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MAVERICK [17]
2 years ago
8

Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is attached to the more massive b

lock, and the blocks are pushed together with the spring between them (Fig. P9.9). A cord initially holding the blocks together is burned; after that happens, the block of mass 3mmoves to the right with a speed of 2.00 m/s. (a) What is the velocity of the block of mass m? (b) Find the system’s original elastic potential energy, taking m = 0.350 kg. (c) Is the original energy in the spring or in the cord? (d) Explain your answer to part (c). (e) Is the momentum of the system conserved in the bursting-apart process? Explain how that is possible considering (f) there are large forces acting and (g) there is no motion beforehand and plenty of motion afterward?
Please check my work and if incorrect please provide detailed response.
a.) conservation of energy
3m*2+m*v=0
v=-6(m/s) (left)

b.) Energy is conserved
K=((0.350kg)(6^2))/2+((3)(0.350kg)(2^2))/2
K=8.4j
c.) The cord has stored no potential energy in the spring.
d.) There is no potential energy in the spring because the potential energy in the spring has transferred to kinetic energy of the two masses.

e/f/g.)Yes. The two blocks of masses m and 3m are in one isolated system (attached to one another with a spring and pushed together) and the mass of the spring is less than both blocks and is only used for storing potential energy.
Physics
1 answer:
ivanzaharov [21]2 years ago
4 0

Answer:

Explanation:

a ) Conservation of momentum is followed

m₁ v₁  = m₂ v₂

3m x 2 = m  v

v = 6 m/s

Total kinetic energy

= 1/2 x .35 x 6 ² + 1/2 x 1.05 x 2 ²

= 8.4 J

This energy must be stored as elastic energy in the spring which was released as kinetic energy on burning the cord.

Yes , the conservation of momentum will be followed  in the bursting apart process. Only internal forces have been involved in the process. Two equal and opposite internal forces are created by spring which creates motion and generates kinetic energy.

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The net torque on the square plate is 2.72 N-m.

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Using formula of torque

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Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

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We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

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Put the value into the formula

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