Respuesta :
Answer:
a) The speed is 6.3 cm/s
b) The maximum compression of the spring is 7.137x10⁻⁴m
c) The speed is 3.373x10⁻²m/s
Explanation:
Given data:
m₁ = 460 g = 0.46 kg
m₂ = 131 g = 0.131 kg
u₁ = 8.1 cm/s = 0.081 m/s
u₂ = 0
k = 1320
a) Applying the conservation of the momentum:
[tex]m_{1} *u_{1} =v*(m_{1} +m_{2} )\\v=\frac{m_{1} *u_{1}}{m_{1} +m_{2}} =\frac{0.46*8.1}{0.46+0.131}=6.3cm/s[/tex]
b) v = 6.3 cm/s = 0.063 m/s
The change in kinetic energy is:
[tex]delta-Ek=\frac{1}{2} (m_{1} +m_{2} )v^{2} -\frac{1}{2} m_{1} u_{1} ^{2} =\frac{1}{2} (v^{2} *(m_{1} +m_{2})-m_{1} u_{1} ^{2} )\\delta-Ek=\frac{1}{2} (0.063^{2} *(0.46+0.131)-(0.46*0.081^{2} ))=-3.362x10^{-4} J[/tex]
The change of kinetic energy is equal to the work done:
[tex]-\frac{1}{2} kx^{2} =-3.362x10^{-4} \\kx^{2} =6.724x10^{-4} \\x=\sqrt{\frac{6.724x10^{-4}}{k} } =\sqrt{\frac{6.724x10^{-4}}{1320} } =7.137x10^{-4}m[/tex]
c) The conservation of energy is:
[tex]\frac{1}{2} (m_{1} +m_{2})v^{2} =\frac{1}{2} kx^{2} \\v=\sqrt{\frac{kx^{2} }{m_{1} +m_{2}} } =\sqrt{\frac{1320*(7.137x10^{-4})^{2} }{0.46+0.131} } =3.373x10^{-2} m/s[/tex]
