Answer:
Explanation:
Force constant of spring K = 21 N /m
we shall find the common velocity of putty-block system from law of conservation of momentum .
Initial momentum of putty
= 5.3 x 10⁻² x 8.97
= 47.54 x 10⁻² kg m/s
If common velocity after collision be V
47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V
V = .937 m/s
If x be compression on hitting the putty
1/2 k x² = 1/2 m V²
21 x² = ( 5.3x 10⁻² + .454) x .937²
x² = .0212
x = .1456 m
14.56 cm
The distance the putty-block system compress the spring is 0.15 meter.
Given the following data:
To determine how far (distance) the putty-block system compress the spring:
First of all, we would solver for the initial momentum of the putty.
[tex]P_p = mass \times velocity\\\\P_p = 5.30\times 10^{-2}\times 8.97\\\\P_p = 47.54 \times 10^{-2} \;kgm/s[/tex]
Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:
[tex]P_p = (M_b + M_p)V\\\\47.54\times 10^{-2} = (0.454 + 5.30\times 10^{-2})V\\\\47.54\times 10^{-2} = 0.507V\\\\V = \frac{0.4754}{0.507}[/tex]
Velocity, V = 0.94 m/s
To find the compression distance, we would apply the law of conservation of energy:
[tex]U_E = K_E\\\\\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\kx^2 =M_{bp}v^2\\\\x^2 = \frac{M_{bp}v^2}{k} \\\\x^2 = \frac{(0.454 + 5.30\times 10^{-2}) \times 0.94^2}{21}\\\\x^2 = \frac{(0.507 \times 0.8836)}{21}\\\\x^2 = \frac{(0.4480)}{21}\\\\x=\sqrt{0.0213}[/tex]
x = 0.15 meter
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