Question

Question 1 Here is the situation. Assume there is no friction. The spring has a force constant of 100 N/m. A 20 g bullet is shot into the 3 kg cart generating the speed Vcart. The maximum compression of the spring is 0.12 m. 20 g Vlaunch 3 kg Vcart www. Xsprig a. What was the speed of the cart with the bullet in it? b. What was the speed of the bullet before hitting the cart?

Answer 1

a.

When the cart fully compresses the spring and comes to a stop, its kinetic energy is completely converted into elastic potential energy of the spring.

$KE_{f}=PE_{spring}$

$\frac{1}{2}(m_{b}+m_{cart})v_{cart}^{2}=\frac{1}{2}kx^{2}$

$(m_{b}+m_{cart})v_{cart}^{2}=kx^{2}$

$v_{cart}=\sqrt{\frac{kx^{2}}{(m_{b}+m_{cart})}}$

$v_{cart}=\sqrt{\frac{100N/m*(0.12m)^{2}}{(0.02kg+3kg)}}$

$v_{cart}=\sqrt{\frac{100*0.12^{2}}{(0.02+3)}}$

ANSWER: ${\color{Red} v_{cart}=0.6905m/s}$

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b.

During collision momentum is conserved

Initial momentum = Final momentum

$m_{b}u_{b}+m_{cart}u_{cart}=(m_{b}+m_{cart})v_{cart}$

$0.02kg*u_{b}+3kg*0m/s=(0.02kg+3kg)*0.6905m/s$

$0.02*u_{b}+3*0=(0.02+3)*0.6905$

$0.02*u_{b}=3.02*0.6905$

$u_{b}=\frac{3.02*0.6905}{0.02}$

ANSWER: ${\color{Red} u_{b}=104.2655m/s}$

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