In the (Figure 1) position-versus-time graph for a collision between two carts o

Question

In the (Figure 1) position-versus-time graph for a collision between two carts on a low-friction track. Cart 1 has an inertia of 1.0 kg; cart 2 has an inertia of 4.0 kg.

Part A: What are the initial and final velocities of each cart?

Part B: What is the change in the velocity of each cart?

Part C: Do the values you calculated in the previous part satisfy the equation mumsvsxvux

Part D: Does cart 1 have a nonzero acceleration? If so, when, and what is the sign of the acceleration?

Part E: Does cart 2 have a nonzero acceleration? If so, when, and what is the sign of the acceleration?

Explanation / Answer

Part - A The slope of a diplacement-time plot gives the velocity of object at any instant.

Cart 1: Before collision, the slope of x-t curve is constant and x(t) increases from -0.3 m to +0.3 m as t value goes from 0 s to 1 s. Thus,

vi,1 = dx/dt (before collision) = [0.3 - (-0.3)]/1 = 0.6 m/s

vf,1 = dx/dt (after collision) = (-0.1 - 0.3)/(2-1) = -0.4 m/s

Cart 2: Before collision, the slope of x-t curve is constant and x(t) decreases from +0.5 m to +0.4 m as t value goes from 0 s to 1 s. Thus,

vi,2 = dx/dt (before collision) = [0.4 - (0.5)]/1 = - 0.1 m/s

vf,2 = dx/dt (after collision) = [0.55 - (0.4)]/1 = 0.15 m/s

Thus initial and final velocities of Cart 1 are 0.6 m/s and -0.4 m/s respectively; initial and final velocities of Cart 2 are -0.1 m/s and 0.15 m/s respectively.

Part - B

Change in velocity of Cart 1 = (- 0.4 - 0.6) m/s = -1 m/s

Change in velocity of Cart 2 = (0.15 - (- 0.1)) m/s = 0.25 m/s

Part - C

Mass of Cart 1 = m1 = 1 kg and Mass of Cart 2 = m2 = 4 kg.

Thus, the momentum equation, m1(vf,1 - vi,1) + m2(vf,2 - vi,2) = 0 is satisfied.

Part - D

Yes. Cart 1 indeed has a non-zero acceleration between t = 1 s and t = 1.2 s when collision occurs. During this time, x-t curve doesn't have a constant slope - it rather decreases. And hence, d2x/dt2 during this period is negative and so is acceleration.

Part - E

Similar to the above case, Cart 2 has a non-zero acceleration between t = 1 s and t = 1.2 s but slope increases in this time interval which means d2x/dt2 during this period is positive and so is acceleration.

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