Block A, with a mass of 4 kg, is moving with a speed of 2.0 m/s while block B, with a mass of 8 kg, is moving in the opposite direction with a speed of 3m/s. The center of mass of the two block

Engineering

Physics

Center of Mass

Momentum and Energy Conservation for System of Particles

Question

Block A, with a mass of 4 kg, is moving with a speed of 2.0 m/s while block B, with a mass of 8 kg, is moving in the opposite direction with a speed of 3m/s. The center of mass of the two block - system is moving with a velocity of :

1.3m/s in the same direction as A

2.7m/s in the same direction as A

1.0m/s in the same direction as B

1.3m/s in the same direction as B

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The velocity of the center of mass (v_cm) for a two-particle system is given by the formula:

$v_{c m} = \frac{m_{A} v_{A} + m_{B} v_{B}}{m_{A} + m_{B}}$

Let's define the direction of block A's motion as positive. Therefore, v_A = +2.0 m/s and v_B = -3.0 m/s. Substituting the values (m_A=4 kg, m_B=8 kg):

$v_{c m} = \frac{(4 \times 2.0) + (8 \times (- 3.0))}{4 + 8} = \frac{8 - 24}{12} = \frac{- 16}{12} \approx - 1.3 m / s$

The negative result means the center of mass velocity is 1.3 m/s in the direction opposite to our chosen positive direction (the direction of A), which is the same direction as B.

Final Answer: 1.3m/s in the same direction as B