Question

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harmonic motion with amplitude 0.19 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 68 m/s and the mass of the bullet is 1.45 gA) Find the speed of the block immediately before the collision.B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), what is the new amplitude B?ωC) The collision occurred at the equilibrium position.How long will it take for the block to reach maximum amplitude after the collision?

Answer 1

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

                                                       = $\sqrt\frac{21}{0.1}$

                                                       = 14.49 rad/sec

A) Speed of the block immediately before the collision:

displacement of Simple Harmonic  Motion is given as:

                                $x = A sin (\omega t + \phi)$

Differentiating this to find speed of the block immediately before the collision:

                    $v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}$

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ $v= A\omega_{o}$

                                       $v=(.19)(14.49)\\v= 2.75 ms^{-1}$

B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:

S.H.M after collision is given as :

                              $x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

C) Time taken by the block to reach maximum amplitude after the collision:

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$