IB Physics SL Theme A: Space, time and motion questions

Answer: A. Displacement is final position minus initial position; distance is the length of the route. From S(0,0) to F(3,4), the displacement components are (3−0, 4−0) = (3, 4) m. The route length is 6 + 4 + 4 + 1 = 15 m.

B: Interchanges the east and north components.
C: Reverses the change in position, using start minus finish.
D: Omits the 4 m segment from Q to R when adding the route lengths.
Remember: displacement depends on the initial and final positions; distance includes the whole path.

Answer: B. For all three blocks, the resultant force is $50-2.0-5.0-3.0=40\ \text{N}$ so their acceleration is $40/(2.0+3.0+5.0)=4.0\ \text{m s}^{-2}$ to the right. For C, the contact force from B satisfies $F-3.0=5.0(4.0)$ giving $F=23\ \text{N}$ to the right. The force exerted by C on B is equal in magnitude and opposite in direction: $23\ \text{N}$ to the left.
A: uses $m_Ca=20\ \text{N}$ as the contact force, omitting friction on C.
C: gives the force on C by B, not the equal-and-opposite force on B.
D: treats the resultant force on all three blocks as the force at the B–C contact.
Remember: interaction forces act on different bodies and are equal in magnitude and opposite in direction.

Answer: D. Work on each segment is the force component along that segment multiplied by its length. The east and north components are $10.0\cos30.0^\circ=8.66\ \text{N}$ and $10.0\sin30.0^\circ=5.00\ \text{N}$ Thus $W=(8.66)(6.0)+(5.00)(8.0)=91.96\ \text{J}$ which is $92\ \text{J}$ to two significant figures. A: subtracts the northward contribution instead of adding it, giving $52-40=12\ \text{J}$ B: includes only the work during the northward displacement, $40\ \text{J}$ C: includes only the work during the eastward displacement, $52\ \text{J}$ Remember: add the work contributions from successive displacements, including their signs.