A 1.2 kg block rests on a frictionless surface and is attached to ahorizontal sp

Question

A 1.2 kg block rests on a frictionless surface and is attached to ahorizontal spring of constant k = 24 N/m. The block isoscillating with amplitude 13 cm and with phase constant ?= -/2. A block of mass 0.80 kg is moving from theright at 1.7 m/s. It strikes the first block when the latter is atthe rightmost point in its oscillation. The collision is completelyinelastic, and the two blocks stick together.

Determine the frequency of the resulting motion.
??? Hz
Determine the amplitude of the resulting motion.
??? cm
Determine the phase constant (relative to the originalt = 0) of the resulting motion.
??? rad

Explanation / Answer

x = A cos ( t - /2) amax = - 2A     at t =/2    the rightmost point of the motion sincea = -  2 A cos ( t -/2) (M + m) V = m v   since the large block is atrest when struck V =.8 / 2 * 1.7 = .68   the speed of the blocksafter the collision Ek = 1/2 * 2 * .682 = .462 J justafter the collision Es = 1/2 k x2 = 12 * .132 =.203 J   the energy stored in the spring E = .665 J    the new total energy E = 1/2 k A2     where A = thenew amplitude A = (2 * .665 / 24) = .235 m f = 1 / (2 ) * (k / (m + M)) = .551 /sec    the new frequency of oscillation cos ( t - ) = x / A = .13 / .235 = .553 t - = 56.4 deg = .985 rad Since t - /2    wasoriginally 0 at the rightmost point of motion the phase has increased by .985 rad (the masses are now movingleft with speed .68 m/s)

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