Questions: Block B has a mass of 2.70 kg and is moving to the left at a speed of 4.80 m/s. Block A has a mass of 3.60 kg and is moving to the right. The two blocks undergo a perfectly inelastic collision. What should be the velocity of Block A in order to have the two blocks remain at rest after the collision?

Solution Steps

We need to find the velocity of Block A such that after a perfectly inelastic collision with Block B, both blocks come to rest. In a perfectly inelastic collision, the two objects stick together after the collision.

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Since the blocks come to rest after the collision, the total momentum after the collision is zero.

Let \( m_A = 3.60 \, \text{kg} \) be the mass of Block A, \( v_A \) be the velocity of Block A, \( m_B = 2.70 \, \text{kg} \) be the mass of Block B, and \( v_B = -4.80 \, \text{m/s} \) be the velocity of Block B (negative because it is moving to the left).

The equation for conservation of momentum is: \[ m_A \cdot v_A + m_B \cdot v_B = 0 \]

Substitute the known values into the momentum equation: \[ 3.60 \cdot v_A + 2.70 \cdot (-4.80) = 0 \]

Simplify and solve for \( v_A \): \[ 3.60 \cdot v_A - 12.96 = 0 \] \[ 3.60 \cdot v_A = 12.96 \] \[ v_A = \frac{12.96}{3.60} \] \[ v_A = 3.6000 \, \text{m/s} \]

Final Answer