1.001 In the raft frame, the boat does a roundtrip at constant speed. This means that the two legs are equal. Since the raft covers l in 2Δt, v=l∕(2Δt)=3 km∕h.
1.002 The idea is to write |v| as a function of v0, v1 and v2, where |v|=d∕(Δt0+2Δt1). Use that v0=d∕(2Δt0) and v1=αd∕(2Δt1) and plug it in |v| so it becomes |v|^(−1)=1∕(2v0)+ɑ∕v1. ɑ is the fraction of d∕2 and its value is needed. The ratio v2∕v1 is (1−ɑ)∕ɑ. Isolate ɑ=1∕(1+v2∕v1) and plug it to |v| to get |v|=2v0(v1+v2)∕(v1+v2+2v0).
1.003 Draw a v−t graph. The result is an isosceles trapezium of area 500 m (total displacement). Let τ be the duration of each tilted interval and T=25−2τ the horizontal (constant speed) interval. The area can be expressed as 5τ(25+T)∕2. From here, τ is solved, giving only one meaningful solution (5 s). Then, T=15 s.
1.004 a) |v|≈2∕20=0.1 m∕s. b) For vmax, there seems to be 5 vertical steps with maximum slope. These steps correspond to 2 horizontal steps. This corresponds to a ratio of 5∕2=2.5 vertical∕horizontal. In order to translate this to speed, take into account that a single square has a slope of 0.1 m∕s. Then, 2.5×0.1 = 0.25 m∕s. c) The mean motion from t=0 to a given t can be represented by a straight line from t=0 to t. If this mean motion is to match the instant speed of the point, this line needs to be tangent to the s−t curve. This can only happen at a single point, which is approximately t0=16 s.
1.005 Draw a quadrilateral with O (coordinate origin), P1 (initial position of 1), P (collision point at t=τ) and P2 (initial position of 2). The diagonal P1P2 is both vec(r2)−vec(r1) and (vec(v1)−vec(v2))τ. This means these two vectors are parallel, and their unit vector versions are equal.
1.006 This problem can be solved with the cosine theorem: v'²=v²+v0²+2v×v0×cos(ϕ), so v'≈40 km∕h (don't give more precision than the input data). The sine theorem can be used to get φ, with sin(φ')∕15=sinφ∕40, giving φ'≈19°. What I find interesting here is the fact that (in Galilean relativity), each point in a velocity diagram is an entity, and going to that point means going to the perspective of that entity's rest frame. To illustrate that, draw the velocity vectors such that vec(v')=vec(v)−vec(v0). The bases of vec(v) and vec(v0) represents the entity of the paper's rest frame. From this point, both the ship and the wind are seen moving, with a mutual angle of π−φ. From the tip of vec(v0), we are at the rest frame of the ship, so vec(v') is how this ship sees the wind. And −vec(v0) would be how the ship sees the paper's rest frame. Finally, from the tips of vec(v) and vec(v') we see the rest frame of the wind, and −vec(v) and −vec(v') would be how the wind sees the paper's frame and the ship's frame, respectively. In my opinion, this is THE way of seeing Galilean relativity. Once a velocity diagram is drawn, no matter how many points or dimensions, a vector from point (entity) A to a point (entity) B is the velocity of how A sees B.
1.007 It is essential to draw a velocity diagram. There are four entities (points of the v−diagram, as explained in the previous problem): ground (O), swimmer 1 (s1), s2 and river (r). The vectors: vec(v0) goes from O to r; vec(v1) goes from O to s1; vec(v2) from O to s2. And the last two, both with magnitude v': one from r to s2 and the other from r to s1. The statement says that v' is the speed of both swimmers with respect to water, so the r entity must be the tail of the arrows with length v'. Once this is clear, the rest is easy. Let τ be the time for s1 to arrive, and d the width of the river. Then, τ=d∕v1. Let τ1 and Τ2 be the swimming and walking time of s2, respectively. Then, τ1=d∕v' and τ2=d×v0∕(u×v'). Finally, as τ=τ1+τ2, we can solve for u and get u = v0∕(a−1), where 1∕a = sqrt(1−(v0∕v')²). With numbers, u = 3 km∕h.
1.008 Since both ships go on a roundtrip, a velocity diagram for every leg (1 and 2) is needed. Each diagram has four entities: buoy, water, A and B. In diagram 1, ηv0 is what connects water to both A and B. Water and buoy are connected with v0. vA is what connects buoy with A and vB is what connects buoy with B. If we see that the angle between water−buoy and water−B is θ such that cosθ=1∕η, then vB is found to be vB=v0×sqrt(η²−1). Then, τB1=L∕vB. For A it's easier: τA1=L∕vA, where vA=v0(τ+1). For the second leg, B takes the same amount of time, by symmetry, while A takes τA2=L∕vA', where vA'=v0(τ−1). Putting all together, we get τ2∕τ1=η∕sqrt(η²−1)=6∕sqrt(11)≈1.8 (since η=6∕5).
1.009 A velocity diagram is needed, with three vertices: ground, water and boat. Ground−water is v0, and water−boat is v0∕n. Geometrically, it is clear that vb (ground−boat), needs to be perpendicular to water−boat. For that, choose a value of n in the drawing (n=2 is given), and with a compass draw an arc from water and radius v0∕n. Now try to connect ground with any point of this arc. The maximum angle β (angle of vb wrt the stream) happens when vb is ⟂ v0∕n. Once we see this, the sine theorem provides sinβ=1∕n. If n=2, β=30º.
1.010 This problem can be easily solved with simple kinematics, but there is a more interesting way where time is not needed until the final step. We go to 1's frame, which is a free falling coordinate system. An observer in free fall would observe 1 and 2 with constant velocities, as if we set g=0. In 1's frame, 1 is at rest and 2 moves to the bottom−right direction. From the velocity diagram we calculate this velocity, which is sqrt(2)v0×sqrt(1−sinθ) (a direct calculation applying the cosine theorem). For the distance, we simply multiply by the ellapsed time. For the values given, v=25 m∕s and θ=60°, d≈22 m.
1.011 Going to a freefall frame, for example 1's, we see how 2 moves horizontally, with constant speed, and being at a distance d=(v1+v2)t. We need to find t for which the two velocities are perpendicular. The horizontal components will not change, and the vertical ones will be identical in both particles, with value −gt. Applying a dot product, (−v1.−gt)×(v2,−gt)=0, we get t=sqrt(v1v2)∕g. Then, d=(v1+v2)×sqrt(v1v2)∕g=7×sqrt(12)∕g≈2.47≈2.5 m.
1.012 Since the symmetry is never lost, the triangle will remain equilateral, and the centroid will not move. The radial speed is vr=v×cos(30º)=v×√3∕2 (constant because the triangle is always equilateral). The radius is r=a∕√3. Then, the time to collapse is r∕vr=2a∕(3v). We cannot apply the same idea to the area, since it does not decrease uniformly with time. And the triangle's side l? It decreases uniformly, but beware dl∕dt is not v! If we draw the triangle before and after a dt, we can calculate the side of the latter. We can draw a triangle with sides a−dl, vdt and a−vdt, the last two forming 60º. Applying the cosine theorem and dropping second order terms like dl² and dt², we get dl∕dt=3v∕2. Once we know this, the time to collapse can also be calculated as a divided by 3v∕2. The latter method does not need any centroid properties.
1.013 The velocity diagram is a triangle with three vertices: rest, A and B. Rest and B are connected by vec(u), and rest and A are connected by vec(v). While vec(u) is along the x direction, vec(v) is along the direction that connects positions A and B. This means that the latter is a radial direction. In the velocity diagram, the vector from A to B is the velocity of B as seen from A. If we project this vector along the radial direction we get a radial speed v−ucosα. The vector from B to A is the velocity of A as seen from B. If we project this vector along the x direction we get an x speed vcosα−u. Now we need to integrate these two speeds between t=0 and t=τ. The integral of the radial speed needs to give (final distance − initial distance) = 0 − l = −l. The integral of the x speed needs to give (final x distance − initial x distance) = (0 − 0) = 0. These are two equations with an unknown to eliminate: ∫cos(α)dt (the integral is from t=0 to t=τ). We get that τ=lv∕(v²-u²).
1.016 vec(r1)=(x1,0)=(−l1+v1t,0) and vec(r2)=(0,y2)=(0,−l2+v2t), so the mutual distance d(t)=|(l1−v1t,−l2+v2t)|. The time τ that minimises d also minimises d², so we differentiate d² wrt time and find τ=(v1×l1+v2×l2)∕(v1²+v2²). Since the second derivative is always positive, τ gives a minimum d. We just substitute τ in d to find the minimum distance = |v1l2−v2l1|∕sqrt(v1²+v2²). There is another way to approach this problem. Notice that the centroid of the two positions vec(r1) and vec(r2), which is (x1,y2)∕2, is equidistant from both particles and also from the origin. The distance d between the particles is, then, twice the distance r from the centroid to the origin. This is interesting, since, considering both particles with the same mass, the centroid is also the centre of mass (CM) of the system, which travels along a straight line. This distance r is minimum when the velocity and the position of CM are ⟂. Notice the total angular momentum of the system (wrt the origin), with both particles directed towards the origin, is vec(0), and this value must remain constant. The angular momentum of CM is also constant, since CM moves with constant velocity. Its value (consider the CM has mass=2) is 2×r_min×sqrt(v1²+v2²)∕2=r_min×sqrt(v1²+v2²)=|−l1v2+l2v1|/2 (the initial value). Then, d_min=2×r_min=|−l1v2+l2v1|∕sqrt(v1²+v2²).