The position-versus-time (x-t) graph below represents the motion of a particle w

Question

The position-versus-time (x-t) graph below represents the motion of a particle with mass m = 2.0 kg. It first undergoes horizontal simple harmonic motion for one period T (A-B), then it moves along a straight line on a rough surface (B - C). In point C it collides inelastically perpendicular with a wall and bounces back to D. (So, the x- t graph in the interval A-B is sinusoidal, and in the interval B-C-D is parabolic.) Calculate the maximum speed in the oscillating interval A-B. Calculate the maximum acceleration in the interval A-B. Knowing that only the friction decelerates the particle, calculate the magnitude of the deceleration in the interval B-C-D and the respective kinetic coefficient of friction. Find the speeds before and after the collision, and use them to calculate the kinetic energy lost during the inelastic collision in point C. Sketch the graphs in the velocity vs. time (vx-t) and then in the acceleration vs. time (ax-t) representations. Use the provided frames.

Explanation / Answer

a) In the region A-B the displacement function is 5sin(2pi x t/4) = 5sin(pi x t/2)

The velocity function in the region A-B is the derivative of the above function = 2.5 x pi x cos(pi x t/2)

Maximum value of the velocity function is obtained when t = 0 and is equal to = 2.5 x pi x 1 = 7.854 m/s

b) The acceleration function in the region A-B is the derivative of the above velocity function and is equal to

= -1.25 x pi2 x sin(pi x t/2)

Maximum value of the acceleration function is obtained when t = 1 and is equal to = -1.25 x pi2 x 1 = 12.337 m/s2

c) Velocity at t =4 s, v = 7.854 m/s

Now distance travelled in the B-C region, S = 15 m

Time taken for this displacement, t = 7 - 4 s = 3 s

Now, S =vt + 0.5 x a x t2

or, 7.854 x 3 + 0.5 x a x 32 = 15

or, a = -1.903 m/s2

uk x 2 x 10 = 2 x 1.903

or, the coefficient of kinetic friction, uk = 0.1903

d) Speed before collision = u + at = 7.854 - 1.903 x3 = 2.145 m/s

Let the position D be at a x value of 13 m

Speed after collision ( in nearly ) = 15-13 m/s = 2 m/s

change in KE = 0.5 x 2 x (2.1452 - 22) = 0.6 J

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