A block of mass 0.3 kg is attached to a spring of spring constant 26 N/m on a fr
ID: 2003791 • Letter: A
Question
A block of mass 0.3 kg is attached to a spring of spring constant 26 N/m on a frictionless track. The block moves in simple harmonic motion with amplitude 0.1 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately before it strikes the block is 67 m/s and the mass of the bullet is 3.84 g. Find the speed of the block immediately before the collision. Answer in units of m/s. If the simple harmonic motion after the collision is described by x = B sin(omega t + phi), what is the new amplitude B? Answer in units of m. The collision occurred at the equilibrium position. How long will it take for the block to reach maximum amplitude after the collision? Answer in units of s.Explanation / Answer
Energy of the spring system Ep = ½kx² = ½ * 26N/m * (0.1m)² = 0.13 J
At equilibrium, this energy is kinetic:
Ek = 0.13J = ½mv² = ½ * 0.3kg * v²
v = sqrt(3.13 m²/s²) = 1.76 m/s speed of block immediately before collision
pre-collision momentum p = 0.3kg * 1.76m/s - 0.00384kg * 67m/s
p = 0.563 kg·m/s - 0.062 kg·m/s = 0.2070kg·m/s
post-collision momentum = 0.207 kg·m/s = (0.3kg + 0.00384kg) * v
v = 0.681 m/s post-collision velocity
So the max energy of the system E = ½mv² = ½ * 0.30384kg * (0.681m/s)²
E = 0.07045 J
and therefore the amplitude is found from
Ep = 0.070J = ½kx² = ½ * 26N/m * x²
x = 0.07733 m = B new amplitude
Collision occurred at equilibrium:
x = 0 = Bsin(t + )
Call this t = 0; then = 0 and
x = 0.0773*sin(t)
Further, = sqrt(k/m) = sqrt(26N/m / 0.3kg) = sqrt(80 kg·m/s²/m/kg) = 9.30 rad/s
x = 0.077*sin(9.30t)
so x = 0.077 when
sin(9.30t) = 1
9.30t = arcsin(1) = 1.571 make sure your calculator is in radians!
t = 1.571/9.30= 0.16 s answer
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