Question

An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12 J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be

• A. 2 m/s
• B. 4 m/s
• C. 1 m/s
• D. 8 m/s

Answer 1

Answer: (A)

2 m/s

Answer 2

The problem involves the release of a compressed spring between two blocks on a smooth horizontal table. This scenario can be analyzed using the principles of conservation of momentum and conservation of energy.

1. Conservation of Momentum: Since the system is isolated horizontally (smooth table), the total initial momentum (zero, as blocks are at rest) equals the total final momentum. This gives the relationship $m_A v_A = m_B v_B$.

2. Conservation of Energy: The potential energy stored in the spring is converted into the kinetic energy of the two blocks. This gives the equation $PE_{spring} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2$.

3. These two equations are then solved simultaneously to find the velocity of the larger block ($v_B$).

Detailed Solution:

• Mass of block A (smaller block), $m_A = 1$ kg
• Mass of block B (larger block), $m_B = 2$ kg
• Energy stored in the spring, $PE_{spring} = 12$ J

Applying conservation of momentum:

$0 = m_A v_A + m_B v_B$

$m_A v_A = m_B v_B$

$1 \text{ kg} \times v_A = 2 \text{ kg} \times v_B$

$v_A = 2 v_B$ (Equation 1)

Applying conservation of energy:

$PE_{spring} = KE_A + KE_B$

$12 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2$

$12 = \frac{1}{2} (1) v_A^2 + \frac{1}{2} (2) v_B^2$

$12 = \frac{1}{2} v_A^2 + v_B^2$ (Equation 2)

Substitute Equation 1 ($v_A = 2 v_B$) into Equation 2:

$12 = \frac{1}{2} (2 v_B)^2 + v_B^2$

$12 = \frac{1}{2} (4 v_B^2) + v_B^2$

$12 = 2 v_B^2 + v_B^2$

$12 = 3 v_B^2$

$v_B^2 = \frac{12}{3}$

$v_B^2 = 4$

$v_B = \sqrt{4}$

$v_B = 2 \text{ m/s}$