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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From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm
ball drops 45m under g=10m/s/s
45=1/2x10xt^2 ... application of kinematic equaion from rest
90/10=t^2
t=3
24.0 m in 3 secs => 8m/s no air resistance
