(pnp223) Patel homework 23 Turner (92160) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Consider the general case of the collision of a mass m1 = m with mass m2 = 2 m along a frictionless horizontal surface. Denote the initial and the final center of mass momenta to be pi = p1 + p2 and cm pf = p + p 2 1 cm and the initial and final kinetic energies to be Ki = K1 + K2 and
1
Consider a perfectly inelastic head-on collision with initial velocity for m1 of v1 but with m2 motionless. Find the final speed of the system. 1. vCM = 2. vCM 3. vCM 4. vCM 1 4 2 = 7 1 = 5 2 = 3 v1 v1 v1 v1
5. vCM = v1 6. vCM = 7. vCM 8. vCM 3 v1 5 1 = v1 correct 3 3 = v1 4
Kf = K1 + K2 .
v1 m
v2 2m
9. vCM = 2 v1 For an elastic collision which pair of statements is correct? 1. pi > pf , CM CM 2. pi > pf , CM CM 3. pi CM < pf CM , and Ki < Kf and Ki > Kf and Ki < Kf and Ki = Kf and Ki > Kf and Ki = Kf and Ki = Kf correct and Ki < Kf and Ki > Kf 1 v1 2 Explanation: 10. vCM = Let : v2 = 0 .
In a collision the linear momentum is always conserved, so m1 v1 + m2 0 = (m1 + m2 ) vf vCM = vf = m v1 m1 v1 = v1 = . m1 + m2 3m 3
4. pi < pf , CM CM 5. pi < pf , CM CM 6. pi > pf , CM CM 7. pi CM = pf CM ,
8. pi = pf , CM CM 9. pi = pf , CM CM
003 (part 3 of 3) 10.0 points Now consider an elastic head-on collision with initial velocity for m1 of v1 but with m2 motionless. Find the final speed of m2 .
f 1. v2 =
Explanation: Momentum and energy are always conserved in elastic collisions. 002 (part 2 of 3) 10.0 points
1 v1 5
f 2. v2 = 2 v1 f 3. v2 =
3 v1 4
Patel (pnp223) homework 23 Turner (92160) 4.
f v2
2
=
f 5. v2 = f 6. v2 = f 7. v2 =
2 v1 correct 3 1 v1 3 3 v1 5 1 v1 2
After a one-dimensional elastic collision, each body changes the direction of its momentum with the magnitude of the momentum conserved. Note: In the CM frame the total momentum is still zero if each velocity changes sign, so the final velocity of m2 in the CM frame is 1 CM v2,f = vCM = v1 3
f lab and the velocity v2,f = v2 in the lab frame is lab CM v2,f = v2,f + vCM 1 2 1 = v1 + v1 = v1 . 3 3 3
f 8. v2 = v1
2 v1 7 1 f 10. v2 = v1 4 Explanation:
f 9. v2 =
Let :
v2 = 0 .
In an elastic collision
f f v1 - 0 = v2 - v1
004 (part 1 of 3) 10.0 points Masses of 22 kg and 5.9 kg hang freely from a light, inextensible cord that passes over a light, frictionless pulley with a radius of 6.4 cm. Initially their centers of mass are a vertical distance 1.5 m apart.
and conservation of momentum gives
f f m1 v1 + m2 0 = m1 v1 + m2 v2 .
6.4 cm
Multiplying the first equation by m1 ,
f f m1 v1 = m1 v2 - m1 v1
22 kg 1.5 m 5.9 kg
m1 v1 = and adding,
f m1 v1
+
f m2 v2
2 m1 v1 = 2 m v1 =
f v2 =
f (m1 + m2 ) v2 f 3 m v2
2 v1 . 3
What is the magnitude of the acceleration of the two masses when they pass each other? The acceleration of gravity is 9.8 m/s2 . Correct answer: 5.6552 m/s2 . Explanation: Let : R = 6.4 cm , m1 = 5.9 kg , m2 = 22 kg , and h = 1.5 m .
Alternate Solution: The CM velocity is given by vCM =
i mi vi i mi
=
m v1 1 m1 v1 = = v1 , m1 + m2 3m 3
so in the CM frame, m2 has an initial velocity 1 CM v2 = v2 - vCM = -vCM = - v1 . 3
Consider the free body diagrams
Patel (pnp223) homework 23 Turner (92160)
3
22 kg m2 g
5.9 kg m1 g
a
a
006 (part 3 of 3) 10.0 points What is the tension in the cord when they pass each other? Correct answer: 91.1857 N. Explanation: T = m1 (g + a) = (5.9 kg) (9.8 m/s2 + 5.6552 m/s2 ) = 91.1857 N .
T
Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m2 and motion upward as positive for m1 . Apply Newton's second law to m1 and m2 respectively: F1 : F2 : T - m1 g = m1 a m2 g - T = m2 a (1) (2)
T
keywords: 007 10.0 points A triangular wedge 6 m high, 12 m base length, and with a 13 kg mass is placed on a frictionless table. A small block with a 8 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below.
Adding these equations, m2 g - m1 g = m1 a + m2 a m2 - m1 g m1 + m2 22 kg - 5.9 kg = 22 kg + 5.9 kg = 5.6552 m/s . 005 (part 2 of 3) 10.0 points What is the acceleration of the center of mass of the two blocks when they pass each other? Correct answer: 3.26339 m/s2 . Explanation: The center of mass accelerates downward. With down as positive a2 = a and a1 = -a, we have m2 a - m1 a m2 - m1 a = m2 + m1 m2 + m1 22 kg - 5.9 kg = (5.6552 m/s2 ) 22 kg + 5.9 kg = 3.26339 m/s2 .
2
8k
g
6m
a=
13 kg (9.8 m/s2 ) 12 m
Xwedge
8k
g
12 m
13 kg
aCM =
All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far Xwedge does the wedge slide to the right? Correct answer: 4.57143 m. Explanation: Let : M = 13 kg ,
6m
Patel (pnp223) homework 23 Turner (92160) m = 8 kg , L = 12 m , H = 6 m. and therefore and m L m+M (8 kg) (12 m) = (8 kg) + (13 kg)
4
Xwedge =
Consider the wedge and the block as a twobody The system. external forces acting on this system -- the weight of the wedge, the weight of the block and the normal force from the table -- are all vertical, hence the net horizontal momentum of the system is conserved,
wedge block Px + Px = constant .
= 4.57143 m . 008 (part 1 of 2) 10.0 points In the "slingshot effect," the transfer of energy in an elastic collision is used to boost the energy of a space probe so that it can escape from the solar system. All speeds are relative to an inertial frame in which the center of the sun remains at rest. A space probe moves at 9.6 km/s toward Saturn, which is moving at 9.2 km/s toward the probe. Because of the gravitational attraction between Saturn and the probe, the probe swings around Saturn and heads back (nearly) in the opposite direction with speed vf . 9.6 km/s 9.2 km/s vf Figure: The trajectory is a parabola. Assuming this collision to be a onedimensional elastic collision with the mass of Saturn much greater than that of the probe, find vf . Correct answer: 28 km/s. Explanation: Let : vprobe = vi = 9.6 km/s and vSaturn = vS = 9.2 km/s .
Furthermore, we start from rest = centerof-mass is not moving, and therefore the X coordinate of the center-of-mass will remain constant while the wedge slides to the right and the block slides down and to the left, Xcm = m Xblock + M Xwedge = constant . m+M
Note: Only the X coordinate of the centerof-mass is a constant of motion; i.e., the Ycm accelerates downward because the Py component of the net momentum is not conserved. Constant Xcm means Xcm = 0 and therefore m Xblock + M Xwedge = 0 . Note that this formula does not depend on where the wedge has its own center-of-mass; as long as the wedge is rigid, its overall displacement Xwedge is all we need to know. Finally, consider the geometry of the problem: By the time the block slides all the way down, its displacement relative to the wedge is equal to the wedge length L, or rather -L because the block moves to the left of the wedge. In terms of displacements relative to the inertial frame of the table, this means Xblock - Xwedge = -L . Consequently, 0 = m Xblock + M Xwedge = m (-L + Xwedge ) + M Xwedge
Because Saturn is so much more massive than the space probe, the center-of-mass velocity is vcm = vSaturn = 9.2 km/s . The velocity of approach is uapp = vSaturn - vprobe uapp = -9.2 km/s - 9.6 km/s = -18.8 km/s ,
Patel (pnp223) homework 23 Turner (92160) and the relative velocity of recession for an elastic collision is urec = -uapp = 18.8 km/s . Thus the velocity of recession relative to the center of mass is v = urec + vcm = 18.8 km/s + 9.2 km/s = 28 km/s . 009 (part 2 of 2) 10.0 points By what factor is the kinetic energy of the probe increased? Correct answer: 8.50694. Explanation: Consider the ratio of the final and initial kinetic energies: 1 2 M vrec Kf 2 = 1 Ki 2 M vi 2 2 vrec = vi = 28 km/s 9.6 km/s
2
5
What is the speed of the two blocks immediately after the collision? Correct answer: 9.26316 m/s. Explanation:
Let : v12 m1 m2 v1 g
= speed of blocks after collision = 480 kg , = 90 kg , = 11 m/s , and = 9.8 m/s2 .
The initial momentum of block 1 is m1 v1 . When m1 sticks with m2 , the resulting momentum p12 is equal to that initial momentum; thus, p12 (m1 + m2 ) v12 = m1 v1 . Therefore, the speed of the combined masses is v12 = m1 v1 m1 + m2 (480 kg) = (11 m/s) (480 kg) + (90 kg) = 9.26316 m/s .
= 8.50694 . The energy comes from an immeasurably small slowing of Saturn. 010 (part 1 of 2) 10.0 points A(n) 480 kg mass is sliding on a horizontal frictionless surface with a speed of 11 m/s when it collides with a 90 kg mass initially at rest, as shown in the figure. The masses stick together and slide up a frictionless track at 60 from horizontal. The acceleration of gravity is 9.8 m/s2 . 011 (part 2 of 2) 10.0 points 570 kg 9.8 m/s2 1 m2 2 1 (m1 + m2 ) v12 = v2 2 2 (m1 + m2 ) 1
To what maximum height h above the horizontal surface will the masses slide? Correct answer: 4.37786 m. Explanation: The kinetic energy K of the combined system is K=
11 m/s 60 90 kg 480 kg
9.8 m/s2
h
Patel (pnp223) homework 23 Turner (92160) At the maximum height, all the kinetic energy is converted to potential energy, and so (m1 + m2 ) g h = K = Solving for h, we get m1 v2 h= 1 2 g m1 + m2 (11 m/s)2 = 2 (9.8 m/s2 )
2
6
m2 1 v2 2 (m1 + m2 ) 1
(480 kg) (480 kg) + (90 kg) = 4.37786 m .
2
012 10.0 points Why is the wobbly motion of a single star an indication that the star has one or more planets orbiting around it? 1. The light emitted from the star is scattered by planets. 2. The motion of a star is affected by the gravity of planets. 3. Planets near the star distorted the timespace, which affects the star. 4. The star is revolving about the center of mass of planets. correct Explanation: The wobbly motion of a star is an indication that it is revolving about a center of mass that is not at its geometric center, implying that there is some other mass nearby to pull the center of mass away from the star's center. This is the way in which astronomers have discovered that planets exist around stars other than our own.