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fenix001 [56]
3 years ago
9

A block of mass 0.1 kg is attached to a spring of spring constant 22 N/m on a frictionless track. The block moves in simple harm

onic motion with amplitude 0.29 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 41 m/s and the mass of the bullet is 1.23 g.
A) Find the speed of the block immediately before the collision.Answer in units of m/sB) If the simple harmonic motion after the collision is described by x = B sin(? t + f), what is the new amplitude B?Answer in units of mC) 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
Physics
1 answer:
Tasya [4]3 years ago
6 0

Answer:

(A) v = 14.8m/s

Explanation:

(A) V = sqrt(k/m) × A = sqrt(22/0.1) × 0.29 =14.8m/s.

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Explanation:

The given data is as follows.

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Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

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Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

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If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

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So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

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