Question

1. a graph of the net force f exerted on an object as a function of x position is shown for the object of mass m as it travels a horizontal distance 3d. which expression represents the change in the kinetic energy of the object?

a) 3fd b) 3.5fd c) 4.5fd d) 6fd

1. a student plans to conduct an experiment in which the momentum of a two-object system can be determined immediately before and after a collision takes place. the student slides block x at an unknown constant speed toward an identical block, block y, that is initially at rest, as shown in the figure. there is negligible friction between the blocks and the surface.

the student conducts a second experiment in which the magnitude of the force exerted on block x from block y is measured over the time in which the collision takes place. a graph of the force as a function of time is shown. in a third experiment, the student creates a collision such that the force exerted on block x from block y is constant for 0.020 s.
which of the following constant forces, if exerted on block x from block y, would produce the same change in momentum as is shown by the graph?
a) 100 n b) 200 n c) 400 n d) 800 n

1. object x of mass m₀ travels to the right with a velocity v₀. object y of mass 2m₀ moves to the left at 2v₀, as shown in the figure. the objects collide and then stick together. what is the change in kinetic energy of the two-object system from immediately before the collision to immediately after the collision?

a) the kinetic energy increases by 6m₀v₀².
b) the kinetic energy increases by 3m₀v₀²
c) the kinetic energy decreases by 6m₀v₀².
d) the kinetic energy decreases by 3m₀v₀²

Explanation:

Step1: Calculate work (area under F-x graph)

The graph has 3 segments:

1. \(F\) from \(0\) to \(d\): area = \(F \times d = Fd\)
2. \(F\) to \(2F\) from \(d\) to \(2d\): area = average force × width = \(\frac{F+2F}{2} \times d = 1.5Fd\)
3. \(2F\) from \(2d\) to \(3d\): area = \(2F \times d = 2Fd\)

Total area = \(Fd + 1.5Fd + 2Fd = 4.5Fd\)

Step2: Relate work to kinetic energy change

By work-energy theorem, \(\Delta KE = \text{work done} = 4.5Fd\)

Step1: Estimate impulse (area under F-t graph)

The graph has ~8 squares; each square ≈ \(100N \times 0.001s = 0.1Ns\)? No—count grid points: peak ~800N, time 0.01s, approximate area ≈ \(400N \times 0.01s = 4Ns\)

Step2: Set constant force impulse equal

Impulse \(J = F \times t = F \times 0.020s = 4Ns\) → \(F = \frac{4}{0.020} = 200N\)

Step1: Find final velocity via momentum conservation

Let right = positive. Initial momentum: \(m_0v_0 - 2m_0(2v_0) = m_0v_0 - 4m_0v_0 = -3m_0v_0\)
Final mass = \(3m_0\), so \(3m_0v_f = -3m_0v_0\) → \(v_f = -v_0\)

Step2: Calculate KE before and after

Initial KE: \(\frac{1}{2}m_0v_0^2 + \frac{1}{2}(2m_0)(2v_0)^2 = 0.5m_0v_0^2 + 4m_0v_0^2 = 4.5m_0v_0^2\)
Final KE: \(\frac{1}{2}(3m_0)(v_0)^2 = 1.5m_0v_0^2\)

Step3: Compute KE change

\(\Delta KE = 1.5m_0v_0^2 - 4.5m_0v_0^2 = -3m_0v_0^2\) (decrease by \(3m_0v_0^2\))

Answer:

C) 4.5Fd