# A mass 1.2 kg of is attached to a spring and placed on a fri

A mass 1.2 kg of is attached to a spring and placed on a friction-less air track. Initially the mass is resting in equilibrium. A 10-gram bullet is fired into the mass to initiate oscillatory motion. The bullet hits the mass from the left side at a speed of 120 m/s at t=0 and gets lodged inside of it. After that, the system starts oscillating harmonically with a period of 15 seconds.  momentum conservation to find the initial velocity of the mass right after it was hit by the bullet. Initial Momentum = m*v = 0.010 * 120 = 1.2 Final momentum = Initial momentum = m * v v = Initial momentum / m = 1.2 / (1.2 + 0.010) = 0.992 m/s What was the kinetic energy of the bullet? What was the kinetic energy of the mass right after it was hit by the bullet? Where did the difference go?Bullet Kinetic energy: Ek = 1/2 * m * v^2 = 1/2 * 0.010 * 120^2 = 72 J Mass after bullet strick: Ek = 1/2 * mtotal * v^2 = 1/2 * 1.21 * 0.992^2 = 0.595 JFind the amplitude and phase constant of the resulting oscillations. Write explicit expressions for the position x(t) and velocity v(t) of the mass as a function of time.

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