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Liono4ka [1.6K]
3 years ago
15

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The

block rests on a frictionless surface. A 5.90×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.
Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97  m /s) ²  / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m  ⇒ 2.59 cm

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c) 61.82 N

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<h3>a) Impulse delivered to the ball</h3>

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Substituting (10) in (6):

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Finding t:

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According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

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