn + mc
¶µ
L vnc
¶ +0
⇒
x=
mc L mn + mc
2. (6 points) Letting your answer to the above problem be D in the +x direction, what is the displacement of the cart relative to the ground at the time Natalie reaches the right end? . . . . . . . . . . . . . . . . . . . . . . . If Natalie had infinite mass, she would not move with respect to the ground (D = 0), and the cart would be displaced −L beneath her feet. Only one of the choices satisfies this condition. D−L
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3. (8 points) What is the speed that a 2 kg ball needs at point A to clear the potential barrier and reach point B with a speed 1 m/s? . . . . . . . . . . . . . At point B, the total mechanical energy of the ball would be Emech = K + U = 12 mv 2 + U =
1 2
2
(2 kg) (1 m/s) + 0 J = 1 J
where the potential energy U is obtained from the graph. At point A, however, the potential energy is 2 J. At the total mechanical energy, is only 1 J, this would require the kinetic energy at A to be negative. The ball cannot clear the potential barrier and reach point B with a speed 1 m/s.
4. (8 points) The block of mass m rests on a frictionless surface, as shown. When it is pushed against the spring, compressing it a distance d, and released, it reaches a maximum height h above its release point. If a block with half the mass were pushed against the spring to compress it twice the distance, to what maximum height would it rise? . . . . . . . . . . . . . . . . . . . . . . . Since the energy stored in a spring is proportional to the square of the compression, doubling the compression stores four times the energy. The gravitational potential energy is linearly proportional to both the height and the mass, getting four times the energy with half the mass will require the block to rise to a height 8h
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√ 5. (8 points) A 2 kg object is traveling along the −x axis with a velocity of + π m/s. When it reaches the point x = −1 m, it becomes subject to the “semicircular” force shown in the figure, which reaches a maximum value of 1.0 N. If the object reaches the point x = +1 m, what is its speed there? . . . . . . . . . . . . . . . . . . . . . . . Since there are no potential energies involved, the work done by the force is the change in kinetic energy. Z ~ = 1 mv 2 − 1 mv 2 W = ∆K ⇒ F~ · ds f i 2 2 Solve for vf .
r vf =
2W + vi2 m
Remember that the work is the area under the graph of force as a function of position, or π/2 J. So s vf =
¢2 2 (π/2 J) ¡√ + π m/s = 2 kg
r
3π m/s 2
6. (8 points) Identical forces push identical blocks across level surfaces. Force S pushes block S a distance d across a smooth, frictionless, surface. Force R pushes block R the same distance d across a rough surface with friction. Which force, if either, does more work on its block? Which force, if any, provides more impulse to its block? . . . . . . . . . . . . . . . . . . . . . . . R ~ The force pushing the Since both forces act for the same distance, both do the same work W = F~ · ds. R block on the rough surface will require more time, however, so that force provides more impulse J~ = F~ dt. Both forces do the same work. Force R provides more impulse.
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7. (4 points) A rubber ball is thrown at toward the floor, striking it with a speed v at an angle θ. It bounces, leaving the floor at the same speed v and at an angle θ as shown. In what direction does the impulse from the floor on the ball point? . . . . . . . . . . . . . . . . . . . . . . .
Remember that impulse on an object is the change in momentum of the object, J~ = ∆~ p = p~f − p~i = m (~vf − ~vi ) Finding the vector difference graphically shows that the change in momentum, and thus the impulse, is In the +y direction.
8. (4 points) A putty ball is thrown at toward the floor, striking it with a speed v at an angle θ. It sticks to the floor and remains there, as shown. In what direction does the impulse from the floor on the ball point? . . . . . . . . . . . . . . . . . . . . . . . Remember that impulse on an object is the change in momentum of the object, J~ = ∆~ p = p~f − p~i = m (~vf − ~vi ) Since the final momentum of the object is zero, the change in momentum is the opposite of the initial momentum, so the impulse is Somewhere in quadrant II.
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Quiz #4
Solutions
G Universal Gravitational Constant g Earth’s Gravitational Acceleration Unless otherwise directed, problems take place on Earth, pulleys and ropes are ideal, and drag should be neglected. Any integrals in free-response problems must be evaluated. Questions about magnitudes will state so explicitly. I. (16 points) Block A, with mass mA , is at rest on a frictionless table and firmly attached to a spring with Hooke’s Law constant k. The spring is at its natural or relaxed length, with the other end anchored to the wall. Block B, with mass mB , is sliding toward block A with speed v0 . What is the maximum compression of the spring after block B hits and sticks to block A? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . There are no net external forces on the blocks during their collision, so momentum is conserved. P~i = P~f
⇒
mA vAi + mB vBi = mA vAf + mB vBf
Block A is initially at rest, vBi = v0 , and the two blocks have the same velocity after they stick together, so 0 + mB v0 = (mA + mB ) vf
⇒
vf =
mB v0 mA + mB
Use the Work-Energy Theorem as the spring compresses. h Wext = ∆K + ∆U =
2 1 2 mvf
i h i 2 2 − 12 mvi2 + 12 k (∆sf ) − 21 k (∆si )
No work is done by external forces on the system consisting of the blocks and the spring. The mass m is the sum of the block masses mA + mB . This initial compression of the spring is zero, and the speed of the blocks is zero at maximum compression. h 0= 0−
1 2
i h i 2 (mA + mB ) vi2 + 12 k (∆sf ) − 0)2
µ ⇒
∆s2f =
mA + mB k
¶ vi2
But the initial speed of the blocks as the spring compresses is just the final speed of the blocks after the collision. µ ¶µ ¶2 mA + mB mB v0 m B v0 ∆s2f = ∆sf = p ⇒ k mA + mB k (mA + mB )
Quiz #4 Solutions Page 1 of 6
II. (16 points) Two blocks, with masses m1 and m2 , are connected by an ideal cord passing over an ideal pulley, as shown. They are held at rest, with block m2 a distance d above the floor, and block m1 on a frictionless surface and connected to a spring with Hooke’s Law constant k. If the spring is at its natural or relaxed length in this configuration, what is the speed of block m2 just before it hits the floor after being released? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . . Use the Work-Energy Theorem. Choose a system consisting of both blocks, the spring, and the Earth. h Wext = ∆K +∆U =
2 2 1 1 2 m1 v1f − 2 m1 v1i
i h i h i h i 2 2 2 2 + 12 m2 v2f − 12 m2 v2i + 12 k (∆sf ) − 21 k (∆si ) + m2 ghf −m2 ghi
No work is done by external forces on the system. The initial speeds of both blocks is zero. The final speeds of both blocks is the same, so v1f = v2f = vf . If we let the final height of block be zero, then its initial height is d. The spring is initially unstretched, and is finally stretched a distance d. h 0=
2 1 2 m1 vf
i h i h i h i − 0 + 12 m2 vf2 − 0 + 12 kd2 − 0 + 0 − m2 gd
Solve for vf .
s m2 gd − 21 kd2 =
1 2
(m1 + m2 ) vf2
⇒
vf =
2m2 gd − kd2 m1 + m2
1. (6 points) In the problem above, how does the magnitude of the tension in the cord compare to the gravitational force on the block m2 as it descends? . . . . . . . . . . . . . . . . . . . . . . . As the block initially accelerates downward, the net force must be downward, and tension must initially be less than the gravitational force on block 2. However, the force applied by the spring increases as the spring stretches, so the tension is not constant. The tension magnitude is initially less than m2 g, and changes as the block descends.
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III. (16 points) Natalie (mass mn ) is standing at the left of a cart with length L and mass mc . The cart has frictionless wheels and is at rest on a frictionless track. Suddenly, Natalie starts running along the cart with a speed vnc relative to the cart. How far will Natalie have traveled relative to the ground when she reaches the right end of the cart? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . Use constant-acceleration kinematics in the reference frame of the cart to find how much time is required for Natalie to reach the end of the cart. Let Natalie travel in the positive direction, starting at x = 0. Note that Natalie’s acceleration is zero. 2
x = x0 + v0 ∆t + 21 a (∆t)
⇒
L = 0 + vnc ∆t + 0
⇒
∆t =
L vnc
There are no net external forces on the system consisting of Natalie and the cart, so momentum is conserved. Natalie and the cart are both initially stationary in the reference frame of the ground. P~i = P~f
⇒
0 = mn vng + mc vcg
where vng is the velocity of Natalie with respect to the ground, and vcg is the velocity of the cart with respect to the ground. Relate the velocity of Natalie with respect to the ground to her velocity with respect to the cart. vng = vnc + vcg ⇒ vcg = vng − vnc Substitute to eliminate the velocity of the cart with respect to the ground, and find the velocity of Natalie with respect to the ground. 0 = mn vng + mc (vng − vnc ) = mn vng + mc vng − mc vnc
⇒
vng =
mc vnc mn + mc
Use constant-acceleration kinematics to find how far Natalie travels at this velocity with respect to the ground, knowing that she travels for the time found above. µ 2
x = x0 + v0 ∆t + 12 a (∆t) = 0 +
mc vnc mn + mc
¶µ
L vnc
¶ +0
⇒
x=
mc L mn + mc
2. (6 points) Letting your answer to the above problem be D in the +x direction, what is the displacement of the cart relative to the ground at the time Natalie reaches the right end? . . . . . . . . . . . . . . . . . . . . . . . If Natalie had infinite mass, she would not move with respect to the ground (D = 0), and the cart would be displaced −L beneath her feet. Only one of the choices satisfies this condition. D−L
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3. (8 points) What is the speed that a 2 kg ball needs at point A to clear the potential barrier and reach point B with a speed 1 m/s? . . . . . . . . . . . . . At point B, the total mechanical energy of the ball would be Emech = K + U = 12 mv 2 + U =
1 2
2
(2 kg) (1 m/s) + 0 J = 1 J
where the potential energy U is obtained from the graph. At point A, however, the potential energy is 2 J. At the total mechanical energy, is only 1 J, this would require the kinetic energy at A to be negative. The ball cannot clear the potential barrier and reach point B with a speed 1 m/s.
4. (8 points) The block of mass m rests on a frictionless surface, as shown. When it is pushed against the spring, compressing it a distance d, and released, it reaches a maximum height h above its release point. If a block with half the mass were pushed against the spring to compress it twice the distance, to what maximum height would it rise? . . . . . . . . . . . . . . . . . . . . . . . Since the energy stored in a spring is proportional to the square of the compression, doubling the compression stores four times the energy. The gravitational potential energy is linearly proportional to both the height and the mass, getting four times the energy with half the mass will require the block to rise to a height 8h
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√ 5. (8 points) A 2 kg object is traveling along the −x axis with a velocity of + π m/s. When it reaches the point x = −1 m, it becomes subject to the “semicircular” force shown in the figure, which reaches a maximum value of 1.0 N. If the object reaches the point x = +1 m, what is its speed there? . . . . . . . . . . . . . . . . . . . . . . . Since there are no potential energies involved, the work done by the force is the change in kinetic energy. Z ~ = 1 mv 2 − 1 mv 2 W = ∆K ⇒ F~ · ds f i 2 2 Solve for vf .
r vf =
2W + vi2 m
Remember that the work is the area under the graph of force as a function of position, or π/2 J. So s vf =
¢2 2 (π/2 J) ¡√ + π m/s = 2 kg
r
3π m/s 2
6. (8 points) Identical forces push identical blocks across level surfaces. Force S pushes block S a distance d across a smooth, frictionless, surface. Force R pushes block R the same distance d across a rough surface with friction. Which force, if either, does more work on its block? Which force, if any, provides more impulse to its block? . . . . . . . . . . . . . . . . . . . . . . . R ~ The force pushing the Since both forces act for the same distance, both do the same work W = F~ · ds. R block on the rough surface will require more time, however, so that force provides more impulse J~ = F~ dt. Both forces do the same work. Force R provides more impulse.
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7. (4 points) A rubber ball is thrown at toward the floor, striking it with a speed v at an angle θ. It bounces, leaving the floor at the same speed v and at an angle θ as shown. In what direction does the impulse from the floor on the ball point? . . . . . . . . . . . . . . . . . . . . . . .
Remember that impulse on an object is the change in momentum of the object, J~ = ∆~ p = p~f − p~i = m (~vf − ~vi ) Finding the vector difference graphically shows that the change in momentum, and thus the impulse, is In the +y direction.
8. (4 points) A putty ball is thrown at toward the floor, striking it with a speed v at an angle θ. It sticks to the floor and remains there, as shown. In what direction does the impulse from the floor on the ball point? . . . . . . . . . . . . . . . . . . . . . . . Remember that impulse on an object is the change in momentum of the object, J~ = ∆~ p = p~f − p~i = m (~vf − ~vi ) Since the final momentum of the object is zero, the change in momentum is the opposite of the initial momentum, so the impulse is Somewhere in quadrant II.
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