Physics 2211 A Quiz #1 Solutions Summer 2007 Unless otherwise
Physics 2211 A Summer 2007
Quiz #1
Solutions
gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A Europan (inhabitant of Jupiter’s moon Europa) throws an ice cube straight up at 7.9 m/s. The ice cube reaches its maximum height above its release point 6.1 s later. What is the maximum height of the ice cube above its release point? (On Europa.) . . . . . . . . . . . . . . . . . . . . . . . Once it leaves the Europan’s “hands”, the ice cube moves only under the influence of gravity, and so undergoes constant acceleration (free fall). The acceleration of gravity on Europa is not known, but it can be calculated. Note that the velocity of the ice cube at its maximum height is zero. v = v0 + a (∆t)
⇒
a=
v − v0 ∆t
⇒
a=−
v0 ∆t
Substituting this into the constant acceleration expression for position
2
x = x0 + v0 (∆t) + 21 a (∆t) ⇓ ³ v ´ 0 2 x = x0 + v0 (∆t) + 12 − (∆t) t ⇓ x = x0 + v0 (∆t) + 12 (−v0 ) (∆t) ⇓ x = x0 + 12 v0 (∆t) Choosing the origin to be at the release point and positive to be up, x0 = 0 m and v0 = +7.9 m/s, so x = 0m +
1 2
(+7.9 m/s) (6.1 s) = 24 m
Quiz #1 Solutions Page 1 of 5
II. (16 points) A particle’s acceleration is described by the function ax (t) = 5.0t3 where ax is in m/s2 when t is in s. The particle’s velocity at t0 = 0 s is v0 = +3.0 m/s. What is the particle’s velocity at t = 2.0 s? . . . . . . . . . . . . . . . . . . . . . . . Acceleration is the time rate of velocity change, so a=
dv dt
Solving for velocity, Z dv = a dt
⇒
v=
Z dv =
Z a dt =
5.0t3 dt = 5.0
t4 +C 4
where C is a constant of integration. That constant can be found using the given velocity of +3.0 m/s at time t0 = 0 s. v0 = 5.0
(0 s)4 + C = +3.0 m/s 4
⇒
C = +3.0 m/s
So at time t = 2.0 s, v = 5.0
(2.0 s)4 + 3.0 m/s = 23 m/s 4
1. (6 points) What are the units of the “5.0” in the expression ax (t) = 5.0t3 ? . . . . . . . . . . . . . . . . . . . . . . . If “something” times seconds cubed yields meters per second squared, then the “something” must have units of meters per second to the fifth. m = (?) s3 s2
⇒
m ³m´ 3 = 5 s s2 s
Quiz #1 Solutions Page 2 of 5
III. (16 points) An oasis is 24 km directly east of a camel. Unfortunately, the camel walks 24 km in a direction 30◦ south of east, and then 16 km directly north. How far is the camel from the oasis now? . . . . . . . . . . . . . . . . . . . . . . .
From the sketch, after walking 24 km in a direction 30◦ south of east, and then 16 km directly north, the camel is 4 km north of the oasis, and 3.2 km west. These are the legs of a right triangle whose hypoteneuse is q 2
2
(4 km) + (3.2 km) = 5.1 km
2. (6 points) In what direction should the camel now walk to reach the oasis? . . . . . . . . . . . . . . . . . . . . . . . Since the camel is north and west of the oasis, it must direct its travels south of east to arrive there. If the angle south of east is φ, then 4 km tan φ = 3.2 km
µ ⇒
φ = tan
−1
4 km 3.2 km
¶ = 51◦
So the camel must head 51◦ south of east.
Quiz #1 Solutions Page 3 of 5
3. (10 points) The figure shows the velocity of a particle moving in one dimension as a function of time. At which points, if any, does the particle change its direction of travel? . . . . . . . . . . . . . . . . . . . . . . . When an object moves in one dimension, the sign of its velocity indicates its direction. The sign of the velocity changes when the velocity vs. time curve crosses the time axis, namely, at points 3 and 5.
4. (10 points) The figure shows the velocity of a truck (dashed line) and a car (solid line) as a function of time as they move along a straight road. The axes cross at the origin. The two vehicles are side by side at time t = 0. What is true at time t = T ? . . . . . . . . . . . . . . . . . . . . . . . The distance travelled is the area under the given position vs. time curve. At time t = T , the area under the truck curve is twice that under the car curve. Neither area is zero. Thus, the truck and car will both have moved, but the truck will have travelled farther than the car.
Quiz #1 Solutions Page 4 of 5
5. (10 points) The figure shows the position of two objects, A and B, as a function of time as they move along a straight line. At what time, if any, do the objects have the same speed? . . . . . . . . . . . . . . . . . . . . . . . The velocity is the slope of the given position vs. time graph. The speed is the magnitude of that slope. The two curves are parallel, that is, have the same slope, at some time between 2 and 3 seconds.
6. (10 points) A freight train of empty hopper cars is accelerating (speeding up) along a straight and level track. It begins to rain, and as the hopper cars fill with water, the magnitude of the acceleration decreases with time. How does the speed of the train vary as the acceleration magnitude decreases? . . . . . . . . . . . . . . . . . . . . . . . Since we know the train is speeding up before it begins to rain, we know that the velocity and acceleration vectors are parallel. Let that be the positive direction. As long as the acceleration magnitude is decreasing, the acceleration cannot become zero or negative (since the magnitude would then begin to increase). Thus, the acceleration and velocity vectors must remain parallel, and the speed increases.
Quiz #1 Solutions Page 5 of 5
Quiz #1
Solutions
gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A Europan (inhabitant of Jupiter’s moon Europa) throws an ice cube straight up at 7.9 m/s. The ice cube reaches its maximum height above its release point 6.1 s later. What is the maximum height of the ice cube above its release point? (On Europa.) . . . . . . . . . . . . . . . . . . . . . . . Once it leaves the Europan’s “hands”, the ice cube moves only under the influence of gravity, and so undergoes constant acceleration (free fall). The acceleration of gravity on Europa is not known, but it can be calculated. Note that the velocity of the ice cube at its maximum height is zero. v = v0 + a (∆t)
⇒
a=
v − v0 ∆t
⇒
a=−
v0 ∆t
Substituting this into the constant acceleration expression for position
2
x = x0 + v0 (∆t) + 21 a (∆t) ⇓ ³ v ´ 0 2 x = x0 + v0 (∆t) + 12 − (∆t) t ⇓ x = x0 + v0 (∆t) + 12 (−v0 ) (∆t) ⇓ x = x0 + 12 v0 (∆t) Choosing the origin to be at the release point and positive to be up, x0 = 0 m and v0 = +7.9 m/s, so x = 0m +
1 2
(+7.9 m/s) (6.1 s) = 24 m
Quiz #1 Solutions Page 1 of 5
II. (16 points) A particle’s acceleration is described by the function ax (t) = 5.0t3 where ax is in m/s2 when t is in s. The particle’s velocity at t0 = 0 s is v0 = +3.0 m/s. What is the particle’s velocity at t = 2.0 s? . . . . . . . . . . . . . . . . . . . . . . . Acceleration is the time rate of velocity change, so a=
dv dt
Solving for velocity, Z dv = a dt
⇒
v=
Z dv =
Z a dt =
5.0t3 dt = 5.0
t4 +C 4
where C is a constant of integration. That constant can be found using the given velocity of +3.0 m/s at time t0 = 0 s. v0 = 5.0
(0 s)4 + C = +3.0 m/s 4
⇒
C = +3.0 m/s
So at time t = 2.0 s, v = 5.0
(2.0 s)4 + 3.0 m/s = 23 m/s 4
1. (6 points) What are the units of the “5.0” in the expression ax (t) = 5.0t3 ? . . . . . . . . . . . . . . . . . . . . . . . If “something” times seconds cubed yields meters per second squared, then the “something” must have units of meters per second to the fifth. m = (?) s3 s2
⇒
m ³m´ 3 = 5 s s2 s
Quiz #1 Solutions Page 2 of 5
III. (16 points) An oasis is 24 km directly east of a camel. Unfortunately, the camel walks 24 km in a direction 30◦ south of east, and then 16 km directly north. How far is the camel from the oasis now? . . . . . . . . . . . . . . . . . . . . . . .
From the sketch, after walking 24 km in a direction 30◦ south of east, and then 16 km directly north, the camel is 4 km north of the oasis, and 3.2 km west. These are the legs of a right triangle whose hypoteneuse is q 2
2
(4 km) + (3.2 km) = 5.1 km
2. (6 points) In what direction should the camel now walk to reach the oasis? . . . . . . . . . . . . . . . . . . . . . . . Since the camel is north and west of the oasis, it must direct its travels south of east to arrive there. If the angle south of east is φ, then 4 km tan φ = 3.2 km
µ ⇒
φ = tan
−1
4 km 3.2 km
¶ = 51◦
So the camel must head 51◦ south of east.
Quiz #1 Solutions Page 3 of 5
3. (10 points) The figure shows the velocity of a particle moving in one dimension as a function of time. At which points, if any, does the particle change its direction of travel? . . . . . . . . . . . . . . . . . . . . . . . When an object moves in one dimension, the sign of its velocity indicates its direction. The sign of the velocity changes when the velocity vs. time curve crosses the time axis, namely, at points 3 and 5.
4. (10 points) The figure shows the velocity of a truck (dashed line) and a car (solid line) as a function of time as they move along a straight road. The axes cross at the origin. The two vehicles are side by side at time t = 0. What is true at time t = T ? . . . . . . . . . . . . . . . . . . . . . . . The distance travelled is the area under the given position vs. time curve. At time t = T , the area under the truck curve is twice that under the car curve. Neither area is zero. Thus, the truck and car will both have moved, but the truck will have travelled farther than the car.
Quiz #1 Solutions Page 4 of 5
5. (10 points) The figure shows the position of two objects, A and B, as a function of time as they move along a straight line. At what time, if any, do the objects have the same speed? . . . . . . . . . . . . . . . . . . . . . . . The velocity is the slope of the given position vs. time graph. The speed is the magnitude of that slope. The two curves are parallel, that is, have the same slope, at some time between 2 and 3 seconds.
6. (10 points) A freight train of empty hopper cars is accelerating (speeding up) along a straight and level track. It begins to rain, and as the hopper cars fill with water, the magnitude of the acceleration decreases with time. How does the speed of the train vary as the acceleration magnitude decreases? . . . . . . . . . . . . . . . . . . . . . . . Since we know the train is speeding up before it begins to rain, we know that the velocity and acceleration vectors are parallel. Let that be the positive direction. As long as the acceleration magnitude is decreasing, the acceleration cannot become zero or negative (since the magnitude would then begin to increase). Thus, the acceleration and velocity vectors must remain parallel, and the speed increases.
Quiz #1 Solutions Page 5 of 5
