Two blocks of mass m_1 = 6.0 kg and m_3 = 4.5 kg, connected by a rod of mass m_2

Question

Two blocks of mass m_1 = 6.0 kg and m_3 = 4.5 kg, connected by a rod of mass m_2 = 1.5 kg, are sitting on a very low-friction surface, and you push to the left on the right block (mass m_1) with a constant force of (-29, 0, 0) N. See figure below. (As usual, +x is to the right and +y is up. Ignore effects due to friction between the objects.) What is the acceleration a^rightarrow = dv^rightarrow/dt of the system of blocks plus rod? A^rightarrow = m/s^2 What is the force F_3^rightarrow exerted by the rod on the block of mass m_3? (This is approximately equal to the compression force in the rod near its left end.) F_3^rightarrow = N What is the force F_2^rightarrow exerted by the block of mass m_1 on the rod? This is approximately equal to the compression force in the rod near its right end.) F_2^rightarrow = N Which of the following statements about this system are true? (Select all that apply.) F_3^rightarrow = F_2^rightarrow = F^rightarrow because all the blocks are connected. The rates of change of momentum of block 1 and block 3 are equal. The atomic bonds inside the rod are compressed. Suppose that instead of pushing on the right block (mass m_1), you pull to the left on the left block (mass m_3) with a constant force of (-29, 0, 0) N. Draw a diagram illustrating this situation. Now what is the force F_3^rightarrow exerted by the rod on the block of mass m_3? F_3^rightarrow = N Which of the following statements about this system are true? (Select all that apply.) The acceleration of the three-object system is the same as it was in part (a). The direction of F_3^rightarrow on block 3 by the rod is the same as it was in part (a). The atomic bonds inside the rod are stretched. The rates of change of momentum of block 1 and block 3 are equal.

Explanation / Answer

(a) Look at the diagram, you find that the three masses are connected to each other.

Therefore, acceleration of the system, a = F / (m1 +m2 +m3) = -29 / (6.0+4.5+1.5) = -2.42 m/s^2

(b) Force exerted by the rod on m3,

F3 = m3*a = 4.5*(-2.42) = -10.89 N.

Force exerted by the block m1 on the rod,

F2 = (m2+m3)*a = (1.5+4.5)*2.42 = 14.52 N

Second option, 'the rate of change of momentum of block 1 and block 3 are equal.'

(c) In this case -

F3 = (m1+m2)*a = (6.0+1.5)*2.42 = 18.15 N

Options (1) and (4) are correct.

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