1. A mass, m 1 = 9 kg, is at rest on a frictionless horizontal surface and conne

Question

1. A mass,m1 = 9 kg, is at rest on a frictionless horizontal surface and connected to a wall by a spring with k = 56 N/m, as shown in the figure. A second mass,m2 = 6 kg, is moving to the right at v0 = 18 m/s. The two masses collide and stick together.

a. How long (how much time) will it take after the collision to reach this maximum compression?

2. A massive object of m = 7.8 kg oscillates with simple harmonic motion. Its position as a function of time varies according to the equation x(t) = 4.6 sin(%u03C0t/4.6 + %u03C0/6).

a. At which time after t = 0 s is the kinetic energy first at a maximum?

Explanation / Answer

A) Time period of oscillations = T = 2*(pi)*sqrt(m/k) = 2*(3.14)*sqrt(15/56) = 3.25 secs

Time taken by the system to stop = T/4 = 0.81 secs [Ans]

B) x(t) = 4.6 sin(pi* t/4.6 + pi/6).

V(t) = dx/dt = pi* cos(pi* t/4.6 + pi/6)

There fore

KE = 0.5 m v^2 = 0.5*7.8 * (pi* cos(pi* t/4.6 + pi/6))^2

= 3.9 *(pi)^2 *(cos(pi* t/4.6 + pi/6))^2  

= 3.9 *(pi)^2 *(( cos(2(pi* t/4.6 + pi/6)) +1)/2)

= 38.45 ( ( cos(2(pi* t/4.6 + pi/6)) +1)/2)

KE = 19.22( cos(2(pi* t/4.6 + pi/6)) +1)

therefore d(KE)dt = -19.22*sin(2(pi* t/4.6 + pi/6))*2*pi/4.6 = 0

or sin(2(pi* t/4.6 + pi/6)) = 0

or (2(pi* t/4.6 + pi/6)) = n*pi [n is a integer]

t = (n/2 *pi - pi/6) * 4.6/pi

t = (n/2-1/6)*4.6

Putting n=1 we get :

t = (1/2-1/6) * 4.6 sec = 1.533 sec [Ans

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